Properties of Triangles

If one angle of a triangle be $60^\circ$, the area $10 \sqrt{3} \; cm^2$ and the perimeter $20 \; cm$, find the lengths of the sides.

Let the sides of the triangle be $a$, $b$ and $c$.

Let $\;$ $A = 60^\circ$

Given: $\;$ Area of $\triangle ABC = \Delta = 10 \sqrt{3} \; cm^2$

Now, $\;$ $\Delta = \dfrac{1}{2}b c \sin A$

i.e. $\;$ $10 \sqrt{3} = \dfrac{1}{2}b c \sin \left(60^\circ\right)$

$\implies$ $bc = \dfrac{20 \sqrt{3}}{\sqrt{3} / 2}$ $\implies$ $bc = 40$ $\;\;\; \cdots \; (1)$

Given: $\;$ Perimeter of $\triangle ABC = 20 \; cm$

Perimeter of $\triangle ABC$ $= a + b + c$

$\therefore \;$ $a + b + c = 20$ $\implies$ $b + c = 20 - a$ $\;\;\; \cdots \; (2)$

By cosine rule, we have,

$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$

$\begin{aligned} i.e. \; a^2 & = b^2 + c^2 - 2 b c \cos A \\\\ & = b^2 + c^2 - 2 b c \cos \left(60^\circ\right) \;\;\; \left[\because \; A = 60^\circ\right] \\\\ & = b^2 + c^2 - 2 \times b c \times \dfrac{1}{2} \\\\ & = b^2 + c^2 - bc \\\\ & = \left(b^2 + c^2 + 2 b c\right) - 3 b c \\\\ i.e. \; a^2 & = \left(b + c\right)^2 - 3 b c \;\;\; \cdots \; (3) \end{aligned}$

In view of equations $(1)$ and $(2)$, equation $(3)$ becomes,

$a^2 = \left(20 - a\right)^2 - 3 \times 40$

i.e. $\;$ $a^2 = 400 + a^2 - 40 a - 120$

i.e. $\;$ $40 a = 280$ $\implies$ $a = 7$

Substituting the value of $a$ in equation $(2)$ we get,

$b + c = 20 - 7 = 13$ $\implies$ $b = 13 - c$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(1)$ becomes,

$\left(13 - c\right)c = 40$

i.e. $\;$ $c^2 - 13 c + 40 = 0$

i.e. $\;$ $\left(c - 8\right) \left(c - 5\right) = 0$

$\implies$ $c = 8$ $\;$ or $\;$ $c = 5$

Substituting the value of $c$ in equation $(1)$ we get,

When $c = 8$, $\;$ $b = \dfrac{40}{c} = 5$

When $c = 5$, $\;$ $b = \dfrac{40}{c} = 8$

$\therefore \;$ The lengths of the sides of the triangle are:

$a = 7 \; cm$, $b = 5 \; cm$, $c = 8 \; cm$ $\;$ or $\;$ $a = 7 \; cm$, $b = 8 \; cm$, $c = 5 \; cm$

Properties of Triangles

A workman is told to make a triangular enclosure of sides $50$, $41$ and $21$ m respectively. Having made the first side one meter too long, what length must he make the other two sides in order to enclose the prescribed area with the prescribed length of fencing?

Actual lengths of sides: $\;$ $a = 50 \; m$, $\;$ $b = 41 \; m$, $\;$ $c = 21 \; m$

Actual perimeter $= a + b + c = 50 + 41 + 21 = 112 \; m$

Actual semi-perimeter $= s = \dfrac{a + b + c}{2} = 56 \; m$ $\;\;\; \cdots \; (1a)$

Actual area $= \Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

i.e. $\;$ $\Delta = \sqrt{56 \left(56 - 50\right) \left(56 - 41\right) \left(56 - 21\right)} = 420 \; m^2$ $\;\;\; \cdots \; (1b)$

Incorrect length $= a_1 = 51 \; m$

Let the remaining lengths be $b_1$ and $c_1$

New semi-perimeter $= s_1 = \dfrac{a_1 + b_1 + c_1}{2} = \dfrac{51 + b_1 + c_1}{2}$ $\;\;\; \cdots \; (2a)$

Given: $\;$ $s_1 = s$

$\therefore \;$ We have from equations $(1a)$ and $(2a)$

$\dfrac{51 + b_1 + c_1}{2} = 56$ $\implies$ $b_1 + c_1 = 61$ $\;\;\; \cdots \; (3)$

New Area $= \Delta_1 = \sqrt{s_1 \left(s_1 - a_1\right) \left(s_1 - b_1\right) \left(s_1 - c_1\right)}$

i.e. $\;$ $\Delta_1 = \sqrt{s \left(s - a_1\right) \left(s - b_1\right) \left(s - c_1\right)}$ $\;\;\;$ $\left[\because \; s_1 = s\right]$

Substituting the values of $s$ and $a_1$ we have,

$\Delta_1 = \sqrt{56 \left(56 - 51\right) \left(56 - b_1\right) \left(56 - c_1\right)}$

i.e. $\;$ $\Delta_1 = \sqrt{56 \times 5 \left(56 - b_1\right) \left(56 - c_1\right)}$ $\;\;\; \cdots \; (2b)$

Also given: $\;$ $\Delta_1 = \Delta$

$\therefore \;$ We have from equations $(1b)$ and $(2b)$

$\sqrt{56 \times 5 \left(56 - b_1\right) \left(56 - c_1\right)} = 420$

i.e. $\;$ $\sqrt{70 \left(56 - b_1\right) \left(56 - c_1\right)} = 210$

i.e. $\;$ $70 \left(56 - b_1\right) \left(56 - c_1\right) = \left(210\right)^2$

i.e. $\;$ $\left(56 - b_1\right) \left(56 - c_1\right) = 630$

i.e. $\;$ $3136 - 56 \left(b_1 + c_1\right) + b_1 c_1 = 630$

Substituting the value of $\left(b_1 + c_1\right)$ from equation $(3)$ we have,

$b_1 c_1 = 630 - 3136 + \left(56 \times 61\right) = 910$ $\;\;\; \cdots \; (4)$

Now, $\;$ $\left(b_1 - c_1\right)^2 = \left(b_1 + c_1\right)^2 - 4 b_1 c_1$ $\;\;\; \cdots \; (5)$

In view of equations $(3)$ and $(4)$, equation $(5)$ becomes

$\left(b_1 - c_1\right)^2 = \left(61\right)^2 - \left(4 \times 910\right) = 81$

$\implies$ $b_1 - c_1 = 9$ $\;\;\; \cdots \; (6)$

Adding equations $(5)$ and $(6)$ we have,

$2b_1 = 70$ $\implies$ $b_1 = 35$

Substituting the value of $b_1$ in equation $(3)$ gives, $c_1 = 61 - 35 = 26$

$\therefore \;$ The new lengths are $35 \; m$ and $26 \; m$

Properties of Triangles

Find the area of $\triangle ABC$ when $a = \sqrt{3}$, $b = \sqrt{2}$ and $c = \dfrac{\sqrt{6} + \sqrt{2}}{2}$

Given: $\;$ $a = \sqrt{3}$, $\;$ $b = \sqrt{2}$, $\;$ $c = \dfrac{\sqrt{6} + \sqrt{2}}{2}$

By cosine rule,

$\begin{aligned} \cos C & = \dfrac{a^2 + b^2 - c^2}{2 a b} \\\\ & = \dfrac{3 + 2 - \left(\dfrac{6 + 2 + 4 \sqrt{3}}{4}\right)}{2 \times \sqrt{3 \sqrt{2}}} \\\\ & = \dfrac{5 - \left(2 + \sqrt{3}\right)}{2 \sqrt{6}} \\\\ & = \dfrac{3 - \sqrt{3}}{2 \sqrt{6}} \\\\ & = \dfrac{\sqrt{3} - 1}{2 \sqrt{2}} \;\;\; \left[\text{dividing numerator and denominator by } \sqrt{3}\right] \end{aligned}$

$\implies$ $C = \cos^{-1} \left(\dfrac{\sqrt{3} - 1}{2 \sqrt{2}}\right) = 75^\circ$

Now, $\;$ area of $\triangle ABC$ $= \Delta = \dfrac{1}{2}a b \sin C$

$\begin{aligned} i.e. \; \Delta & = \dfrac{1}{2} \times \sqrt{3} \times \sqrt{2} \times \sin \left(75^\circ\right) \\\\ & = \dfrac{1}{2} \times \sqrt{6} \times \dfrac{\left(\sqrt{3} + 1\right)}{2 \sqrt{2}} \\\\ & = \dfrac{\sqrt{3} \left(\sqrt{3} + 1\right)}{4} \\\\ & = 1.183 \; \text{sq units} \end{aligned}$

Properties of Triangles

To get the distance of a point $A$ from a point $B$, a line $BC$ and the angles $ABC$ and $BCA$ are measured and found to be $287 \; cm$ and $55^\circ 32'10''$ and $51^\circ 8' 20''$ respectively. Find the distance $AB$.

Given: $\;$ $\angle ABC = B = 55^\circ 32' 10''$, $\;$ $\angle BCA = C = 51^\circ 8' 20''$, $\;$ $BC = 287 \; cm$

In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$

$\begin{aligned} \therefore \; A & = 180^\circ - \left(B + C\right) \\\\ & = 180^\circ - \left(55^\circ 32' 10'' + 51^\circ 8' 20''\right) \\\\ & = 73^\circ 19' 30'' \end{aligned}$

In $\triangle ABC$, by sine rule, $\;$ $\dfrac{AB}{\sin C} = \dfrac{BC}{\sin A}$

$\therefore \;$ $AB = \dfrac{BC \; \sin C}{\sin A}$

$\begin{aligned} i.e. \; AB & = \dfrac{287 \times \sin \left(51^\circ 8' 20''\right)}{\sin \left(73^\circ 19' 30''\right)} \\\\ & = \dfrac{287 \times 0.7787}{0.9579} \\\\ & = 233.31 \; cm \end{aligned}$

Properties of Triangles

If the angles of a triangle be as $5 : 10 : 21$, and the side opposite the smaller angle be $3 \;$ cm, find the other sides.

Given: $\;$ $A : B : C = 5 : 10 : 21$

Let $k$ be the constant of proportionality.

Then, $\;$ $A = 5k$, $\;$ $B = 10 k$, $\;$ $C = 21 k$

In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$

i.e. $\;$ $5 k + 10 k + 21 k = 180^\circ$

i.e $\;$ $36 k = 180^\circ$ $\implies$ $k = 5^\circ$

$\implies$ $A = 5 k = 25^\circ$, $\;$ $B = 10 k = 50^\circ$, $\;$ $C = 21 k = 105^\circ$

In $\triangle ABC$, by sine rule

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

i.e. $\;$ $b = \dfrac{a \; \sin B}{\sin A} = \dfrac{3 \times \sin \left(50^\circ\right)}{\sin \left(25^\circ\right)} = \dfrac{3 \times 0.7760}{0.4226} = 5.4378 \;$ cm

and $\;$ $c = \dfrac{a \; \sin C}{\sin A} = \dfrac{3 \times \sin \left(105^\circ\right)}{\sin \left(25^\circ\right)} = \dfrac{3 \times 0.9659}{0.4226} = 6.8568 \;$ cm

Properties of Triangles

The base angles of a triangle are $22.5^\circ$ and $112.5^\circ$. Prove that the base is equal to twice the height of the triangle.

Let the base angles be $B = 22.5^\circ$ and $C = 112.5^\circ$

Draw $AD \perp BC$ extended.

In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$

$\therefore \;$ $A = 180^\circ - \left(B + C\right) = 180^\circ - \left(22.5^\circ + 112.5^\circ\right) = 45^\circ$

In $\triangle ABC$, by sine rule,

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = k$ (say), where $k$ is the constant of proportionality.

$\implies$ $b = \dfrac{a \sin B}{\sin A} = \dfrac{a \sin \left(22.5^\circ\right)}{\sin \left(45^\circ\right)}$ $\;\;\; \cdots \; (1)$

Now, in right $\triangle ACD$,

$\begin{aligned} AD & = AC \; \sin \left(\angle ACD\right) \\\\ & = b \; \sin \left(180^\circ - \angle ACB\right) \\\\ & = b \; \sin \left(\angle ACB\right) \\\\ & = b \; \sin \left(112.5^\circ\right) \\\\ & = \dfrac{a \times \sin \left(22.5^\circ\right) \times \sin \left(112.5^\circ\right)}{\sin \left(45^\circ\right)} \;\; \left[\text{by equation (1)}\right] \\\\ & = \dfrac{a \times \left(\dfrac{\sqrt{2 - \sqrt{2}}}{2}\right) \times \left(\dfrac{\sqrt{2 + \sqrt{2}}}{2}\right)}{\dfrac{1}{\sqrt{2}}} \\\\ & = \dfrac{\sqrt{2} \times a \times \sqrt{2}}{4} \\\\ & = \dfrac{a}{2} \end{aligned}$

$\implies$ $a = 2 \; AD$

i.e. $\;$ The base of $\triangle ABC$ is equal to twice its height.

Hence proved.

Properties of Triangles

If $A = 45^\circ$, $B = 75^\circ$ and $C = 60^\circ$, prove that $a + c \sqrt{2} = 2b$

Given: $\;$ $A = 45^\circ$, $\;$ $B = 75^\circ$, $\;$ $C = 60^\circ$

By sine rule,

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = k$

where $k$ is a constant of proportionality.

