In the adjoining figure, ∆ABC is a right-angled triangle in which


In the adjoining figure, $\triangle A B C$ is a right-angled triangle in which $\angle B=90^{\circ}, \angle A=30^{\circ}$ and $A C=20 \mathrm{~cm}$.


(i) BC,

(ii) AB.




From the given right-angled triangle, we have:

$\frac{B C}{A C}=\sin 30^{\circ}$

$\Rightarrow \frac{B C}{20}=\frac{1}{2}$

$\Rightarrow B C=\frac{20}{2}=10 \mathrm{~cm}$

Also, $\frac{A B}{A C}=\cos 30^{\circ}$

$\Rightarrow \frac{A B}{20}=\frac{\sqrt{3}}{2}$

$\Rightarrow A B=\left(20 \times \frac{\sqrt{3}}{2}\right)=10 \sqrt{3} \mathrm{~cm}$

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


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