Question

In the adjoining figure, $△ABC$ is right-angled at $B$ and $\angle A={45}^{\circ }$. If $AC=3\sqrt{2}cm$,
Find (i) $BC$,(ii) $AB$.

Answer 1

From right angled $△ABC$,

(i) $\frac{BC}{AC}=\sin {45}^{\circ }$

$\frac{BC}{3\sqrt{2}}=\frac{1}{\sqrt{2}}$

Or $BC=3$

(ii) By Pythagoras theorem

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

$=(3\sqrt{2}{)}^2-(3{)}^2$

$=18-9$

$=9$

$AB=3cm$.