$\therefore \;$ We have, $\;$ $a = k \sin A = k \sin 45^\circ = \dfrac{k}{\sqrt{2}}$ $\;\;\; \cdots \; (1a)$

$b = k \sin B = k \sin 75^\circ = \dfrac{k \left(\sqrt{3} + 1\right)}{2 \sqrt{2}}$ $\;\;\; \cdots \; (1b)$

$c = k \sin C = k \sin 60^\circ = \dfrac{\sqrt{3} k}{2}$ $\;\;\; \cdots \; (1c)$

Now, from equations $(1a)$ and $(1c)$

$\begin{aligned} a + c \sqrt{2} & = \dfrac{k}{\sqrt{2}} + \dfrac{\sqrt{3} \times \sqrt{2} \; k}{2} \\\\ & = \dfrac{2 k + 2 \sqrt{3} \; k}{2 \sqrt{2}} \\\\ & = \dfrac{2 k \left(\sqrt{3} + 1\right)}{2 \sqrt{2}} \\\\ & = 2b \;\;\; \left[\text{in view of equation (1b)}\right] \end{aligned}$

$\therefore \;$ $a + c \sqrt{2} = 2 b$

Hence proved.

Properties of Triangles

If $\cos A = \dfrac{17}{22}$ and $\cos C = \dfrac{1}{14}$, find the ratio of $a : b : c$.

Given: $\;$ $\cos A = \dfrac{17}{22}$, $\;$ $\cos C = \dfrac{1}{14}$

Now, $\;$ $\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\dfrac{17}{22}\right)^2} = \dfrac{\sqrt{195}}{22}$ $\;\;\; \cdots \; (1a)$

and $\;$ $\sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - \left(\dfrac{1}{14}\right)^2} = \dfrac{\sqrt{195}}{14}$ $\;\;\; \cdots \; (1b)$

In $\triangle ABC$, $\;$ $A + B + C = \pi$

i.e. $\;$ $B = \pi - \left(A + C\right)$

$\begin{aligned} \therefore \; \sin B & = \sin \left[\pi - \left(A + C\right)\right] \\\\ & = \sin \left(A + C\right) \\\\ & = \sin A \cos C + \cos A \sin C \\\\ & = \dfrac{\sqrt{195}}{22} \times \dfrac{1}{14} + \dfrac{17}{22} \times \dfrac{\sqrt{195}}{14} \\\\ & = \dfrac{9 \sqrt{195}}{154} \;\;\; \cdots \; (1c) \end{aligned}$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

$\begin{aligned} \implies a : b : c & = \sin A : \sin B : \sin C \\\\ & = \dfrac{\sqrt{195}}{22} : \dfrac{9 \sqrt{195}}{154} : \dfrac{\sqrt{195}}{14} \;\; \left[\text{by equations (1a), (1b), (1c)}\right] \\\\ & = \dfrac{1}{22} : \dfrac{9}{154} : \dfrac{1}{14} \\\\ & = \dfrac{7}{154} : \dfrac{9}{154} : \dfrac{11}{154} \end{aligned}$

$\therefore \;$ $a : b : c = 7 : 9 : 11$

Properties of Triangles

Given $A = 10^\circ$, $a = 2308.7$, $b = 7903.2$, find the smaller value of $c$.

Given: $\;$ $A = 10^\circ$, $\;$ $a = 2308.7$, $\;$ $b = 7903.2$

Now, $\;$ $b \sin A = 7903.2 \times \sin \left(10^\circ\right) = 1371.9955 < a$

$\implies$ There are two values of $B$, $\;$ $0^\circ < B < 90^\circ$ (first quadrant) and $90^\circ < B < 180^\circ$ (second quadrant)

By sine rule in $\triangle ABC$, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

$\implies$ $\sin B = \dfrac{b \sin A}{a} = \dfrac{7903.2 \times \sin \left(10^\circ\right)}{2308.7} = 0.5943$

$\implies$ $B = \sin^{-1} \left(0.5943\right)$

i.e. $\;$ $B = 36^\circ 27' 45.72''$ $\;$ or $\;$ $B = 143^\circ 32' 14''$

Now, in $\triangle ABC$, $A + B + C = 180^\circ$ $\implies$ $C = 180^\circ - \left(A + B\right)$

When $\;$ $B = 36^\circ 27' 45.72''$, $\;$ $C = 180^\circ - \left(10^\circ + 36^\circ 27' 45.72''\right) = 133^\circ 32' 14''$

When $\;$ $B = 143^\circ 32' 14''$, $\;$ $C = 180^\circ - \left(10^\circ + 143^\circ 32' 14''\right) = 26^\circ 27' 46''$

$\therefore \;$ Smaller value of $c$ is obtained for $C = 26^\circ 27' 46''$

By sine rule we have,

$c = \dfrac{a \sin C}{\sin A} = \dfrac{2308.7 \times \sin \left(26^\circ 27' 46''\right)}{\sin \left(10^\circ\right)} = \dfrac{2308.7 \times 0.4456}{0.1736} = 5926.02$

Properties of Triangles

Two straight roads intersect at an angle of $30^\circ$. From the point of junction two pedestrians $A$ and $B$ start at the same time, $A$ walking along one road at the rate of $5$ miles per hour and $B$ walking uniformly along the other road. At the end of $3$ hours they are $9$ miles apart. Show that there are two rates at which $B$ may walk to fulfill this condition and find them.

Let $R_1$ and $R_2$ be the two roads intersecting at point $O$.

Let $A$ walk along road $A$ from the point $O$ at a rate of $5$ miles per hour.

$\therefore \;$ In $3$ hours, distance covered by $A$ $= 5 \times 3 = 15$ miles

Let $B$ walk along road $R_2$ covering a distance of $OB$ miles in $3$ hours.

After $3$ hours, distance between $A$ and $B$ $= AB = 9$ miles

In the figure, $OB = a$, $\;$ $OA = b = 15$, $\;$ $AB = o = 9$, $\;$ $\angle BOA = O = 30^\circ$

Now, $\;$ $b \sin O = 15 \sin \left(30^\circ\right) = \dfrac{15}{2} < o$

$\implies$ There are two values of $B$, $\;$ $0^\circ < B < 90^\circ$ (first quadrant) and $90^\circ < B < 180^\circ$ (second quadrant)

i.e. $\;$ $B$ can walk with two rates towards $A$.

By sine rule, in $\triangle OAB$, $\;$ $\dfrac{a}{\sin A} = \dfrac{o}{\sin O} = \dfrac{b}{\sin B}$

$\implies$ $\sin B = \dfrac{b \sin O}{o} = \dfrac{15 \sin \left(30^\circ\right)}{9} = \dfrac{15}{9} \times \dfrac{1}{2} = 0.8333$

$\implies$ $B = \sin^{-1} \left(0.8333\right)$

i.e. $\;$ $B = 56^\circ 26' 21.12''$ $\;$ or $\;$ $B = 123^\circ 33' 39''$

Now, in $\triangle OAB$, $\;$ $O + A + B = 180^\circ$ $\implies$ $A = 180^\circ - \left(O + B\right)$

When $\;$ $B = 56^\circ 26' 21.12''$, $\;$ $A = 180^\circ - \left(30^\circ + 56^\circ 26' 21.12''\right) = 93^\circ 33' 39''$

When $\;$ $B = 123^\circ 33' 39''$, $\;$ $A = 180^\circ - \left(30^\circ + 123^\circ 33' 39''\right) = 26^\circ 26' 21''$

Also, by sine rule in $\triangle OAB$, $\;$ $a = \dfrac{o \sin A}{\sin O}$

When $\;$ $A = 93^\circ 33' 39''$,

$a = \dfrac{9 \times \sin \left(93^\circ 33' 39''\right)}{\sin \left(30^\circ\right)} = 9 \times 0.9981 \times 2 = 17.9658$ miles

When $\;$ $A = 26^\circ 26' 21''$,

$a = \dfrac{9 \times \sin \left(26^\circ 26' 21''\right)}{\sin \left(30^\circ\right)} = 9 \times 0.4452 \times 2 = 8.0136$ miles

$\therefore \;$ Distance covered by $B$ in $1$ hour $= \dfrac{17.9658}{3} = 5.99$ miles $\;$ or $\;$ $= \dfrac{8.0136}{3} = 2.67$ miles

$\therefore \;$ Rate at which $B$ walks is $5.99$ miles/hour or $2.67$ miles/hour

Properties of Triangles

In $\triangle ABC$, if $a = 9$, $b = 12$ and $A = 30^\circ$, find $c$.

Given: $\;$ $a = 9$, $\;$ $b = 12$, $\;$ $A = 30^\circ$

Now, $\;$ $b \sin A = 12 \times \sin 30^\circ = \dfrac{12}{2} = 6 < a$

$\implies$ There are two values of $B$, $\;$ $0^\circ < B < 90^\circ$ (first quadrant) and $90^\circ < B < 180^\circ$ (second quadrant)

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

$\implies$ $\sin B = \dfrac{b \sin A}{a}$

i.e. $\;$ $\sin B = \dfrac{12 \times \sin 30^\circ}{9} = \dfrac{12}{9 \times 2} = \dfrac{2}{3}$

i.e. $\;$ $B = \sin^{-1} \left(\dfrac{2}{3}\right)$

i.e. $\;$ $B = 41.8103^\circ = 41^\circ 48' 37.08''$ $\;$ or $\;$ $B = 138.1897^\circ = 138^\circ 11' 22.92''$

By projection formula, $\;$ $c = a \cos B + b \cos A$

When $\;$ $B = 41^\circ 48' 37.08''$,

$c = 9 \cos \left(41^\circ 48' 37.08''\right) + 12 \cos \left(30^\circ\right)$

i.e. $c = 9 \times 0.7454 + 12 \times 0.8660 = 17.10$

When $\;$ $B = 138^\circ 11' 22.92''$,

$c = 9 \cos \left(138^\circ 11' 22.92''\right) + 12 \cos \left(30^\circ\right)$

i.e. $c = 9 \times \left(- 0.7454\right) + 12 \times 0.8660 = 3.68$

Properties of Triangles

1. If $B = 30^\circ$, $c = 150$ and $b = 50 \sqrt{3}$,
   • prove that of the two triangles which satisfy the data, one will be isosceles and the other right angled.
   • Find the greater value of the third side.
2. Would be solution have been ambiguous if $B = 30^\circ$, $c = 150$ and $b = 75$?

1. Given: $\;$ In $\triangle ABC$, $\;$ $B = 30^\circ$, $\;$ $c = 150$, $\;$ $b = 50 \sqrt{3}$
   
   
   • Now, $\;$ $c \sin B = 150 \times \sin 30^\circ = \dfrac{150}{2} = 75 < b$
     
     $\implies$ There are two values of $C$, $0^\circ < C < 90^\circ$ (first quadrant) and $90^\circ < C < 180^\circ$ (second quadrant)
     
     By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
     
     $\implies$ $\sin C = \dfrac{c \sin B}{b} = \dfrac{150 \times \sin 30^\circ}{50 \sqrt{3}} = \dfrac{\sqrt{3}}{2}$
     
     i.e. $\;$ $C = \sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)$
     
     $\implies$ $C = 60^\circ$ $\;$ or $\;$ $C = 120^\circ$
     
     In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$
     
     $\implies$ $A = 180^\circ - \left(B + C\right)$
     
     When $\;$ $C = 60^\circ$, $\;$ $A = 180^\circ - \left(30^\circ + 60^\circ\right) = 90^\circ$
     
     When $\;$ $C = 120^\circ$, $\;$ $A = 180^\circ - \left(30^\circ + 120^\circ\right) = 30^\circ$
     
     $\therefore \;$ There are two possible triangles with
     
     $A = 90^\circ$, $\;$ $B = 30^\circ$, $\;$ $C = 60^\circ$ $\;$ i.e. right angled triangle with right angle $A$
     
     or, $\;$ $A = 30^\circ$, $\;$ $B = 30^\circ$, $\;$ $C = 120^\circ$ $\;$ i.e. isosceles triangle with $\angle A = \angle B$
   
   
   • The greater value of the third side `$a$' is obtained in the triangle with $A = 90^\circ$, $\;$ $B = 30^\circ$, $\;$ $C = 60^\circ$
     
     By sine rule, $\;$ $a = \dfrac{c \sin A}{\sin C}$
     
     i.e. $\;$ $a = \dfrac{150 \times 1}{\sqrt{3} / 2} = 100 \sqrt{3}$
2. When $\;$ $B = 30^\circ$, $\;$ $c = 150$, $\;$ $b = 75$
   
   Now, $\;$ $c \sin B = 150 \times \sin 30^\circ = \dfrac{150}{2} = 75 = b$
   
   $\implies$ The solution of $\triangle ABC$ is not ambiguous.
   
   There is only ONE right angled triangle with $C = 90^\circ$

Properties of Triangles

If $\;$ $2b = 3a$ and $\tan^2 A = \dfrac{3}{5}$, prove that there are two values to the third side, one of which is double the other.

Given: $\;$ $2b = 3a$ $\implies$ $a = \dfrac{2 b}{3}$ $\;\;\; \cdots \; (1)$

and $\;$ $\tan^2 A = \dfrac{3}{5}$

i.e. $\;$ $1 + \tan^2 A = \sec^2 A = 1 + \dfrac{3}{5} = \dfrac{8}{5}$

i.e. $\;$ $\sec A = \dfrac{2\sqrt{2}}{\sqrt{5}}$ $\implies$ $\cos A = \dfrac{\sqrt{5}}{2 \sqrt{2}}$ $\;\;\; \cdots \; (2)$

In $\triangle ABC$, by cosine rule, $\;$ $\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$

i.e. $\;$ $\dfrac{\sqrt{5}}{2 \sqrt{2}} = \dfrac{b^2 + c^2 - \dfrac{4 b^2}{9}}{2 b c}$ $\;\;\;$ [by equations $(1)$ and $(2)$]

i.e. $\;$ $\dfrac{\sqrt{5}}{\sqrt{2}} b c = \dfrac{5}{9} b^2 + c^2$

i.e. $\;$ $c^2 - \left(\dfrac{\sqrt{5}}{\sqrt{2}}b\right)c + \dfrac{5}{9}b^2 = 0$

i.e. $\;$ $c = \dfrac{\dfrac{\sqrt{5}}{\sqrt{2}}b \pm \sqrt{\dfrac{5b^2}{2} - \dfrac{20 b^2}{9}}}{2}$

i.e. $\;$ $c = \dfrac{\dfrac{\sqrt{5}}{\sqrt{2}}b \pm \dfrac{\sqrt{5}}{3 \sqrt{2}}b}{2}$

i.e. $\;$ $c = \dfrac{\sqrt{5}}{2 \sqrt{2}} b \pm \dfrac{\sqrt{5}}{6 \sqrt{2}}b$

i.e. $\;$ $c = \dfrac{4 \sqrt{5}}{6 \sqrt{2}}b$ $\;$ or $\;$ $c = \dfrac{2 \sqrt{5}}{6 \sqrt{2}}b$

i.e. $\;$ $c = 2 \left(\dfrac{\sqrt{5}}{3 \sqrt{2}}b\right)$ $\;$ or $\;$ $c = \dfrac{\sqrt{5}}{3 \sqrt{2}}b$

$\therefore \;$ There are two values to the third side $c$, one of which is double the other.

