Question:
In the adjoining figure, ∆ABC is right-angled at B and ∠A = 30°. If BC = 6 cm, find
(i) AB,
(ii) AC.
Solution:
From the given right-angled triangle, we have:
$\frac{B C}{A B}=\tan 30^{\circ}$
$\Rightarrow \frac{6}{A B}=\frac{1}{\sqrt{3}}$
$\Rightarrow A B=6 \sqrt{3} \mathrm{~cm}$
Also, $\frac{B C}{A C}=\sin 30^{\circ}$
$\Rightarrow \frac{6}{A C}=\frac{1}{2}$
$\Rightarrow A C=(2 \times 6)=12 \mathrm{~cm}$
$\therefore A B=6 \sqrt{3} \mathrm{~cm}$ and $A C=12 \mathrm{~cm}$
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