$Δ ABC$ is an isosceles triangle such that $AB = AC.$

The vertical angle $BAC = 2θ$

Triangle is inscribed in the circle with center $O$ and radius $a.$

Draw $AM$ perpendicular to $BC.$

Since, $Δ ABC$ is an isosceles triangle, the circumcenter of the circle will lie on the perpendicular from $A$ to $BC.$

Let $O$ be the circumcenter.

\[BOC=2\times 2\theta =4\theta \text{ }\left( Using\text{ }central\text{ }angle\text{ }theorem \right)\]

\[COM\text{ }=\text{ }2\theta \] (Since, $Δ OMB$ and $Δ OMC$ are congruent triangles)

\[OA\text{ }=\text{ }OB\text{ }=\text{ }OC\text{ }=\text{ }a\text{ }\left( radius\text{ }of\text{ }the\text{ }circle \right)\]

In $Δ OMC,$

\[CM\text{ }=\text{ }asin2\theta \]

\[OM\text{ }=\text{ }acos2\theta \]

\[BC\text{ }=\text{ }2CM\] (Perpendicular from the center bisects the chord)

\[BC\text{ }=\text{ }2asin2\theta \]

Height of \[\Delta \text{ }ABC\text{ }=\text{ }AM\text{ }=\text{ }AO\text{ }+\text{ }OM\]

\[AM\text{ }=\text{ }a\text{ }+\text{ }acos2\theta \]