- 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.
- Would be solution have been ambiguous if $B = 30^\circ$, $c = 150$ and $b = 75$?
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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. $\;$ $a = \dfrac{150 \times 1}{\sqrt{3} / 2} = 100 \sqrt{3}$
- Now, $\;$ $c \sin B = 150 \times \sin 30^\circ = \dfrac{150}{2} = 75 < b$
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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 $C = 90^\circ$