Question

In right triangle $ABC$, $\angle B={90}^o$, $AB=3cm$ and $AC=6cm$. Find $\angle C$ and $\angle A$.

Answer 1

$\sin C=\frac{AB}{AC}=\frac{3}{6}=\frac{1}{2}$

$C={\sin }^{-1}\left[\frac{1}{2}\right]={30}^{\circ }$

$\angle B+\angle C+\angle A={180}^{\circ }$

${90}^{\circ }+30+\angle A={180}^{\circ }$

$\angle A+={60}^{\circ }$