Question

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

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

Answer 1

Answer: (A)

right angled

Answer 2

Given, $A-B=60^{\circ}$

By sine rule ,
$\frac{2 a}{\sin A}=\frac{a}{\sin B}$
$\Rightarrow \,\,\,\,\,\, \sin A-2 \sin B=0$
$\Rightarrow \,\,\,\,\,\,\, \sin \left(60^{\circ}+B\right)-2 \sin B=0$
$\Rightarrow \,\,\,\,\,\,\, \frac{\sqrt{3}}{2} \cos B+\frac{1}{2} \sin B-2 \sin B=0$
$\Rightarrow \,\,\,\,\,\,\,\frac{\sqrt{3}}{2} \cos B-\frac{3}{2} \sin B=0$
$\Rightarrow \,\,\,\,\,\,\,\sqrt{3}\left(\frac{1}{2} \cos B-\frac{\sqrt{3}}{2} \sin B\right)=0$
$\Rightarrow \,\,\,\,\,\,\, \sqrt{3}\left[\cos \left(60^{\circ}+B\right)\right]=0$
$\Rightarrow \,\,\,\,\,\,\, 60^{\circ}+B=90^{\circ}$
$\Rightarrow \,\,\,\,\,\,\,B=30^{\circ}$
$\Rightarrow \,\,\,\,\,\,\,A=90^{\circ}$