Find a formula for sinA\sin AsinA, in terms of bbb and ccc only. (N.B. it's an isosceles triangle) Solution sinA=hc\sin A = \dfrac{h}{c}sinA=chh=c2−(b2)2=c2−b24=4c2−b24=4c2−b22\begin{align*}h &= \sqrt{{c^2}-\left(\dfrac{b}{2}\right)^2} \\[1em]&= \sqrt{{c^2}-\dfrac{{b^2}}{4}} \\[1em]&= \sqrt{\dfrac{4c^2-b^2}{4}} \\[1em]&= \dfrac{\sqrt{4c^2-b^2}}{2}\end{align*}h=c2−(2b)2=c2−4b2=44c2−b2=24c2−b2So sinA=4c2−b22c\sin A = \dfrac{\sqrt{4c^2-b^2}}{2c}sinA=2c4c2−b2.