Hence proved.

Properties of Triangles

If $\;$ $a = 2$, $c = \sqrt{3} + 1$, and $A = 45^\circ$, solve the triangle.

Given: $\;$ $a = 2$, $\;$ $c = \sqrt{3} + 1$, $\;$ $A = 45^\circ$

Now, $\;$ $c \sin A = \left(\sqrt{3} + 1\right) \sin 45^\circ = \dfrac{\sqrt{3} + 1}{\sqrt{2}} = 1.932 < a$

$\implies$ There are two values of $C$, $0^\circ < C < 90^\circ$ and $90^\circ < C < 180^\circ$

In $\triangle ABC$, by sine rule, $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

$\implies$ $\dfrac{2}{\sin 45^\circ} = \dfrac{\sqrt{3} + 1}{\sin C}$

i.e. $\;$ $\sin C = \dfrac{\left(\sqrt{3} + 1\right) \sin 45^\circ}{2} = \dfrac{\sqrt{3} + 1}{2 \sqrt{2}}$

$\implies$ $C = 75^\circ$ $\;$ or $\;$ $105^\circ$

In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$ $\implies$ $B = 180^\circ - \left(A + C\right)$

When $\;$ $C = 75^\circ$, $\;$ $B = 180^\circ - \left(45^\circ + 75^\circ\right) = 60^\circ$

When $\;$ $C = 105^\circ$, $\;$ $B = 180^\circ - \left(45^\circ + 10^\circ\right) = 30^\circ$

Also, by sine rule,

$b = \dfrac{a \sin B}{\sin A}$

When $\;$ $B = 60^\circ$, $\;$ $b = \dfrac{2 \times \sin 60^\circ}{\sin 45^\circ} = \dfrac{2 \times \dfrac{\sqrt{3}}{2}}{\dfrac{1}{\sqrt{2}}} = \sqrt{6}$

When $\;$ $B = 30^\circ$, $\;$ $b = \dfrac{2 \times \sin 30^\circ}{\sin 45^\circ} = \dfrac{2 \times \dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}} = \sqrt{2}$

$\therefore \;$ $B = 60^\circ$, $\;$ $C = 75^\circ$, $\;$ $b = \sqrt{6}$

or, $\;$ $B = 30^\circ$, $\;$ $C = 105^\circ$, $\;$ $b = \sqrt{2}$

Properties of Triangles

If $\;$ $\tan \phi = \left(\dfrac{a - b}{a + b}\right) \cot \left(\dfrac{C}{2}\right)$, $\;$ prove that $\;$ $c = \left(a + b\right) \left[\dfrac{\sin \left(\dfrac{C}{2}\right)}{\cos \phi}\right]$

Given: $\;$ $\tan \phi = \left(\dfrac{a - b}{a + b}\right) \cot \left(\dfrac{C}{2}\right)$ $\;\;\; \cdots \; (1)$

By cosine rule, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

$\begin{aligned} i.e. \; c^2 & = a^2 + b^2 - 2 a b \cos C \\\\ & \left[\text{Note: }\sin^2 \left(\dfrac{\theta}{2}\right) + \cos^2 \left(\dfrac{\theta}{2}\right) = 1 \right. \\ & \left. \hspace{1.2cm} \cos \theta = \cos^2 \left(\dfrac{\theta}{2}\right) - \sin^2 \left(\dfrac{\theta}{2}\right) \right] \\\\ & = \left(a^2 + b^2\right) \left[\sin^2 \left(\dfrac{C}{2}\right) + \cos^2 \left(\dfrac{C}{2}\right)\right] - 2 a b \left[\cos^2 \left(\dfrac{C}{2}\right) - \sin^2 \left(\dfrac{C}{2}\right)\right] \\\\ & = \sin^2 \left(\dfrac{C}{2}\right) \left[a^2 + b^2 + 2 a b\right] + \cos^2 \left(\dfrac{C}{2}\right) \left[a^2 + b^2 - 2 a b\right] \\\\ & = \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) + \left(a - b\right)^2 \cos^2 \left(\dfrac{C}{2}\right) \\\\ & = \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) \left[1 + \left(\dfrac{a - b}{a + b}\right)^2 \times \dfrac{\cos^2 \left(\dfrac{C}{2}\right)}{\sin^2 \left(\dfrac{C}{2}\right)}\right] \\\\ & = \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) \left[1 + \left(\dfrac{a - b}{a + b}\right)^2 \cot^2 \left(\dfrac{C}{2}\right)\right] \\\\ & = \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) \left[1 + \tan^2 \phi\right] \;\;\; \left[\text{by equation (1)}\right] \\\\ & = \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) \sec^2 \phi \\\\ \therefore \; c & = \left(a + b\right) \sin \left(\dfrac{C}{2}\right) \sec \phi \\\\ i.e. \; c & = \left(a + b\right) \left[\dfrac{\sin \left(\dfrac{C}{2}\right)}{\cos \phi}\right] \end{aligned}$

Hence proved.

Properties of Triangles

The sides of a triangle are $9$ and $3$, and the difference of the angles opposite to them is $90^\circ$. Find the base and the angles.

Let the given sides of the triangle be $a = 9$ and $b = 3$.

Difference of angles opposite to sides $a$ and $b$ is $A - B = 90^\circ$

Now, $\;$ $\tan \left(\dfrac{A - B}{2}\right) = \left(\dfrac{a - b}{a + b}\right) \cot \left(\dfrac{C}{2}\right)$

i.e. $\;$ $\tan \left(\dfrac{90^\circ}{2}\right) = \left(\dfrac{9 - 3}{9 + 3}\right) \cot \left(\dfrac{C}{2}\right)$

i.e. $\;$ $\cot \left(\dfrac{C}{2}\right) = \dfrac{12 \times \tan \left(45^\circ\right)}{6} = 2$

i.e. $\;$ $\dfrac{C}{2} = \cot^{-1} \left(2\right) = 26.5651^\circ$

$\implies$ $C = 53.1302^\circ = 53^\circ 7' 48.72''$

By cosine rule, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

i.e. $\;$ $c^2 = a^2 + b^2 - 2 a b \cos C$

i.e. $\;$ $c^2 = 9^2 + 3^2 - 2 \times 9 \times 3 \times \cos \left(53^\circ 7' 48.72''\right)$

i.e. $\;$ $c^2 = 57.6054$ $\implies$ $c = 7.5898$

In $\triangle ABC$, $\;$ $A + B = 180^\circ - C$

i.e. $A + B = 180^\circ - 53^\circ 7' 48.72'' = 126^\circ 52' 11''$ $\;\;\; \cdots \; (1a)$

Given: $\;$ $A - B = 90^\circ$ $\;\;\; \cdots \; (1b)$

Adding equations $(1a)$ and $(1b)$ we get,

$2 A = 216^\circ 52' 11''$ $\implies$ $A = 108^\circ 26' 6''$

Substituting the value of $A$ in equation $(1b)$ gives

$B = A - 90^\circ = 108^\circ 26' 6'' - 90^\circ = 18^\circ 26' 6''$

$\therefore \;$ $A = 108^\circ 26' 6''$, $\;$ $B = 18^\circ 26' 6''$, $\;$ $C = 53^\circ 7' 48.72''$, $\;$ $c = 7.5898$

Properties of Triangles

If in any $\triangle ABC$, $a = 5$, $b = 4$ and $\cos \left(A - B\right) = \dfrac{31}{32}$, then show that the third side of the triangle is $c = 6$.

In any $\triangle ABC$, by law of tangents,

$\tan \left(\dfrac{A - B}{2}\right) = \dfrac{a - b}{a + b} \cot \left(\dfrac{C}{2}\right)$ $\;\;\; \cdots \; (1)$

Now, by definition,

$\cot \left(\dfrac{C}{2}\right) = \dfrac{\sin C}{1 - \cos C}$ $\;\;\; \cdots \; (2a)$,

$\tan \left(\dfrac{A - B}{2}\right) = \dfrac{1 - \cos \left(A - B\right)}{\sin \left(A - B\right)}$ $\;\;\; \cdots \; (2b)$

and $\;$ $\sin \left(A - B\right) = \sqrt{1 - \cos^2 \left(A - B\right)}$ $\;\;\; \cdots \; (2c)$

$\therefore \;$ In view of equations $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ becomes

$\dfrac{1 - \cos \left(A - B\right)}{\sqrt{1 - \cos^2 \left(A - B\right)}} = \left(\dfrac{a - b}{a + b}\right) \left(\dfrac{\sin C}{1 - \cos C}\right)$ $\;\;; \cdots \; (3)$

Given: $\;$ $a = 5$, $\;$ $b = 4$, $\;$ $\cos \left(A - B\right) = \dfrac{31}{32}$ $\;\;\; \cdots \; (4)$

$\therefore \;$ In view of equation $(4)$ equation $(3)$ becomes,

$\dfrac{1 - \dfrac{31}{32}}{\sqrt{1 - \dfrac{961}{1024}}} = \left(\dfrac{5 - 4}{5 + 4}\right) \left(\dfrac{\sin C}{1 - \cos C}\right)$

i.e. $\;$ $\dfrac{\dfrac{1}{32}}{\dfrac{\sqrt{63}}{32}} = \dfrac{1}{9} \times \left(\dfrac{\sqrt{1 - \cos^2 C}}{1 - \cos C}\right)$

i.e. $\;$ $\dfrac{\sqrt{1 - \cos^2 C}}{1 - \cos C} = \dfrac{9}{\sqrt{63}}$

i.e. $\;$ $\dfrac{1 - \cos^2 C}{\left(1 - \cos C\right)^2} = \dfrac{81}{63}$

i.e. $\;$ $\dfrac{\left(1 + \cos C\right) \left(1 - \cos C\right)}{\left(1 - \cos C\right)^2} = \dfrac{81}{63}$

i.e. $\;$ $\dfrac{1 + \cos C}{1 - \cos C} = \dfrac{81}{63}$ $\;\;\;$ provided $\;\;$ $\left(1 - \cos C\right) \neq 0$

i.e. $\;$ $63 + 63 \cos C = 81 - 81 \cos C$

i.e. $\;$ $144 \cos C = 18$

$\implies$ $\cos C = \dfrac{1}{8}$ $\;\;\; \cdots \; (5)$

By cosine rule, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$ $\;\;\; \cdots \; (6)$

In view of equations $(4)$ and $(5)$ equation $(6)$ becomes,

$\dfrac{1}{8} = \dfrac{25 + 16 - c^2}{2 \times 5 \times 4}$

i.e. $\;$ $5 = 41 - c^2$

i.e. $\;$ $c^2 = 36$ $\implies$ $c = 6$

Properties of Triangles

If $b = 91$, $c = 125$ and $\tan \left(\dfrac{A}{2}\right) = \dfrac{17}{6}$, find the value of $a$.

By definition, $\;$ $\tan\left(\dfrac{A}{2}\right) = \sqrt{\dfrac{\left(s - b\right) \left(s - c\right)}{s \left(s - a\right)}}$ $\;\;\; \cdots \; (1)$

where $\;$ $s = \dfrac{a + b + c}{2}$ $\;$ is the semi-perimeter of $\triangle ABC$.

Given:

$b = 91$, $\;$ $c = 125$, $\;$ $\tan \left(\dfrac{A}{2}\right) = \dfrac{17}{6}$ $\;\;\; \cdots \; (2)$

Here, $\;$ $s = \dfrac{a + 91 + 125}{2} = \dfrac{a}{2} + 108$ $\;\;\; \cdots \; (3)$

Now, from equation $(1)$ we have,

$\tan^2 \left(\dfrac{A}{2}\right) = \dfrac{\left(s - b\right) \left(s - c\right)}{s \left(s - a\right)}$

i.e. $\left(\dfrac{17}{6}\right)^2 = \dfrac{\left(\dfrac{a}{2} + 108 - 91\right) \left(\dfrac{a}{2} + 108 - 125\right)}{\left(\dfrac{a}{2} + 108\right) \left(\dfrac{a}{2} + 108 - a\right)}$ $\;\;$ [by equations $(2)$ and $(3)$]

i.e. $\;$ $\dfrac{289}{36} = \dfrac{\left(\dfrac{a}{2} + 17\right) \left(\dfrac{a}{2} - 17\right)}{\left(\dfrac{a}{2} + 108\right) \left(108 - \dfrac{a}{2}\right)}$

i.e. $\;$ $\dfrac{289}{36} = \dfrac{\dfrac{a^2}{4} - 289}{11664 - a^2}$

i.e. $\;$ $289 \left(11664 - \dfrac{a^2}{4}\right) = 36 \left(\dfrac{a^2}{4} - 289\right)$

i.e. $\;$ $3370896 - \dfrac{289 a^2}{4} = \dfrac{36 a^2}{4} - 10404$

i.e. $\;$ $3381300 = \dfrac{325 a^2}{4}$

i.e. $\;$ $a^2 = 41616$

$\implies$ $a = 204$

Properties of Triangles

If $a = 2$, $b = 1 + \sqrt{3}$ and $C = 60^\circ$, solve the triangle.

