# Area of Isosceles Triangle/Proof 1

## Theorem

Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.

Let $\theta$ be the angle of the apex $A$.

Let $r$ be the length of a leg of $\triangle ABC$.

Then the area $\mathcal A$ of $\triangle ABC$ is given by:

$\mathcal A = \dfrac 1 2 r^2 \sin \theta$

## Proof

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \mathcal A$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac 1 2 b h$$ $$\displaystyle$$ $$\displaystyle$$ Area of Triangle in Terms of Side and Altitude $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac 1 2 b \left({r \cos \dfrac \theta 2}\right)$$ $$\displaystyle$$ $$\displaystyle$$ by definition of cosine $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac 1 2 2 \left({r \sin \dfrac \theta 2}\right) \left({r \cos \dfrac \theta 2}\right)$$ $$\displaystyle$$ $$\displaystyle$$ by definition of sine $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac 1 2 r^2 \sin \theta$$ $$\displaystyle$$ $$\displaystyle$$ Double Angle Formula for Sine

$\blacksquare$