Question

In $△ABC$ right angle is at $B.AB=3cmAC=6cm$ determine $\angle BAC$ and $\angle ACB$.

Answer 1

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

$\sin \angle BCA=\sin 30°$

$\angle BCA=30°$

$\angle ACB=30°$

$\therefore \angle ABC+\angle BAC+\angle ACB=180°$ [sum of all angles in a triangle in $180°$]

$\angle BAC+90°+30°=180°$

$\angle BAC=60°$