Question

In a $△$ ABC, the tangent of half the difference of two angles is one - third the tangent of half the sum of the two angles. The ratio of the sides opposite the angles is

• A. 2:3
• B. 1:3
• C. 2:1
• D. 3:4

Answer 1

The correct option is C

2:1

$tan\left(\frac{A-B}{2}\right)=\frac{1}{3}tan\left(\frac{A+B}{2}\right)=\frac{1}{3}cot\frac{C}{2}$
$\Rightarrow$ From Law of Tangents $\frac{a-b}{a+b}=\frac{1}{3}\Rightarrow 3a-3b=a+b\Rightarrow 2a=4b$
$\Rightarrow \frac{a}{b}=2$