Question

In triangle ABC $\angle B={90}^{\circ }$, AB $=$ 3, AC $=$ 5 and BD is perpendicular to AC then find BD.

Answer 1

$△ABC$ is right angled $△,\angle B=90°$
$BD\perp AC$
$AB=3,AC=5$
By Pythagoras theorem,
${AB}^2+{BC}^2={AC}^2$
$\Rightarrow {BC}^2={AC}^2-{AB}^2=5^2-3^2$
$\Rightarrow BC=4$
In $△ABC$,
$\angle B=90°,AB=3,BC=4,AC=5$
$\sin C=\frac{AB}{AC}=\frac{3}{5}\to 1$
In $△BDC$
$\angle BDC=90°,BC=4$
$\sin C=\frac{BD}{BC}\to 2$
Comparing $1$ and $2$
$\frac{3}{5}=\frac{BD}{4}\Rightarrow BD=3.4$