Question

In $△$ABC, right angled at B, BC = 5cm, $\angle BAC$ = ${30}^{\circ }$, find the length of the sides AB and AC.

Answer 1

We are given

$\angle BAC$= ${30}^{\circ }$, i.e., $\angle A$= ${30}^{\circ }$ and BC = 5cm

Now sin A = $\frac{BC}{AC}$ or sin ${30}^{\circ }$ = $\frac{5}{AC}$

or $\frac{5}{AC}$ = $\frac{1}{2}$ ......[$\because sin{30}^{\circ }=\frac{1}{2}$]

or AC = $2\times 5$ or 10cm

To find AB, we have,

$\frac{AB}{AC}$ = cos A

or $\frac{AB}{10}$ = cos ${30}^{\circ }$

or $\frac{AB}{10}$ = $\frac{\sqrt{3}}{2}$

$\therefore AB$ = $\frac{\sqrt{3}}{2}\times 10$ or 5$\sqrt{3}$cm

Hence, AB = 5$\sqrt{3}$cm and AC = 10cm