Given: $\;$ $a = 2$, $\;$ $b = 1 + \sqrt{3}$, $\;$ $C = 60^\circ$

In $\triangle ABC$, by cosine rule,

$\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

$\begin{aligned} \implies c^2 & = a^2 + b^2 - 2 a b \cos C \\\\ & = \left(2\right)^2 + \left(1 + \sqrt{3}\right)^2 - 2 \times 2 \times \left(1 + \sqrt{3}\right) \cos 60^\circ \\\\ & = 4 + 1 + 3 + 2 \sqrt{3} - 2 - 2 \sqrt{3} \end{aligned}$

i.e. $\;$ $c^2 = 6$ $\implies$ $c = \sqrt{6}$

By sine rule,

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

$\therefore \;$ We have, $\;$ $\dfrac{2}{\sin A} = \dfrac{\sqrt{6}}{\sin 60^\circ}$

i.e. $\;$ $\sin A = \dfrac{2 \sin 60^\circ}{\sqrt{6}} = 2 \times \dfrac{\sqrt{3}}{2} \times \dfrac{1}{\sqrt{6}} = \dfrac{1}{\sqrt{2}}$

$\implies$ $A = \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) = 45^\circ$

and $\;$ $\dfrac{1 + \sqrt{3}}{\sin B} = \dfrac{\sqrt{6}}{\sin 60^\circ}$

i.e. $\;$ $\sin B = \dfrac{\left(1 + \sqrt{3}\right) \sin 60^\circ}{\sqrt{6}} = \dfrac{\left(1 + \sqrt{3}\right) \sqrt{3}}{2 \sqrt{6}} = \dfrac{1 + \sqrt{3}}{2 \sqrt{2}}$

$\implies$ $B = \sin^{-1} \left(\dfrac{1 + \sqrt{3}}{2 \sqrt{2}}\right) = 75^\circ$

$\therefore \;$ $c = \sqrt{6}$, $\;$ $A = 45^\circ$, $\;$ $B = 75^\circ$

Properties of Triangles

If $a = 2b$ and $C = 120^\circ$, find the values of $A$, $B$ and the ratio of $c$ to $a$.

In $\triangle ABC$,

$\tan \left(\dfrac{A - B}{2}\right) = \dfrac{a - b}{a + b} \cot \left(\dfrac{C}{2}\right)$

$\begin{aligned} i.e. \; \tan \left(\dfrac{A - B}{2}\right) & = \dfrac{2b - b}{2b + b} \cot \left(\dfrac{120^\circ}{2}\right) \\\\ & = \dfrac{1}{3} \cot 60^\circ \\\\ & = \dfrac{1}{3 \sqrt{3}} = 0.1925 \end{aligned}$

$\implies$ $\dfrac{A - B}{2} = \tan^{-1} \left(0.1925\right) = 10.8962^\circ$

i.e. $\;$ $A - B = 2 \times 10.8962^\circ = 21.7924^\circ$ $\;\;\; \cdots \; (1)$

In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$

i.e. $\;$ $A + B = 180^\circ - C = 180^\circ - 120^\circ = 60^\circ$ $\;\;\; \cdots \; (2)$

Adding equations $(1)$ and $(2)$ we have,

$2 A = 60^\circ + 21.7924^\circ = 81.7924^\circ$

$\implies$ $A = 40.8962^\circ = 40^\circ 53' 46.32''$

Substituting the value of $A$ in equation $(2)$ we have,

$B = 60^\circ - A = 60^\circ - 40^\circ 53' 46.32'' = 19^\circ 6' 14''$

$\therefore \;$ $A = 40^\circ 53' 46.32''$, $\;$ $B = 19^\circ 6' 14''$

Now, in $\triangle ABC$, by cosine rule we have,

$\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

i.e. $\;$ $\cos 120^\circ = \dfrac{a^2 + \dfrac{a^2}{4} - c^2}{2 \times a \times \dfrac{a}{2}}$

i.e. $\;$ $-\dfrac{1}{2} = \dfrac{\dfrac{5 a^2}{4} - c^2}{a^2}$

i.e. $\;$ $- a^2 = \dfrac{5 a^2}{2} - 2 c^2$

i.e. $\;$ $\dfrac{7 a^2}{2} = 2 c^2$

i.e. $\;$ $\dfrac{c^2}{a^2} = \dfrac{7}{4}$

$\implies$ $\dfrac{c}{a} = \dfrac{\sqrt{7}}{2}$

Properties of Triangles

If the angles of a triangle be in A.P. and the lengths of the greatest and the least sides be $24 \; cm$ and $16 \; cm$ respectively, find the length of the third side and the other angles.

Let the angles of $\triangle ABC$ be $A$, $B$ and $C$.

Given: $\;$ $A, \; B, \; C$ are in $A.P.$

Then the least side is `$a$' and the greatest side is `$c$'.

Given: $\;$ Least side $= a = 16 \; cm$; $\;$ greatest side $= c = 24 \; cm$

$\because \;$ $A, \; B, \; C$ are in $A.P.$ $\implies$ $2 B = A + C$ $\;\;\; \cdots \; (1)$

Now, in $\triangle ABC$, $\;$ $A + B + C = 180^\circ$

$\implies$ $A + C = 180^\circ - B$ $\;\;\; \cdots \; (2)$

$\therefore \;$In view of equation $(2)$, equation $(1)$ becomes

$2 B = 180^\circ - B$

i.e. $3 B = 180^\circ$ $\implies$ $B = 60^\circ$

By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$

$\begin{aligned} i.e. \; b^2 & = c^2 + a^2 - 2 c a \cos B \\\\ & = 24^2 + 16^2 - \left(2 \times 24 \times 16 \times \cos 60^\circ\right) \\\\ & = 576 + 256 - 384 = 448 \end{aligned}$

$\therefore \;$ $b = \sqrt{448} = 21.17 \; cm$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

$\therefore \;$ We have, $\;$ $\dfrac{16}{\sin A} = \dfrac{21.17}{\sin 60^\circ}$

i.e. $\;$ $\sin A = \dfrac{16 \times \sqrt{3}}{21.17 \times 2} = 0.6545$

$\implies$ $A = \sin^{-1} \left(0.6545\right) = 40.8817^\circ = 40^\circ 52' 54.12''$

and, $\;$ $\dfrac{21.17}{\sin 60^\circ} = \dfrac{24}{\sin C}$

i.e. $\;$ $\sin C = \dfrac{24 \times \sqrt{3}}{21.17 \times 2} = 0.9818$

$\implies$ $C = \sin^{-1} \left(0.9818\right) = 79.0520^\circ = 79^\circ 3' 7.2''$

$\therefore \;$ $b = 21.17 \; cm$, $\;$ $A = 40^\circ 52' 54.12''$, $\;$ $B = 60^\circ$, $\;$ $C = 79^\circ 3' 7.2''$

Properties of Triangles

If $b = 90$, $c = 70$ and $A = 72^\circ 48' 30''$, find $B$ and $C$.

Given: $\;$ $b = 90$, $\;$ $c = 70$, $\;$ $A = 72^\circ 48' 30''$

By definition,

$\begin{aligned} \tan \left(\dfrac{B - C}{2}\right) & = \dfrac{b - c}{b + c} \cot \left(\dfrac{A}{2}\right) \\\\ & = \dfrac{90 - 70}{90 + 70} \cot \left(\dfrac{72^\circ 48' 30''}{2}\right) \\\\ & = \dfrac{20}{160} \cot \left(36^\circ 24' 15''\right) \\\\ & = \dfrac{1}{8} \times 1.3562 = 0.1695 \end{aligned}$

$\therefore \;$ $\dfrac{B- C}{2} = \tan^{-1} \left(0.1695\right) = 9.6202^\circ$

$\implies$ $B - C = 2 \times 9.6202^\circ = 19.2404^\circ = 19^\circ 14' 25.44''$ $\;\;\; \cdots \; (1)$

Now, in $\triangle ABC$, $\;$ $A + B + C = 180^\circ$

$\implies$ $B + C = 180^\circ - A$

i.e. $\;$ $B + C = 180^\circ - 72^\circ 48' 30'' = 107^\circ 11' 30''$ $\;\;\; \cdots \; (2)$

Adding equations $(1)$ and $(2)$ we have,

$2 B = 126^\circ 25' 55.44''$ $\implies$ $B = 63^\circ 12' 57.78''$

Substituting the value of $B$ in equation $(2)$ gives,

$C = 107^\circ 11' 30'' - 63^\circ 12' 57.78'' = 43^\circ 58' 32''$

$\therefore \;$ $B = 63^\circ 12' 57.78''$, $\;$ $C = 43^\circ 58' 32''$

Properties of Triangles

Find all the angles of $\triangle ABC$ when $a = 25$, $b = 26$ and $c = 27$.

Given: $\;$ $a = 25$, $\;$ $b = 26$, $\;$ $c = 27$

In $\triangle ABC$, $\;$ $\tan \left(\dfrac{A}{2}\right) = \sqrt{\dfrac{\left(s - b\right) \left(s - c\right)}{s \left(s - a\right)}}$, $\;$ $\tan \left(\dfrac{B}{2}\right) = \sqrt{\dfrac{\left(s - c\right) \left(s - a\right)}{s \left(s - b\right)}}$

where $s$ is the semi-perimeter of $\triangle ABC$ $= \dfrac{a + b + c}{2}$

Here, $\;$ $s = \dfrac{25 + 26 + 27}{2} = 39$

$\begin{aligned} \tan \left(\dfrac{A}{2}\right) & = \sqrt{\dfrac{\left(39 - 26\right) \left(39 - 27\right)}{39 \times \left(39 - 25\right)}} \\\\ & = \sqrt{\dfrac{13 \times 12}{39 \times 14}} = \sqrt{\dfrac{2}{7}} = 0.5345 \end{aligned}$

$\implies$ $\dfrac{A}{2} = \tan^{-1} \left(0.5345\right) = 28.1245^{\circ}$

$\implies$ $A = 2 \times 28.1245^{\circ} = 56.2490^{\circ} = 56^{\circ}14'56.4''$

$\begin{aligned} \tan \left(\dfrac{B}{2}\right) & = \sqrt{\dfrac{\left(39 - 27\right) \left(39 - 25\right)}{39 \times \left(39 - 26\right)}} \\\\ & = \sqrt{\dfrac{12 \times 14}{39 \times 13}} = 0.5757 \end{aligned}$

$\implies$ $\dfrac{B}{2} = \tan^{-1} \left(0.5757\right) = 29.9290^{\circ}$

$\implies$ $B = 2 \times 29.9290^{\circ} = 59.858^{\circ} = 59^{\circ}51'28.8''$

$\therefore \;$ $C = 180^{\circ} - \left(56^\circ 14' 56.4'' + 59^\circ 51' 28.8''\right) = 180^\circ - 116^\circ 6' 25.2'' = 63^\circ 53' 35''$

$\therefore \;$ $A = 56^\circ 14' 56.4''$, $\;$ $B = 59^\circ 51' 28.8''$, $\;$ $C = 63^\circ 53' 35''$

Properties of Triangles

The sides of a triangle are $x^2 + x + 1$, $2 x + 1$ and $x^2 - 1$. Show that the greatest angle is $120^{\circ}$.

Let the sides of the triangle be

$a = x^2 + x + 1$, $\;$ $b = 2 x + 1$, $\;$ $c = x^2 - 1$

Then, greatest side is side `$a$' and greatest angle is `$A$'.

By cosine rule, $\;$ $\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$

$\begin{aligned} i.e. \; \cos A & = \dfrac{\left(2 x + 1\right)^2 + \left(x^2 - 1\right)^2 - \left(x^2 + x + 1\right)^2}{2 \left(2 x + 1\right) \left(x^2 - 1\right)} \\\\ & = \dfrac{4 x^2 + 4 x + 1 + x^4 - 2 x^2 + 1 - \left(x^4 + x^2 + 1 + 2 x^3 + 2 x + 2x^2\right)}{2 \left(2x + 1\right) \left(x^2 - 1\right)} \\\\ & = \dfrac{x^4 + 2 x^2 + 4x + 2 - x^4 - 2 x^3 - 3 x^2 - 2 x - 1}{2 \left(2 x^3 - 2 x + x^2 - 1\right)} \\\\ & = \dfrac{- 2 x^3 - x^2 + 2 x + 1}{2 \left(2 x^3 + x^2 - 2 x - 1\right)} \\\\ & = - \dfrac{1}{2} \end{aligned}$

$\implies$ $A = \cos^{-1} \left(-\dfrac{1}{2}\right) = 120^{\circ}$

Properties of Triangles

Find the acute angles of a right angled triangle whose hypotenuse is four times as long as the perpendicular drawn to it from the opposite angle.

Let $ABC$ be a right angled triangle with right angle at $B$ and sides $a, \; b, \; c$ as shown in the figure.

$BP$ is the perpendicular from $B$ to the side $b$.

