Question

In $△ABC$, $ABC={45}^o$, $AM\perp BC$, $AM=4cm$ and $BC=7cm$. Find the length of $AC$.

Answer 1

In $△AMB$, $\angle AMB={90}^o$, $\angle ABM={45}^o$ $\therefore \angle BAM={45}^o$
$\therefore △AMB$ is an isosceles right angled triangle
$\therefore MA=MB=4cm$
$MC=BC-MB=7-4=3cm$
$\therefore MC=3cm$
In $△AMC$, ${AC}^2-{AM}^2+{MC}^2$ [$\because$ Pythagoras theorem]
${AC}^2=4^2+3^2$
${AC}^2=16+9$
${AC}^2=25$
$AC=\sqrt{25}=5cm$