Question

Prove that if the angles of a triangle are ${30}^o$, ${60}^o$ and ${90}^o$, then the side opposite to ${30}^o$ is half of the hypotenuse, and side opposite to ${60}^o$ is $\frac{\sqrt{3}}{2}$ times of hypotenuse.

Answer 1

Given : $ABC$ is a triangle with $\angle A={30}^o$, $\angle B={90}^o$, $\angle C={60}^o$
To Prove : $BC=\frac{1}{2}AC$
and $AB=\frac{\sqrt{3}}{2}AC$
Proof : In right angled $\Delta ABC$
$\sin {30}^o=\frac{BC}{AC}$
$\frac{1}{2}=\frac{BC}{AC}$
$\therefore BC=\frac{1}{2}AC$
Again, in right angled triangle $ABC$
$\sin {60}^o=\frac{AB}{AC}$
$\frac{\sqrt{3}}{2}=\frac{AB}{AC}$
$\therefore AB=\frac{\sqrt{3}}{2}AC$.