Question

In a $△ABC,\angle A={60}^{\circ },\angle B={30}^{\circ }$ and $\angle C={90}^{\circ }.$

Then the ratio $AC:BC:AB$ will be equal to

• A. $1:2:\sqrt{3}$
• B. $1:1:\sqrt{2}$
• C. $1:\sqrt{3}:2$
• D. $\sqrt{3}:2:1$

Answer 1

The correct option is C

$1:\sqrt{3}:2$

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

In $△ABC$, we have

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