Given: $\;$ $b = 4 \; BP$ $\implies$ $BP = \dfrac{b}{4}$

From the figure,

in $\triangle ABC$, $\;$ $\cos C = \dfrac{a}{b}$ $\;\;\; \cdots \; (1)$

in $\triangle BPC$, $\;$ $\sin C = \dfrac{b / 4}{a} = \dfrac{b}{4a}$ $\;\;\; \cdots \; (2)$

From equations $(1)$ and $(2)$,

$\sin C \times \cos C = \dfrac{b}{4a} \times \dfrac{a}{b}$

i.e. $\;$ $\sin C \cos C = \dfrac{1}{4}$

$\therefore \;$ $2 \sin C \cos C = 2 \times \dfrac{1}{4} = \dfrac{1}{2}$

i.e. $\;$ $\sin 2 C = \dfrac{1}{2}$

i.e. $\;$ $2 C = 30^{\circ}$ $\;$ or $\;$ $2 C = 150^{\circ}$

i.e. $C = 15^{\circ}$ $\;$ or $\;$ $C = 75^{\circ}$

When $\;$ $C = 15^{\circ}$, $\;$ $A = 180^{\circ} - \left(B + C\right) = 180^{\circ} - \left(90^{\circ} + 15^{\circ}\right) = 75^{\circ}$

When $\;$ $C = 75^{\circ}$, $\;$ $A = 180^{\circ} - \left(B + C\right) = 180^{\circ} - \left(90^{\circ} + 75^{\circ}\right) = 15^{\circ}$

$\therefore \;$ The angles of the triangle are

$A = 75^{\circ}$, $\;$ $B = 90^{\circ}$, $\;$ $C = 15^{\circ}$ $\;$ or $\;$ $A = 15^{\circ}$, $\;$ $B = 90^{\circ}$, $\;$ $C = 75^{\circ}$

Properties of Triangles

In a right angled triangle $ABC$, where $C$ is the right angle, if $a = 50$ and $B = 75^{\circ}$, find the sides.

Given: $\;$ $\triangle ABC$, with $C = 90^{\circ}$, $B = 75^{\circ}$, $a = 50$

By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$; $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

Here,

$\cos B = \cos 75^{\circ} = \dfrac{c^2 + 50^2 - b^2}{2 \times 50 \times c}$

i.e. $\;$ $\dfrac{\sqrt{3} - 1}{2 \sqrt{2}} = \dfrac{c^2 + 2500 - b^2}{2 \times 50 \times c}$

i.e. $\;$ $\dfrac{\sqrt{3} - 1}{\sqrt{2}} = \dfrac{c^2 + 2500 - b^2}{50 \times c}$ $\;\;\; \cdots \; (1)$

and,

$\cos C = \cos 90^{\circ} = \dfrac{50^2 + b^2 - c^2}{2 \times 50 \times b}$

i.e. $\;$ $2500 + b^2 - c^2 = 0$

i.e. $\;$ $c^2 - b^2 = 2500$ $\;\;\; \cdots \; (2)$

$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes

$\dfrac{\sqrt{3} - 1}{\sqrt{2}} = \dfrac{2500 + 2500}{50 c}$

i.e. $\;$ $\dfrac{\sqrt{3} - 1}{\sqrt{2}} = \dfrac{5000}{50 c}$

i.e. $\;$ $\dfrac{\sqrt{3} - 1}{\sqrt{2}} = \dfrac{100}{c}$

$\implies$ $c = \dfrac{100 \sqrt{2}}{\sqrt{3} - 1} = 193.15$

Substituting the value of $c$ in equation $(2)$ gives

$b^2 = c^2 - 2500 = 193.15^2 - 2500 = 34806.92$

$\implies$ $b = \sqrt{34806.92} = 186.57$

$\therefore \;$ The sides of the triangle are $\;$ $a = 50$, $b = 186.57$, $c = 193.15$

Properties of Triangles

In $\triangle ABC$, if $\theta$ be any angle, then prove that $\;$ $b \cos \theta = c \cos \left(A - \theta\right) + a \cos \left(C + \theta\right)$

In $\triangle ABC$, we have by sine rule,

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\implies$ $\sin A = \dfrac{a}{2 R}$, $\;$ $\sin B = \dfrac{b}{2 R}$, $\;$ $\sin C = \dfrac{c}{2 R}$

By projection formula we have,

$b = c \cos A + a \cos C$

Now,

$\begin{aligned} RHS & = c \cos \left(A - \theta\right) + a \cos \left(C + \theta\right) \\\\ & \left[\text{Note: } \cos \left(\alpha - \beta\right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \right. \\ & \left. \hspace{1.1cm} \cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \right] \\\\ & = c \left(\cos A \cos \theta + \sin A \sin \theta\right) + a \left(\cos C \cos \theta - \sin C \sin \theta\right) \\\\ & = \cos \theta \left(a \cos C + c \cos A\right) + \sin \theta \left(c \sin A - a \sin C\right) \\\\ & = b \cos \theta + \sin \theta \left(c \times \dfrac{a}{2 R} - a \times \dfrac{c}{2 R}\right) \;\; \left[\text{by projection formula and sine rule}\right] \\\\ & = b \cos \theta = LHS \end{aligned}$

Hence proved.

Properties of Triangles

In $\triangle ABC$, if $C = 60^{\circ}$, then prove that $\dfrac{1}{a + c} + \dfrac{1}{b + c} = \dfrac{3}{a + b + c}$

To prove that (T.P.T) $\;$ $\dfrac{1}{a + c} + \dfrac{1}{b + c} = \dfrac{3}{a + b + c}$

i.e. $\;$ T.P.T $\;$ $\dfrac{b + c + a + c}{\left(a + c\right) \left(b + c\right)} = \dfrac{3}{a + b + c}$

i.e. $\;$ T.P.T $\;$ $\left(a + b + 2c\right) \left(a + b + c\right) = 3 \left(a + c\right) \left(b + c\right)$

i.e. $\;$ T.P.T $\;$ $a^2 + ab + ac + ab + b^2 + bc + 2ac + 2bc + 2c^2$

$\hspace{6cm}$ $= 3 ab + 3 ac + 3 bc + 3 c^2$

i.e. $\;$ T.P.T $\;$ $a^2 + b^2 + 2 c^2 + 2 ab + 3 ac + 3 bc = 3 ab + 3 ac + 3 bc + 3 c^2$

i.e. $\;$ T.P.T $\;$ $a^2 + b^2 - c^2 = ab$ $\;\;\; \cdots \; (1)$

By cosine rule, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$ $\;\;\; \cdots \; (2)$

Given: $\;$ $C = 60^{\circ}$

Substituting the value of $C$ in equation $(2)$ we get,

$\cos 60^{\circ} = \dfrac{a^2 + b^2 - c^2}{2 a b}$

i.e. $\;$ $\dfrac{1}{2} = \dfrac{a^2 + b^2 - c^2}{2 a b}$

i.e. $\;$ $a^2 + b^2 - c^2 = a b$ $\;\;\; \cdots \; (3)$

$\therefore \;$ We have from equations $(1)$ and $(3)$,

$\dfrac{1}{a + c} + \dfrac{1}{b + c} = \dfrac{3}{a + b + c}$

Hence proved.

Properties of Triangles

The sides of a triangle are in A.P. and the greatest angle exceeds the least by $90^\circ$. Prove that the sides are proportional to $\sqrt{7} + 1$, $\sqrt{7}$ and $\sqrt{7} -1$.

Let $ABC$ be a triangle with sides $a, \; b, \; c$ and angles $A, \; B, \; C$.

Let $A$ be the least angle and $C$ the greatest angle.

Given: $\;$ $a, \; b, \; c$ are in A.P.

$\implies$ $2 b = a + c$ $\;\;\; \cdots \; (1a)$

and, $\;$ $C = A + 90^{\circ}$ $\;\;\; \cdots \; (1b)$

To prove that: $\;$ $a : b : c = \sqrt{7} + 1 : \sqrt{7} : \sqrt{7} - 1$

In $\triangle ABC$, we have by sine rule,

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

i.e. $\;$ $a = 2 R \sin A$ $\;\; \cdots \; (2a)$, $\;$ $b = 2 R \sin B$ $\;\; \cdots \; (2b)$, $\;$ $c = 2 R \sin C$ $\;\; \cdots \; (2c)$

In view of equations $(2a)$, $(2b)$ and $(2c)$, equation $(1a)$ becomes

$2 \times 2 R \sin B = 2 R \sin A + 2 R \sin C$

i.e. $\;$ $2 \sin B = \sin A + \sin C$ $\;\;\; \cdots \; (3a)$

$\left[\text{Note: } \sin \theta = 2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right) \right.$
$\left. \sin \alpha + \sin \beta = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha - \beta}{2}\right) \right]$

i.e. $\;$ $2 \times 2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) = 2 \sin \left(\dfrac{A + C}{2}\right) \cos \left(\dfrac{A - C}{2}\right)$

i.e. $\;$ $2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) = \sin \left(\dfrac{A + C}{2}\right) \cos \left(\dfrac{A - C}{2}\right)$ $\;\;\; \cdots \; (3b)$

Now, in $\triangle ABC$, $\;$ $A + B + C = \pi$

$\implies$ $A + C = \pi - B$

$\implies$ $\dfrac{A + C}{2} = \dfrac{\pi}{2} - \dfrac{B}{2}$

$\therefore \;$ $\sin \left(\dfrac{A + C}{2}\right) = \sin \left[\dfrac{\pi}{2} - \left(\dfrac{B}{2}\right)\right] = \cos \left(\dfrac{B}{2}\right)$ $\;\;\; \cdots \; (4a)$

From equation $1b$, $\;$ $A - C = - 90^{\circ}$

$\therefore \;$ $\dfrac{A - C}{2} = - 45^{\circ}$

$\therefore \;$ $\cos \left(\dfrac{A - C}{2}\right) = \cos \left(- 45^{\circ}\right) = \dfrac{1}{\sqrt{2}}$ $\;\;\; \cdots \; (4b)$

$\therefore \;$ In view of equations $(4a)$ and $(4b)$, equation $(3b)$ becomes

$2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) = \dfrac{1}{\sqrt{2}} \times \cos \left(\dfrac{B}{2}\right)$

i.e. $\sin \left(\dfrac{B}{2}\right) = \dfrac{1}{2 \sqrt{2}}$ $\;\; \cdots \; (5a)$

From equation $(3a)$ we have,

$\sin A + \sin C = 2 \times 2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right)$ $\;\;\; \cdots \; (6)$

We have from equation $(5a)$,

$\cos \left(\dfrac{B}{2}\right) = \sqrt{1 - \sin^2 \left(\dfrac{B}{2}\right)} = \sqrt{1 - \dfrac{1}{8}} = \dfrac{\sqrt{7}}{2 \sqrt{2}}$ $\;\;\; \cdots \; (5b)$

$\therefore \;$ In view of equations $(5a)$ and $(5b)$ equation $(6)$ becomes

$\sin A + \sin C = 4 \times \dfrac{1}{2 \sqrt{2}} \times \dfrac{\sqrt{7}}{2 \sqrt{2}}$

i.e. $\;$ $\sin A + \sin C = \dfrac{\sqrt{7}}{2}$ $\;\;\; \cdots \; (6a)$

$\left[\text{Note: } \sin \alpha - \sin \beta = 2 \sin \left(\dfrac{\alpha - \beta}{2}\right) \cos \left(\dfrac{\alpha + \beta}{2}\right)\right]$

Now, $\;$ $\sin A - \sin C = 2 \sin \left(\dfrac{A - C}{2}\right) \cos \left(\dfrac{A + C}{2}\right)$ $\;\;\; \cdots \; (7)$

In $\triangle ABC$, $\;$ $\cos \left(\dfrac{A + C}{2}\right) = \cos \left[\dfrac{\pi}{2} - \left(\dfrac{B}{2}\right)\right] = \sin \left(\dfrac{B}{2}\right)$

i.e. $\;$ $\cos \left(\dfrac{A + C}{2}\right) = \dfrac{1}{2 \sqrt{2}}$ $\;\;$ [in view of equation $(5a)$] $\;\;\; \cdots \; (8a)$

From equation $(1b)$, $\;$ $\sin \left(\dfrac{A - C}{2}\right) = \sin \left(- 45^{\circ}\right) = \dfrac{-1}{\sqrt{2}}$ $\;\;\; \cdots \; (8b)$

$\therefore \;$ In view of equations $(8a)$ and $(8b)$ equation $(7)$ becomes

$\sin A - \sin C = 2 \times \left(\dfrac{-1}{\sqrt{2}}\right) \times \dfrac{1}{2 \sqrt{2}}$

i.e. $\;$ $\sin A- \sin C = \dfrac{-1}{2}$ $\;\;\; \cdots \; (7a)$

Adding equations $(6a)$ and $(7a)$ we have,

$2 \sin A = \dfrac{\sqrt{7} - 1}{2}$

$\implies$ $\sin A = \dfrac{\sqrt{7} - 1}{4}$ $\;\;\; \cdots \; (9a)$

Substituting the value of $\sin A$ in equation $(6a)$ we have,

$\sin C = \dfrac{\sqrt{7}}{2} - \dfrac{\sqrt{7} - 1}{4}$

$\implies$ $\sin C = \dfrac{\sqrt{7} + 1}{4}$ $\;\;\; \cdots \; (9b)$

We have from equations $(3a)$, $(9a)$ and $(9b)$,

$\sin B = \dfrac{\sin A + \sin C}{2}$

i.e. $\sin B = \dfrac{\dfrac{\sqrt{7} - 1}{4} + \dfrac{\sqrt{7} + 1}{4}}{2}$

i.e. $\;$ $\sin B = \dfrac{\sqrt{7}}{4}$ $\;\;\; \cdots \; (9c)$

Now, by sine rule, $\;$ $a : b : c = \sin A : \sin B : \sin C$

i.e. $\;$ $a : b : c = \dfrac{\sqrt{7} - 1}{4} : \dfrac{\sqrt{7}}{4} : \dfrac{\sqrt{7} + 1}{4}$ $\;\;$ [by equations $(9a)$, $(9b)$, $(9c)$]

i.e. $\;$ $a : b : c = \sqrt{7} - 1 : \sqrt{7} : \sqrt{7} + 1$

Hence proved.

Properties of Triangles

If $a, \; b, \; c$ are in H.P, prove that $\sin^2 \left(\dfrac{A}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{B}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{C}{2}\right)$ are also in H.P.

Given: $\;$ $a, \; b, \; c$ are in H.P.

