Question

In a $△ABC,\angle A={60}^{\circ },\angle B={30}^{\circ }and\angle C={90}^{\circ }.$
Then find the ratio of the sides AC: BC: AB.

Answer 1

Let us assume the length the of the side opposite to $\angle B$ = x.

In $△ABC$, we have

$sin\left(\angle B\right)=sin{30}^{\circ }=\frac{x}{AB}\Rightarrow \frac{1}{2}=\frac{x}{AB}\Rightarrow AB=2x...\left(i\right)And,tan{30}^{\circ }=\frac{x}{BC}\Rightarrow \frac{1}{\sqrt{3}}=\frac{x}{BC}\Rightarrow BC=\sqrt{3}x....\left(ii\right)$

$\therefore AC:BC:AB=x:x\sqrt{3}:2x=1:\sqrt{3}:2$