The sides of a triangle are in A.P and its area is $\dfrac{3}{5}$ths of an equilateral triangle of the same perimeter. Prove that its sides are in the ratio $3 : 5 : 7$ and find the greatest angle of the triangle.
Let the sides of the triangle be $s_1$, $s_2$ and $s_3$.
Given: $\;$ $s_1$, $s_2$ and $s_3$ are in A.P
$\therefore \;$ Let $s_1 = a - x$, $s_2 = a$, $s_3 = a + x$
Semi-perimeter of the triangle $= s = \dfrac{a - x + a + a + x}{2} = \dfrac{3a}{2}$
Area of the triangle $= \Delta = \sqrt{s \left(s - s_1\right) \left(s - s_2\right) \left(s - s_3\right)}$
i.e. $\;$ $\Delta = \sqrt{\left(\dfrac{3a}{2}\right) \left(\dfrac{3a}{2} - a + x\right) \left(\dfrac{3a}{2} - a\right) \left(\dfrac{3a}{2} - a - x\right)}$
i.e. $\;$ $\Delta = \sqrt{\left(\dfrac{3a}{2}\right) \left(\dfrac{a + 2x}{2}\right) \left(\dfrac{a}{2}\right) \left(\dfrac{a - 2x}{2}\right)}$
i.e. $\;$ $\Delta = \dfrac{a}{4} \sqrt{3 \left(a^2 - 4 x^2\right)}$ $\;\;\; \cdots \; (1)$
Area of an equilateral triangle with perimeter same as that of triangle with sides $s_1$, $s_2$ and $s_3$ is
$\Delta_1 = \dfrac{a^2 \sqrt{3}}{4}$ $\;\;\; \cdots \; (2)$
Given: $\;$ $\Delta = \dfrac{3}{5} \Delta_1$
$\therefore \;$ In view of equations $(1)$ and $(2)$ we have,
$\dfrac{a}{4} \sqrt{3 \left(a^2 - 4 x^2\right)} = \dfrac{3}{5} \times \dfrac{a^2 \sqrt{3}}{4}$
i.e. $\;$ $\sqrt{a^2 - 4x^2} = \dfrac{3 a}{5}$
i.e. $\;$ $a^2 - 4 x^2 = \dfrac{9 a^2}{25}$
i.e. $\;$ $4 x^2 = a^2 - \dfrac{9a^2}{25} = \dfrac{16 a^2}{25}$
i.e. $\;$ $x^2 = \dfrac{4 a^2}{25}$
$\implies$ $x = \dfrac{2 a}{5}$ $\;\;\;$ [negative value for $x$ is rejected as sides of a triangle cannot be negative]
$\therefore \;$ The sides of the triangle are
$s_1 = a - \dfrac{2a}{5}$, $\;$ $s_2 = a$, $\;$ $s_3 = a + \dfrac{2a}{5}$
i.e. $\;$ $s_1 = \dfrac{3a}{5}$, $\;$ $s_2 = a$, $\;$ $s_3 = \dfrac{7a}{5}$
Now, ratio of sides of the triangle is
$s_1 : s_2 : s_3 = \dfrac{3a}{5} : a : \dfrac{7a}{5}$
i.e. $\;$ $s_1 : s_2 : s_3 = 3 : 5 : 7$
Greatest side of the triangle is $s_3 = \dfrac{7a}{5}$
$\therefore \;$ angle opposite to $s_3$ i.e. $S_3$ is the greatest angle.
By cosine rule,
$\begin{aligned}
\cos \left(S_3\right) & = \dfrac{s_1^2 + s_2^2 - s_3^2}{2 s_1 s_2} \\\\
& = \dfrac{\dfrac{9a^2}{25} + a^2 - \dfrac{49 a^2}{25}}{2 \times \dfrac{3a}{5} \times a} \\\\
& = \dfrac{- 15 / 25}{6 / 5} \\\\
& = - \dfrac{1}{2}
\end{aligned}$
$\implies$ $S_3 = \cos^{-1} \left(- \dfrac{1}{2}\right) = 120^\circ$