Question

In $△ABC$, if $m\angle ABC=90,m\angle ACB=45$ and $AC=6$, then area of $△ABC$ is ______

Answer 1

REF.Image.

In $cos{45}^{\circ }=\frac{BC}{AC}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{BC}{6}$

$\Rightarrow BC=3\sqrt{2}$

Area of traingle $=\frac{1}{2}\times BC\times CA\times sin{45}^{\circ }$

$=\frac{1}{2}\times 3\sqrt{2}\times 6\times \frac{1}{\sqrt{2}}$

Area $=9m^2$