$\implies$ $\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$ $\;\;\; \cdots \; (1)$

To prove that (T.P.T) $\;$ $\sin^2 \left(\dfrac{A}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{B}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{C}{2}\right)$ are in H.P.

i.e. $\;$ $T.P.T$ $\;$ $\dfrac{2}{\sin^2 \left(\dfrac{B}{2}\right)} = \dfrac{1}{\sin^2 \left(\dfrac{A}{2}\right)} + \dfrac{1}{\sin^2 \left(\dfrac{C}{2}\right)}$

In $\triangle ABC$,

$\sin \left(\dfrac{A}{2}\right) = \sqrt{\dfrac{\left(s - b\right) \left(s - c\right)}{bc}}$, $\;$ $\sin \left(\dfrac{B}{2}\right) = \sqrt{\dfrac{\left(s - c\right) \left(s - a\right)}{ca}}$, $\;$ $\sin \left(\dfrac{C}{2}\right) = \sqrt{\dfrac{\left(s - a\right) \left(s - b\right)}{ab}}$

where $\;$ $s = \dfrac{a + b + c}{2}$ $\;$ is the semi perimeter of $\triangle ABC$.

$\implies$ $2 s = a + b + c$ $\;\;\; \cdots \; (2)$

$\begin{aligned} \therefore \; \dfrac{1}{\sin^2 \left(\dfrac{A}{2}\right)} + \dfrac{1}{\sin^2 \left(\dfrac{C}{2}\right)} & = \dfrac{bc}{\left(s - b\right) \left(s - c\right)} + \dfrac{ab}{\left(s - a\right) \left(s - b\right)} \\\\ & = \dfrac{b}{\left(s - b\right)} \left[\dfrac{c}{s - c} + \dfrac{a}{s - a}\right] \\\\ & = \dfrac{b}{\left(s - b\right)} \left[\dfrac{cs - ca + as - ca}{\left(s - c\right) \left(s - a\right)}\right] \\\\ & = \dfrac{b s \left(c + a\right) - 2 abc}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \;\;\; \cdots \; (3) \end{aligned}$

We have from equation $(1)$, $\;$ $a + c = \dfrac{2 a c}{b}$

Substituting the value of $\left(a + c\right)$ in equation $(3)$, we have,

$\begin{aligned} \dfrac{1}{\sin^2 \left(\dfrac{A}{2}\right)} + \dfrac{1}{\sin^2 \left(\dfrac{C}{2}\right)} & = \dfrac{bs \times \dfrac{2 a c}{b} - 2 abc}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \\\\ & = \dfrac{2s \times ac - 2abc}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \\\\ & = \dfrac{\left(a + b + c\right)a c - 2 a b c}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \;\; \left[\text{by equation (2)}\right] \\\\ & = \dfrac{ac \left(a + b + c - 2 b\right)}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \\\\ & = \dfrac{ac \left(a + c - b\right)}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \;\;\; \cdots \; (4) \end{aligned}$

We have from equation $(2)$, $\;$ $a + c = 2 s - b$

Substituting the value of $\left(a + c\right)$ in equation $(4)$, we have,

$\begin{aligned} \dfrac{1}{\sin^2 \left(\dfrac{A}{2}\right)} + \dfrac{1}{\sin^2 \left(\dfrac{C}{2}\right)} & = \dfrac{a c \left(2 s - b - b\right)}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \\\\ & = \dfrac{2 a c \left(s - b\right)}{\left(s - a\right) \left(s - b\right) \left(s - c\right)} \\\\ & = 2 \times \dfrac{ac}{\left(s - a\right) \left(s - c\right)} \end{aligned}$

i.e. $\;$ $\dfrac{1}{\sin^2 \left(\dfrac{A}{2}\right)} + \dfrac{1}{\sin^2 \left(\dfrac{C}{2}\right)} = \dfrac{2}{\sin^2 \left(\dfrac{B}{2}\right)}$

$\implies$ $\sin^2 \left(\dfrac{A}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{B}{2}\right)$, $\;$ $\sin^2 \left(\dfrac{C}{2}\right)$ are in H.P.

Hence proved.

Properties of Triangles

If $a^2$, $b^2$ and $c^2$ are in $A.P$, prove that $\cot A$, $\cot B$ and $\cot C$ are also in $A.P$.

Given: $\;$ $a^2, \; b^2, \; c^2$ are in $A.P$

$\implies$ $2 b^2 = a^2 + c^2$

i.e. $\;$ $b^2 = a^2 +c^2 - b^2$ $\;\;\; \cdots \; (1)$

In any $\triangle ABC$,

$\cot A = \dfrac{b^2 + c^2 - a^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$, $\;$ $\cot B = \dfrac{c^2 + a^2 - b^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$, $\;$ $\cot C = \dfrac{a^2 + b^2 - c^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$

Now,

$\begin{aligned} \cot A + \cot C & = \dfrac{b^2 + c^2 - a^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} + \dfrac{a^2 + b^2 - c^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} \\\\ & = \dfrac{2 b^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} \\\\ & = \dfrac{b^2}{2 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} \\\\ & = \dfrac{a^2 + c^2 - b^2}{2 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} \;\; \left[\text{in view of equation (1)}\right] \\\\ & = 2 \times \left(\dfrac{a^2 + c^2 - b^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}\right) \\\\ & = 2 \cot B \end{aligned}$

i.e. $\;$ $\cot A + \cot C = 2 \cot B$

$\implies$ $\cot A, \; \cot B, \; \cot C$ are in $A.P$

Hence proved.

Properties of Triangles

The perpendicular $AD$ to the base of a $\triangle ABC$ divides it into segments such that $BD$, $CD$ and $AD$ are in the ratio of $2$, $3$ and $6$. Prove that the vertical angle of the triangle is $45^\circ$.

Let $ABC$ be the given triangle with base $BC$.

$AD$ is the perpendicular to the base of the triangle.

Given: $\;$ $BD : CD : AD = 2 : 3 : 6$

$\therefore \;$ Let $\;$ $BD = 2 k$, $\;$ $CD = 3 k$, $\;$ $AD = 6 k$ $\;$ where $k$ is the constant of proportionality

To prove that: $\;$ In $\triangle ABC$, $\;$ $\angle A = 45^\circ$

In $\triangle ACD$,

$\begin{aligned} AC & = \sqrt{CD^2 + AD^2} \\\\ & = \sqrt{\left(3 k\right)^2 + \left(6 k\right)^2} \\\\ & = \sqrt{9 k^2 + 36 k^2} \\\\ & = 3 k \sqrt{5} \end{aligned}$

Applying sine rule to $\triangle ACD$, we have

$\dfrac{CD}{\sin \left(\angle DAC\right)} = \dfrac{CA}{\sin \left(\angle CDA\right)}$

i.e. $\;$ $\dfrac{3 k}{\sin \left(\angle DAC\right)} = \dfrac{3 k \sqrt{5}}{\sin 90^{\circ}}$

i.e. $\;$ $\sin \left(\angle DAC\right) = \dfrac{1}{\sqrt{5}}$

$\implies$ $\angle DAC = \sin^{-1} \left(\dfrac{1}{\sqrt{5}}\right)$ $\;\;\; \cdots \; (1)$

In $\triangle ADB$,

$\begin{aligned} AB & = \sqrt{DB^2 + AD^2} \\\\ & = \sqrt{\left(2 k\right)^2 + \left(6 k\right)^2} \\\\ & = \sqrt{4 k^2 + 36 k^2} \\\\ & = 2 k \sqrt{10} \end{aligned}$

Applying sine rule to $\triangle ADB$, we have

$\dfrac{DB}{\sin \left(\angle DAB\right)} = \dfrac{AB}{\sin \left(\angle ADB\right)}$

i.e. $\;$ $\dfrac{2 k}{\sin \left(\angle DAB\right)} = \dfrac{2 k \sqrt{10}}{\sin 90^{\circ}}$

i.e. $\;$ $\sin \left(\angle DAB\right) = \dfrac{1}{\sqrt{10}}$

$\implies$ $\angle DAB = \sin^{-1} \left(\dfrac{1}{\sqrt{10}}\right)$ $\;\;\; \cdots \; (2)$

Now, in $\triangle ABC$,

$\begin{aligned} \angle A & = \angle DAC + \angle DAB \\\\ & = \sin^{-1} \left(\dfrac{1}{\sqrt{5}}\right) + \sin^{-1} \left(\dfrac{1}{\sqrt{10}}\right) \;\; \left[\text{from equations (1) and (2)}\right] \\\\ & \left[\text{Note: } \sin^{-1} x + \sin^{-1} y = \sin^{-1} \left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right)\right] \\\\ & = \sin^{-1} \left(\dfrac{1}{\sqrt{5}} \times \sqrt{1 - \dfrac{1}{10}} + \dfrac{1}{\sqrt{10}} \times \sqrt{1 - \dfrac{1}{5}}\right) \\\\ & = \sin^{-1} \left(\dfrac{1}{\sqrt{5}} \times \dfrac{3}{\sqrt{10}} + \dfrac{1}{\sqrt{10}} \times \dfrac{2}{\sqrt{5}}\right) \\\\ & = \sin^{-1} \left(\dfrac{5}{5 \sqrt{2}}\right) \\\\ & = \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) \end{aligned}$

$\implies$ $\angle A = 45^{\circ}$

Hence proved.

Properties of Triangles

In any $\triangle ABC$, if $\;$ $\tan \left(\dfrac{A}{2}\right) = \dfrac{5}{6}$, $\;$ $\tan \left(\dfrac{B}{2}\right) = \dfrac{20}{37}$, $\;$ find $\;$ $\tan \left(\dfrac{C}{2}\right)$ and prove that in this triangle $\;$ $a + c = 2b$.

Given: $\;$ $\tan \left(\dfrac{A}{2}\right) = \dfrac{5}{6}$, $\;$ $\tan \left(\dfrac{B}{2}\right) = \dfrac{20}{37}$

In $\triangle ABC$, $\;$ $A + B + C = \pi$

i.e. $\;$ $C = \pi - \left(A + B\right)$

i.e. $\;$ $\dfrac{C}{2} = \dfrac{\pi}{2} - \left(\dfrac{A}{2} + \dfrac{B}{2}\right)$

$\therefore \;$ $\tan \left(\dfrac{C}{2}\right) = \tan \left[\dfrac{\pi}{2} - \left(\dfrac{A}{2} + \dfrac{B}{2}\right)\right] = \cot \left(\dfrac{A}{2} + \dfrac{B}{2}\right)$

Now, $\cot \left(\alpha + \beta\right) = \dfrac{1 - \tan \alpha \tan \beta}{\tan \alpha + \tan \beta}$

$\therefore \;$ We have

$\begin{aligned} \tan \left(\dfrac{C}{2}\right) & = \cot \left(\dfrac{A}{2} + \dfrac{B}{2}\right) \\\\ & = \dfrac{1 - \tan \left(\dfrac{A}{2}\right) \times \tan \left(\dfrac{B}{2}\right)}{\tan \left(\dfrac{A}{2}\right) + \tan \left(\dfrac{B}{2}\right)} \\\\ & = \dfrac{1 - \dfrac{5}{6} \times \dfrac{20}{37}}{\dfrac{5}{6} + \dfrac{20}{37}} \\\\ & = \dfrac{122}{305} \end{aligned}$

i.e. $\;$ $\tan \left(\dfrac{C}{2}\right) = \dfrac{2}{5}$

For this given triangle,

To Prove That (TPT) $\;$ $a + c = 2b$

i.e. $\;$ $TPT \;\;$ $\dfrac{a}{2 R} + \dfrac{c}{2 R} = 2 \times \dfrac{b}{2 R}$ $\;\;\;$ [$R$ is the circumradius of $\triangle ABC$]

$\left[\text{Note: Sine rule: } \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2 R \right.$
$\left. \implies \dfrac{a}{2R} = \sin A, \; \dfrac{b}{2 R} = \sin B, \; \dfrac{c}{2 R} = \sin C \right]$

i.e. $\;$ $TPT \;\;$ $\sin A + \sin C = 2 \sin B$

$\left[\text{Note: } \sin \theta = \dfrac{2 \tan \left(\dfrac{\theta}{2}\right)}{1 + \tan^2 \left(\dfrac{\theta}{2}\right)}\right]$

i.e. $\;$ $TPT \;\;$ $\dfrac{2 \tan \left(\dfrac{A}{2}\right)}{1 + \tan^2 \left(\dfrac{A}{2}\right)} + \dfrac{2 \tan \left(\dfrac{C}{2}\right)}{1 + \tan^2 \left(\dfrac{C}{2}\right)} = 2 \times \dfrac{2 \tan \left(\dfrac{B}{2}\right)}{1 + \tan^2 \left(\dfrac{B}{2}\right)}$

i.e. $\;$ $TPT \;\;$ $\dfrac{\tan \left(\dfrac{A}{2}\right)}{1 + \tan^2 \left(\dfrac{A}{2}\right)} + \dfrac{\tan \left(\dfrac{C}{2}\right)}{1 + \tan^2 \left(\dfrac{C}{2}\right)} = \dfrac{2 \tan \left(\dfrac{B}{2}\right)}{1 + \tan^2 \left(\dfrac{B}{2}\right)}$

i.e. $\;$ $TPT \;\;$ $\dfrac{\dfrac{5}{6}}{1 + \dfrac{25}{36}} + \dfrac{\dfrac{2}{5}}{1 + \dfrac{4}{25}} = \dfrac{2 \times \dfrac{20}{37}}{1 + \dfrac{400}{1369}}$

i.e. $\;$ $TPT \;$ $\dfrac{30}{61} + \dfrac{10}{29} = \dfrac{1480}{1769}$

i.e. $\;$ $TPT \;$ $\dfrac{1480}{1769} = \dfrac{1480}{1769}$ $\;$ which is true.

$\therefore \;$ $a + c = 2b$

Hence proved.

Properties of Triangles

The sides of a right angled triangle are $21$ and $28 \; cm$. Find the length of the perpendicular drawn to the hypotenuse from the right angle.

Let $ABC$ be the given triangle with sides $AB = c = 28 \; cm$ and $AC = b = 21 \; cm$ and right angled at $A$.

