Question

In the adjoining figure, $△ABC$ is a right-angled triangle in which $\angle B={90}^{\circ },\angle A={30}^{\circ }$ and $AC=20cm$.
Find (i) $BC$, (ii) $AB$.

Answer 1

Draw a right angled $△ABC$ using given instructions:

Here $\sin 30=\frac{BC}{AC}$

$\frac{1}{2}=\frac{BC}{20}$

Or $BC=10cm$

By Pythagoras theorem:

$(AB{)}^2=(AB{)}^2-(BC{)}^2$

$=(20{)}^2-(10{)}^2$

$=300$

$AB=10\sqrt{3}cm$.