Question

Given two sides b and c of a triangle and its area S = 2bc/5. Find the third side of the triangle.

Answer 1

Given two sides are $b,c$. Let the other side be $a$ and angle opposite to it be $A$.

Given area of the triangle is $S=\frac{2bc}{5}$

$⟹\frac{1}{2}bc\sin A=\frac{2bc}{5}$

$⟹\sin A=4/5$

$⟹\cos A=\sqrt{1-{\sin }^{2}A}=\frac{3}{5}$

Using the cos rule, we have

$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{3}{5}$

$a^2=b^2+c^2-\frac{6}{5}bc$

$a=\sqrt{\frac{5b^2+5c^2-6bc}{5}}$