Question

In the given figure, $\angle B={90}^{\circ }$ and $\angle ADB=x^{\circ }$, then find $\sin \angle CAB$.

Answer 1

In $△CAB$,

$CA=20$ and $AB=12$

Using Pythagoras theorem,
$C{A}^2=A{B}^2+B{C}^2$

${20}^2={12}^2+B{C}^2$

$BC=16$
$\sin \angle CAB=\frac{P}{H}=\frac{BC}{AC}=\frac{16}{20}=\frac{4}{5}$