Question

In an isosceles triangle the vertex angle is equal to c and the area is S. Find the length of the base of the triangle.

Answer 1

Let s be one of the equal sides. Then the area of the triangle is given by,

$\frac{1}{2}{x}^{2}sin\left(c\right)=S$

$⟹x=\sqrt{\frac{2S}{\sin \left(c\right)}}$

The base of the triangle is given by $b=2x\sin \left(\frac{c}{2}\right)=2\sin \left(\frac{c}{2}\right)\sqrt{\frac{2S}{\sin \left(c\right)}}$