Question

If one side of a triangle is double the other and the angles opposite to these sides differ by ${60}^0$, then the triangle is

• A. obtuse angled
• B. acute angled
• C. isosceles
• D. right angled

Answer 1

The correct option is D

right angled

Given, A-B=${60}^0$

By sine rule,

$\frac{2a}{sinA}=\frac{a}{sinB}\Rightarrow sinA-2sinB=0\Rightarrow sin({60}^0+B)-2sinB=0\Rightarrow \frac{\sqrt{3}}{2}cosB+\frac{1}{2}sinB-2sinB=0\Rightarrow \frac{\sqrt{3}}{2}cosB-\frac{3}{2}sinB=0\Rightarrow \sqrt{3}(\frac{1}{2}cosB-\frac{\sqrt{3}}{2}sinB)=0\Rightarrow \sqrt{3}\left[cos\right({60}^0+B\left)\right]=0\Rightarrow {60}^0+B={90}^0\Rightarrow B={30}^0\Rightarrow A={90}^0$

Hence, it is right angled triangle