Question

In the adjoining figure, ∆ABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm.
Find (i) BC, (ii) AB.

Answer 1

From the given right-angled triangle, we have:

$\frac{BC}{AC}=\sin {30}^0$

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

$\Rightarrow BC=\frac{20}{2}=10cm$

Also, $\frac{AB}{AC}=\cos {30}^0$

$\Rightarrow \frac{AB}{20}=\frac{\sqrt{3}}{2}\Rightarrow AB=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}cm$

$\therefore BC=10cm$ and $AB=10\sqrt{3}$ cm