In the adjoining figure, ∆ABC is a right-angled at B and
Question:

In the adjoining figure, $\triangle A B C$ is a right-angled at $B$ and $\angle A=45^{\circ}$. If $A C=3 \sqrt{2} \mathrm{~cm}$,

find

(i) BC,

(ii) AB.

Solution:

From right-angled ∆ABCwe have:

$\frac{B C}{A C}=\sin 45^{\circ}$

$\Rightarrow \frac{B C}{3 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow B C=3 \mathrm{~cm}$

Also, $\frac{A B}{A C}=\cos 45^{\circ}$

$\Rightarrow \frac{A B}{3 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow A B=3 \mathrm{~cm}$

$\therefore B C=3 \mathrm{~cm}$ and $A B=3 \mathrm{~cm}$