Question

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

• A. an isosceles triangle
• B. a right angled triangle
• C. an equilateral triangle
• D. a right angled isosceles triangle

Answer 1

Answer: (B)

a right angled triangle

Using Sine law in the triangle,

$\frac{a}{\sin (\theta -{60}^0)}=\frac{2a}{\sin \theta }$

$\Rightarrow \frac{1}{2}=\frac{\sin \theta \cdot \cos {60}^0-\cos \theta \cdot \sin {60}^0}{\sin \theta }$

$\Rightarrow 1=1-\sqrt{3}\cot \theta$

$\Rightarrow \cot \theta =0$

$\Rightarrow \theta =\frac{\pi }{2}$

Now, one angle is ${90}^0$, one is ${30}^0$ and the last is ${60}^0$

So, its a right-angled triangle only.

Hence, option B