# For the composite geometric shapes given the semi circle has the same area measure as the right angle triangle, and the base of right triangle is congruent to the radius of the semicircle. Calculate the value of angle theta?

$\theta = {57.52}^{o}$
As area of triangle is $\frac{\pi {r}^{2}}{2}$ (it is equal to semi-circle) and base is $r$, it's height is given by $\frac{2 \times a r e a}{b a s e}$ or $\frac{\pi {r}^{2}}{r} = \pi r$.
Now in ∆ABC, as $\tan \theta = \frac{\pi r}{2 r} = \frac{\pi}{2}$
$\theta = {\tan}^{- 1} \left(\frac{\pi}{2}\right) = {57.52}^{o}$