Question

The sides of a triangle are in ratio $1:\sqrt{3}:2$, then angles of the triangle are in the ratio :

• A. $1:2:3$
• B. $1:3:5$
• C. $2:3:4$
• D. $3:2:1$

Answer 1

The correct option is A $1:2:3$

Let the sides of the triangle be a, b and c.

Given, $a:b:c=1:\sqrt{3}:2$

Using the sine rule,

$\frac{a}{\sin A}=$ $\frac{b}{\sin B}=$ $\frac{c}{\sin C}=k$

$a=k^{\prime }$

$b=\sqrt{3}{k}^{\prime }$

$c=2k^{\prime }$

We get,

$\frac{k^{\prime }}{\sin A}=$ $\frac{\sqrt{3}{k}^{\prime }}{\sin B}=$ $\frac{2k^{\prime }}{\sin C}=k$

Dividing by $2k^{\prime },$ we get

$\frac{\frac{1}{2}}{\sin A}=$ $\frac{\frac{\sqrt{3}}{2}}{\sin B}=$ $\frac{1}{\sin C}=\frac{k}{2k^{\prime }}$.

We also observe, $c^2=a^2+b^2$. Hence, the triangle is right angled at $C$.

The above equation modifies to,

$\frac{\frac{1}{2}}{\sin A}=$ $\frac{\frac{\sqrt{3}}{2}}{\sin B}=$ $\frac{1}{\sin 90}=\frac{k}{2k^{\prime }}$.

$\sin A=\frac{1}{2}$, that gives $A={30}^{\circ }$

$\sin B=\frac{\sqrt{3}}{2}$, that gives $B={60}^{\circ }$.

Therefore, $A:B:C=30:60:90=1:2:3$.

Option A is correct.