Question

In the figure $ABC$ is a right angled triangle, $AB=4cm,\angle A={45}^{\circ }$ and $D$ is midpoint of $AC$.Then find the length of $BC,AC$ and $BD$.

Answer 1

From the figure, we have
$\cos {45}^{\circ }=\frac{AB}{AC}$
$\Rightarrow AC=\frac{AB}{\cos {45}^{\circ }}=\frac{4}{1\sqrt{2}}=4\sqrt{2}cm$
Now, $\sin {45}^{\circ }=\frac{BC}{AC}$
$\Rightarrow BC=AC\times \sin {45}^{\circ }=4\sqrt{2}\times \frac{1}{\sqrt{2}}=4cm$
Also, $\sin {45}^{\circ }=\frac{BD}{AB}$
$\Rightarrow BD=AB\times \sin {45}^{\circ }=4\times \frac{1}{\sqrt{2}}=2\sqrt{2}cm$.