Question

In a right- angled triangle, if one acute angle is double the other. then prove that the hypotenuse is double the smallest side.

Answer 1

Let $\angle BAC=p$

then $\angle ACB=2p$

Now in $△ABC$,

$ABC+BCA+ACB={180}^{\circ }$

${90}^{\circ }+2p+p={180}^{\circ }$

$p={30}^{\circ }=\angle ACB$

$\angle ACB=2\left({30}^{\circ }\right)={60}^{\circ }$

Side opposite to the smallest angle is smallest

Hence, $BC$ is smallest

Now

$\cos 2p=\frac{BC}{AC}$

$\cos {60}^{\circ }=\frac{BC}{AC}$

$\frac{1}{2}=\frac{BC}{AC}$

$AC=2BC$. Hence Proved the result.