Hypotenuse of $\triangle ABC$ $= AB = a = \sqrt{b^2 + c^2} = \sqrt{\left(21\right)^2 + \left(28\right)^2} = \sqrt{1225} = 35 \; cm$

In $\triangle ABC$, by sine rule we have $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$

i.e. $\;$ $\dfrac{35}{\sin 90^{\circ}} = \dfrac{21}{\sin B}$

i.e. $\;$ $\sin B = \dfrac{21}{35} = \dfrac{3}{5}$ $\;\;\; \cdots \; (1)$

Let $AP$ be the perpendicular from the right angle to the hypotenuse.

In $\triangle APB$, by sine rule we have $\;$ $\dfrac{AB}{\sin P} = \dfrac{AP}{\sin B}$

i.e. $\;$ $\dfrac{28}{\sin 90^{\circ}} = \dfrac{AP}{3 / 5}$ $\;\;\;$ [by equation $(1)$]

i.e. $\;$ $AP = 28 \times \dfrac{3}{5} = \dfrac{84}{5} = 16.8 \; cm$

Properties of Triangles

In any $\triangle ABC$, prove that $\;$ $a^3 \cos \left(B - C\right) + b^3 \cos \left(C - A\right) + c^3 \cos \left(A - B\right) = 3 a b c$

By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$

$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$

In $\triangle ABC$, $\;$ $A + B + C = \pi$

$\begin{aligned} LHS & = a^3 \cos \left(B - C\right) + b^3 \cos \left(C - A\right) + c^3 \cos \left(A - B\right) \\\\ & = a^2 \times a \cos \left(B - C\right) + b^2 \times b \cos \left(C - A\right) + c^2 \times c \cos \left(A - B\right) \\\\ & = 2 R \left[a^2 \sin A \cos \left(B - C\right) + b^2 \sin B \cos \left(C - A\right) \right. \\ & \hspace{4.5cm} \left. + c^2 \sin C \cos \left(A - B\right)\right] \;\; \left[\text{By sine rule}\right] \\\\ & \left[\text{Note: In } \triangle ABC, \; A + B + C = \pi \right. \\ & \hspace{2.5cm} \left. i.e. \; A = \pi - \left(B + C\right) \right. \\ & \hspace{2.5cm} \left. \therefore \; \sin A = \sin \left[\pi - \left(B + C\right)\right] = \sin \left(B + C\right) \right. \\ & \left. \text{Similarly, } \sin B = \sin \left(C + A\right), \; \sin C = \sin \left(A + B\right) \right] \\\\ & = 2 R \left[a^2 \sin \left(B + C\right) \cos \left(B - C\right) \right. \\ & \hspace{2cm} \left. + b^2 \sin \left(C + A\right) \cos \left(C - A\right) \right. \\ & \hspace{4cm} \left. + c^2 \sin \left(A + B\right) \cos \left(A - B\right) \right] \\\\ & \left[\text{Note: } \sin \alpha \cos \beta = \dfrac{1}{2} \left[\sin \left(\alpha + \beta\right) + \sin \left(\alpha - \beta\right)\right]\right] \\\\ & = 2 R \left\{\dfrac{a^2}{2} \left[\sin \left(B + C + B - C\right) + \sin \left(B + C - B + C\right)\right] \right. \\ & \hspace{2cm} \left. + \dfrac{b^2}{2} \left[\sin \left(C + A + C - A\right) + \sin \left(C + A - C + A\right)\right] \right. \\ & \hspace{3cm} \left. + \dfrac{c^2}{2} \left[\sin \left(A + B + A - B\right) + \sin \left(A + B - A + B\right)\right] \right\} \\\\ & = 2 R \left\{\dfrac{a^2}{2} \left(\sin 2 B + \sin 2 C\right) + \dfrac{b^2}{2} \left(\sin 2 C + \sin 2 A\right) + \dfrac{c^2}{2} \left(\sin 2 A + \sin 2 B\right) \right\} \\\\ & = R \left(b^2 + c^2\right) \sin 2 A + R \left(c^2 + a^2\right) \sin 2 B + R \left(a^2 + b^2\right) \sin 2 C \end{aligned}$

$\begin{aligned} \left[\text{Note: } R \left(a^2 + b^2\right) \sin 2 C \right. & \left. = R \left(a^2 + b^2\right) \times 2 \sin C \cos C \right. \\ \left. \right. & \left. = R \left(a^2 + b^2\right) \times 2 \times \dfrac{c}{2 R} \times \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right) \right. \\ \left. \right. & \left. = \dfrac{c}{2 a b} \left(a^2 + b^2\right) \left(a^2 + b^2 - c^2\right) \right. \\ \left. \right. & \left. = \dfrac{c^2}{2 a b c} \left(a^2 + b^2\right) \left(a^2 + b^2 - c^2\right) \right. \\ \left. \text{Similarly, } \; R \left(b^2 + c^2\right) \sin 2 A \right. & \left. = \dfrac{a^2}{2 a b c} \left(b^2 + c^2\right) \left(b^2 + c^2 - a^2\right) \right. \\ \left. \text{and, } \; R \left(c^2 + a^2\right) \sin 2 B \right. & \left. = \dfrac{b^2}{2 a b c} \left(c^2 + a^2\right) \left(c^2 + a^2 - b^2\right) \right] \end{aligned}$

$\begin{aligned} \therefore \; LHS & = \dfrac{1}{2 a b c} \left[a^2 \left(b^2 + c^2\right) \left(b^2 + c^2 - a^2\right) \right. \\ & \hspace{2cm} \left. + b^2 \left(c^2 + a^2\right) \left(c^2 + a^2 - b^2\right) \right. \\ & \hspace{3cm} \left. + c^2 \left(a^2 + b^2\right) \left(a^2 + b^2 - c^2\right) \right] \\\\ & = \dfrac{1}{2 a b c} \left[a^2 \left(b^4 + c^4 + 2 b^2 c^2 - a^2 b^2 - a^2 c^2\right) \right. \\ & \hspace{2cm} \left. + b^2 \left(c^4 + a^4 + 2 c^2 a^2 - b^2 c^2 - a^2 b^2\right) \right. \\ & \hspace{3cm} \left. + c^2 \left(a^4 + b^4 + 2 a^2 b^2 - a^2 c^2 - b^2 c^2\right) \right] \\\\ & = \dfrac{1}{2 a b c} \left[a^2 b^4 + a^2 c^4 + 2 a^2 b^2 c^2 - a^4 b^2 - a^4 c^2 \right. \\ & \hspace{2cm} \left. + b^2 c^4 + a^4 b^2 + 2 a^2 b^2 c^2 - b^4 c^2 - a^2 b^4 \right. \\ & \hspace{3cm} \left. + a^4 c^2 + b^4 c^2 + 2 a^2 b^2 c^2 - a^2 c^4 - b^2 c^4 \right] \\\\ & = \dfrac{6 a^2 b^2 c^2}{2 a b c} \\\\ & = 3 a b c = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In any $\triangle ABC$, prove that $\;$ $\dfrac{\left(a + b + c\right)^2}{a^2 + b^2 + c^2} = \dfrac{\cot \left(\dfrac{A}{2}\right) + \cot \left(\dfrac{B}{2}\right) + \cot \left(\dfrac{C}{2}\right)}{\cot A + \cot B + \cot C}$

In $\triangle ABC$,

$\cot \left(\dfrac{A}{2}\right) = \sqrt{\dfrac{s \left(s - a\right)}{\left(s - b\right) \left(s - c\right)}}$, $\;$ $\cot \left(\dfrac{B}{2}\right) = \sqrt{\dfrac{s \left(s - b\right)}{\left(s - c\right) \left(s - a\right)}}$, $\;$ $\cot \left(\dfrac{C}{2}\right) = \sqrt{\dfrac{s \left(s - c\right)}{\left(s - a\right) \left(s - b\right)}}$

$\cot A = \dfrac{b^2 + c^2 - a^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$, $\;$ $\cot B = \dfrac{c^2 + a^2 - b^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$, $\;$ $\cot C = \dfrac{a^2 + b^2 - c^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$

where $\;$ $s = \dfrac{a + b + c}{2}$ $\;$ is the semi perimeter of $\triangle ABC$ $\;\;\; \cdots \; (1a)$

$\implies$ $2 s = a + b + c$ $\;\;\; \cdots \; (1b)$

$\therefore \;$ $\cot \left(\dfrac{A}{2}\right) + \cot \left(\dfrac{B}{2}\right) + \cot \left(\dfrac{C}{2}\right)$

$= \sqrt{\dfrac{s \left(s - a\right)}{\left(s - b\right) \left(s - c\right)}} + \sqrt{\dfrac{s \left(s - b\right)}{\left(s - c\right) \left(s - a\right)}} + \sqrt{\dfrac{s \left(s - c\right)}{\left(s - a\right) \left(s - b\right)}}$

$= \sqrt{\dfrac{s^2 \left(s - a\right)^2}{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} + \sqrt{\dfrac{s^2 \left(s - b\right)^2}{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} $
$\hspace{6cm}$ $ + \sqrt{\dfrac{s^2 \left(s - c\right)^2}{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$

$= \dfrac{s \left[s - a + s - b + s - c\right]}{\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$

$= \dfrac{s \left[3 s - \left(a + b + c\right)\right]}{\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$

$= \dfrac{s \left(3 s - 2s\right)}{\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$ $\;\;\;$ [by equation $(1b)$]

$= \dfrac{s^2}{\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$ $\;\;\; \cdots \; (2)$

and,

$\cot A + \cot B + \cot C$

$= \dfrac{b^2 + c^2 - a^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}} + \dfrac{c^2 + a^2 - b^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$
$\hspace{5.5cm}$ $ + \dfrac{a^2 + b^2 - c^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$

$= \dfrac{b^2 + c^2 - a^2 + c^2 + a^2 - b^2 + a^2 + b^2 - c^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$

$= \dfrac{a^2 + b^2 + c^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}$ $\;\; \cdots \; (3)$

$\therefore \;$ We have from equations $(2)$ and $(3)$,

$\begin{aligned} RHS & = \dfrac{\dfrac{s^2}{\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}}{\dfrac{a^2 + b^2 + c^2}{4 \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}}} \\\\ & = \dfrac{4 s^2}{a^2 + b^2 + c^2} \\\\ & = \dfrac{4 \times \dfrac{\left(a + b + c\right)^2}{4}}{a^2 + b^2 + c^2} \;\;\; \left[\text{by equation (1a)}\right] \\\\ & = \dfrac{\left(a + b + c\right)^2}{a^2 + b^2 + c^2} \\\\ & = LHS \end{aligned}$

Hence proved.

Properties of Triangles

In any $\triangle ABC$, prove that $\;$ $a \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B - C}{2}\right) + b \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C - A}{2}\right) + c \sin \left(\dfrac{C}{2}\right) \sin \left(\dfrac{A - B}{2}\right) = 0$

In $\triangle ABC$, $\;$ $A + B + C = \pi$

i.e. $\;$ $A = \pi - \left(B + C\right)$

i.e. $\;$ $\dfrac{A}{2} = \dfrac{\pi}{2} - \left(\dfrac{B + C}{2} \right)$

$\therefore \;$ $\sin \left(\dfrac{A}{2}\right) = \sin \left[\dfrac{\pi}{2} - \left(\dfrac{B + C}{2}\right)\right] = \cos \left(\dfrac{B + C}{2}\right)$

Similarly, $\;$ $\sin \left(\dfrac{B}{2}\right) = \cos \left(\dfrac{C + A}{2}\right)$

and $\;$ $\sin \left(\dfrac{C}{2}\right) = \cos \left(\dfrac{A + B}{2}\right)$

$\begin{aligned} LHS & = a \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B - C}{2}\right) + b \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C - A}{2}\right) \\\\ & \hspace{5cm} + c \sin \left(\dfrac{C}{2}\right) \sin \left(\dfrac{A - B}{2}\right) \\\\ & = a \cos \left(\dfrac{B + C}{2}\right) \sin \left(\dfrac{B - C}{2}\right) + b \cos \left(\dfrac{C + A}{2}\right) \sin \left(\dfrac{C - A}{2}\right) \\\\ & \hspace{5.5cm} + c \cos \left(\dfrac{A + B}{2}\right) \sin \left(\dfrac{A - B}{2}\right) \\\\ & \left[\text{Note: } \sin A \cos B = \dfrac{1}{2} \left\{\sin \left(A + B\right) + \sin \left(A - B\right) \right\}\right] \\\\ & = \dfrac{a}{2} \left[\sin \left(\dfrac{B - C}{2} + \dfrac{B + C}{2}\right) + \sin \left(\dfrac{B - C}{2} - \dfrac{B + C}{2}\right)\right] \\\\ & \hspace{1cm} + \dfrac{b}{2} \left[\sin \left(\dfrac{C - A}{2} + \dfrac{C + A}{2}\right) + \sin \left(\dfrac{C - A}{2} - \dfrac{C + A}{2}\right)\right] \\\\ & \hspace{2cm} + \dfrac{c}{2} \left[\sin \left(\dfrac{A - B}{2} + \dfrac{A + B}{2}\right) + \sin \left(\dfrac{A - B}{2} - \dfrac{A + B}{2}\right)\right] \\\\ & = \dfrac{a}{2} \left[\sin B + \sin \left(- C\right)\right] + \dfrac{b}{2} \left[\sin C + \sin \left(- A\right)\right] + \dfrac{c}{2} \left[\sin A + \sin \left(- B\right)\right] \\\\ & = \dfrac{a}{2} \left[\sin B - \sin C\right] + \dfrac{b}{2} \left[\sin C - \sin A\right] + \dfrac{c}{2} \left[\sin A - \sin B\right] \\\\ & = \dfrac{a \sin B}{2} - \dfrac{a \sin C}{2} + \dfrac{b \sin C}{2} - \dfrac{b \sin A}{2} + \dfrac{c \sin A}{2} - \dfrac{c \sin B}{2} \\\\ & = \dfrac{\sin A}{2} \left(c - b\right) + \dfrac{\sin B}{2} \left(a - c\right) + \dfrac{\sin C}{2} \left(b - a\right) \\\\ & \left[\text{Note: Sine rule: } \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2 R \right. \\ & \hspace{2.75cm} \left. R = \text{circumradius of } \triangle ABC \right. \\ & \hspace{2.75cm} \left. \implies \sin A = \dfrac{a}{2 R}, \; \sin B = \dfrac{b}{2 R}, \; \sin C = \dfrac{c}{2 R} \right] \\\\ & = \dfrac{a}{4 R} \left(c - b\right) + \dfrac{b}{4 R} \left(a - c\right) + \dfrac{c}{4 R} \left(b - a\right) \\\\ & = \dfrac{1}{4 R} \left[a c - a b + a b - b c + b c - a c\right] \\\\ & = 0 = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In any $\triangle ABC$, prove that $\;$ $c^2 = \left(a - b\right)^2 \cos^2 \left(\dfrac{C}{2}\right) + \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right)$

