Question

Find angle ${}^{\prime }{A}^{\prime }$ in degrees from the above figure :

Answer 1

Consider triangle ABC, $\angle ABC={90}^{\circ }$, $AC=10$, $BC=\frac{10}{\sqrt{2}}$ and $\angle CAB=A$
Now, $\sin A=\frac{P}{H}$
$\sin A=\frac{\frac{10}{\sqrt{2}}}{10}$
$\sin A=\frac{1}{\sqrt{2}}$
$A={45}^{\circ }$