$\begin{aligned} RHS & = \left(a - b\right)^2 \cos^2 \left(\dfrac{C}{2}\right) + \left(a + b\right)^2 \sin^2 \left(\dfrac{C}{2}\right) \\\\ & \left[\text{Note: } \cos^2 \left(\dfrac{\theta}{2}\right) = \dfrac{1 + \cos \theta}{2}, \; \sin^2 \left(\dfrac{\theta}{2}\right) = \dfrac{1 - \cos \theta}{2}\right] \\\\ & = \left(a - b\right)^2 \left(\dfrac{1 + \cos C}{2}\right) + \left(a + b\right)^2 \left(\dfrac{1 - \cos C}{2}\right) \\\\ & = \dfrac{1}{2} \left[\left(a - b\right)^2 \left(1 + \cos C\right) + \left(a + b\right)^2 \left(1 - \cos C\right)\right] \\\\ & = \dfrac{1}{2} \left[\left(a^2 + b^2 - 2 a b\right) + \cos C \left(a^2 + b^2 - 2 a b\right) \right. \\ & \hspace{2cm} \left. + \left(a^2 + b^2 + 2 a b\right) - \cos C \left(a^2 + b^2 + 2 a b\right)\right] \\\\ & = \dfrac{1}{2} \left[2 a^2 + 2 b^2 + \cos C \left(- 4 a b\right)\right] \\\\ & = a^2 + b^2 - 2 a b \cos C \\\\ & \left[\text{Note: Cosine rule: } \cos C = \dfrac{a^2 + b^2 - c^2}{2 a b} \right. \\ & \hspace{3cm} \left. \implies c^2 = a^2 + b^2 - 2 a b \cos C\right] \\\\ & = c^2 = LHS \end{aligned}$

Hence proved.

Properties of Triangles

In any $\triangle ABC$, prove that $\;$ $a^2 + b^2 + c^2 = 2 \left(b c \cos A + c a \cos B + a b \cos C\right)$

By cosine rule,

$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

$\begin{aligned} RHS & = 2 \left[b c \cos A + c a \cos B + a b \cos C\right] \\\\ & = 2 \left[b c \times \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) + c a \times \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) + a b \times \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right)\right] \\\\ & = b^2 + c^2 - a^2 + c^2 + a^2 - b^2 + a^2 + b^2 - c^2 \\\\ & = a^2 + b^2 + c^2 = LHS \end{aligned}$

Hence proved.

Properties of Triangles

In any $\triangle ABC$, prove that $\;$ $\left(b + c - a\right)\left[\cot\left(\dfrac{B}{2}\right) + \cot \left(\dfrac{C}{2}\right)\right] = 2 a \cot \left(\dfrac{A}{2}\right)$

In $\triangle ABC$,

$\cot \left(\dfrac{A}{2}\right) = \sqrt{\dfrac{s \left(s - a\right)}{\left(s - b\right) \left(s - c\right)}}$ $\;\; \cdots \; (1a)$,

$\cot \left(\dfrac{B}{2}\right) = \sqrt{\dfrac{s \left(s - b\right)}{\left(s - c\right) \left(s - a\right)}}$ $\;\; \cdots \; (1b)$,

$\cot \left(\dfrac{C}{2}\right) = \sqrt{\dfrac{s \left(s - c\right)}{\left(s - a\right) \left(s - b\right)}}$ $\;\; \cdots \; (1c)$

$s = \dfrac{a + b + c}{2} = $ semi perimeter of $\triangle ABC$

i.e. $\;$ $2 s = a + b + c$ $\;\;\; \cdots \; (2a)$

and $\;$ $2 s - 2a = b + c - a$ $\;$ i.e. $\;$ $2 \left(s - a\right) = b + c - a$ $\;\;\; \cdots \; (2b)$

$\begin{aligned} LHS & = \left(b + c - a\right)\left[\cot\left(\dfrac{B}{2}\right) + \cot \left(\dfrac{C}{2}\right)\right] \\\\ & = 2 \left(s - a\right) \left[\sqrt{\dfrac{s \left(s - b\right)}{\left(s - c\right) \left(s - a\right)}} + \sqrt{\dfrac{s \left(s - c\right)}{\left(s - a\right) \left(s - b\right)}}\right] \\\\ & \hspace{3cm} \left[\text{by equations (1b), (1c) and (2b)}\right] \\\\ & = 2 \left(s - a\right) \left[\sqrt{\dfrac{s \left(s - b\right)^2}{\left(s - a\right) \left(s - b\right) \left(s - c\right)}} + \sqrt{\dfrac{s \left(s - c\right)^2}{\left(s - a\right) \left(s - b\right) \left(s - c\right)}}\right] \\\\ & = 2 \left(s - a\right) \times \sqrt{\dfrac{s}{\left(s - a\right) \left(s - b\right) \left(s - c\right)}} \times \left(s - b + s - c\right) \\\\ & = 2 \left(s - a\right) \times \left(2s - b - c\right) \times \sqrt{\dfrac{s}{\left(s - a\right) \left(s - b\right) \left(s - c\right)}} \\\\ & = 2 \times \left(a + b + c - b - c\right) \times \sqrt{\dfrac{s \left(s - a\right)^2}{\left(s - a\right) \left(s - b\right) \left(s - c\right)}} \;\; [\text{by equation (2a)}] \\\\ & = 2 a \sqrt{\dfrac{s \left(s - a\right)}{\left(s - b\right) \left(s - c\right)}} \\\\ & = 2 a \cot \left(\dfrac{A}{2}\right) = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In $\triangle ABC$, if $a = 287, \; b = 816, \; c = 865$, find the values of $\tan \left(\dfrac{A}{2}\right)$ and $\tan A$.

In any $\triangle ABC$,

$\tan \left(\dfrac{A}{2}\right) = \sqrt{\dfrac{\left(s - b\right) \left(s - c\right)}{s \left(s - a\right)}}$

$\tan A = \dfrac{2 \tan \left(\dfrac{A}{2}\right)}{1 - \tan^2 \left(\dfrac{A}{2}\right)}$

$s = \dfrac{a + b + c}{2} = $ semi perimeter of $\triangle ABC$

Given: $\;$ $a = 287, \; b = 816, \; c = 865$

$s = \dfrac{287 + 816 + 865}{2} = 984$

$\begin{aligned} \tan \left(\dfrac{A}{2}\right) & = \sqrt{\dfrac{\left(984 - 816\right) \left(984 - 865\right)}{984 \left(984 - 287\right)}} \\\\ & = \sqrt{\dfrac{168 \times 119}{984 \times 697}} = \sqrt{\dfrac{7 \times 7}{41 \times 41}} = \dfrac{7}{41} \end{aligned}$

$\begin{aligned} \tan A & = \dfrac{2 \times \dfrac{7}{41}}{1 - \left(\dfrac{7}{41}\right)^2} \\\\ & = \dfrac{2 \times 7 \times 41}{1632} = \dfrac{287}{816} \end{aligned}$

Properties of Triangles

In $\triangle ABC$, if $a = 125, \; b = 123, \; c = 62$, find the sines of half the angles and the sines of the angles.

In any $\triangle ABC$,

$\sin \left(\dfrac{A}{2}\right) = \sqrt{\dfrac{\left(s - b\right) \left(s - c\right)}{b c}}$, $\;$ $\sin \left(\dfrac{B}{2}\right) = \sqrt{\dfrac{\left(s - c\right) \left(s - a\right)}{c a}}$, $\;$ $\sin \left(\dfrac{C}{2}\right) = \sqrt{\dfrac{\left(s - a\right) \left(s - b\right)}{a b}}$

$\sin A = \dfrac{2}{bc} \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, $\;$ $\sin B = \dfrac{2}{ca} \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$, $\;$ $\sin C = \dfrac{2}{ab} \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

$s = \dfrac{a + b + c}{2} = $ semi perimeter of $\triangle ABC$

Given: $\;$ $a = 125, \; b = 123, \; c = 62$

$s = \dfrac{125 + 123 + 62}{2} = 155$

$\begin{aligned} \sin \left(\dfrac{A}{2}\right) & = \sqrt{\dfrac{\left(155 - 123\right) \left(155 - 62\right)}{123 \times 62}} \\\\ & = \sqrt{\dfrac{32 \times 93}{123 \times 62}} = \sqrt{\dfrac{16}{41}} = \dfrac{4}{\sqrt{41}} \end{aligned}$

$\begin{aligned} \sin \left(\dfrac{B}{2}\right) & = \sqrt{\dfrac{\left(155 - 62\right) \left(155 - 125\right)}{62 \times 125}} \\\\ & = \sqrt{\dfrac{93 \times 30}{62 \times 125}} = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5} \end{aligned}$

$\begin{aligned} \sin \left(\dfrac{C}{2}\right) & = \sqrt{\dfrac{\left(155 - 125\right) \left(155 - 123\right)}{125 \times 123}} \\\\ & = \sqrt{\dfrac{30 \times 32}{125 \times 123}} = \sqrt{\dfrac{64}{25 \times 41}} = \dfrac{8}{5 \sqrt{41}} \end{aligned}$

$\begin{aligned} \sin A & = \dfrac{2}{123 \times 62} \sqrt{155 \left(155 - 125\right) \left(155 - 123\right) \left(155 - 62\right)} \\\\ & = \dfrac{1}{123 \times 31} \sqrt{155 \times 30 \times 32 \times 93} \\\\ & = \dfrac{1}{123 \times 31} \times 4 \times 5 \times 6 \times 31 = \dfrac{40}{41} \end{aligned}$

$\begin{aligned} \sin B & = \dfrac{2}{62 \times 125} \sqrt{155 \left(155 - 125\right) \left(155 - 123\right) \left(155 - 62\right)} \\\\ & = \dfrac{1}{31 \times 125} \sqrt{155 \times 30 \times 32 \times 93} \\\\ & = \dfrac{1}{31 \times 125} \times 4 \times 5 \times 6 \times 31 = \dfrac{24}{25} \end{aligned}$

$\begin{aligned} \sin C & = \dfrac{2}{125 \times 123} \sqrt{155 \left(155 - 125\right) \left(155 - 123\right) \left(155 - 62\right)} \\\\ & = \dfrac{2}{125 \times 123} \sqrt{155 \times 30 \times 32 \times 93} \\\\ & = \dfrac{2}{125 \times 123} \times 4 \times 5 \times 6 \times 31 = \dfrac{496}{1025} \end{aligned}$

Properties of Triangles

In $\triangle ABC$, prove that $2 a c \sin \dfrac{1}{2} \left(A - B + C\right) = c^2 + a^2 - b^2$

In $\triangle ABC$, $\;$ $A + B + C = \pi$ $\implies$ $A + C = \pi - B$

$\begin{aligned} LHS & = 2 a c \sin \left(\dfrac{A - B + C}{2}\right) \\\\ & = 2 a c \sin \left(\dfrac{\pi - B - B}{2}\right) \\\\ & = 2 a c \sin \left(\dfrac{\pi}{2} - B\right) \\\\ & = 2 a c \cos B \\\\ & = 2 a c \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) \;\; \left[\text{by cosine rule}\right] \\\\ & = c^2 + a^2 - b^2 = RHS \end{aligned}$

Hence proved.

Properties of Triangles

If $\; A, \; B, \; C \;$ in $\;$ $\triangle ABC$ $\;$ are in A.P. and the sides $\; a, \; b, \; c \;$ are in G.P., then show that $\; a^2, \; b^2, \; c^2 \;$ are in A.P.

$\because \;$ $A, \; B, \; C \;$ are in A.P. $\implies$ $2 B = A + C$ $\;\;\; \cdots \; (1)$

$\because \;$ $a, \; b, \; c \;$ are in G.P. $\implies$ $b^2 = a c$ $\;\;\; \cdots \; (2)$

In $\triangle ABC$, $\;$ $A + B + C = \pi$

i.e. $\;$ $A + C = \pi - B$ $\;\;\; \cdots \; (3)$

In view of equation $(1)$, equation $(3)$ becomes,

$2 B = \pi - B$ $\implies$ $3 B = \pi$ $\implies$ $B = \dfrac{\pi}{3}$ $\; \; \; \cdots \; (4)$

By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$

i.e. $\;$ $b^2 = c^2 + a^2 - 2 c a \cos B$

i.e. $\;$ $b^2 = c^2 + a^2 - 2 b^2 \cos \left(\dfrac{\pi}{3}\right)$ $\;\;\;$ [by equations $(2)$ and $(4)$]

i.e. $\;$ $b^2 = c^2 + a^2 - 2 b^2 \times \dfrac{1}{2}$

i.e. $\;$ $b^2 = c^2 + a^2 - b^2$

i.e. $\;$ $2 b^2 = a^2 + c^2$

$\implies$ $a^2, \; b^2, \; c^2 \;$ are in A.P.