Question

In a $\Delta ABC$, 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 to the two angles is:

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

Answer 1

The correct option is A $1:2$

We have: $\tan \left(\frac{A-B}{2}\right)=\frac{1}{3}\tan \left(\frac{A+B}{2}\right)\cdots \left(i\right)$
We know, using Napier's analogy,
$\tan \left(\frac{A-B}{2}\right)=\left(\frac{a-b}{a+b}\right)\cot \frac{C}{2}\cdots \left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$,
$\frac{1}{3}\tan \left(\frac{A+B}{2}\right)=\left(\frac{a-b}{a+b}\right)\cot \frac{C}{2}$
$\Rightarrow \frac{1}{3}\cot \frac{C}{2}=\left(\frac{a-b}{a+b}\right)\cot \frac{C}{2}$
$\Rightarrow \left(\frac{a-b}{a+b}\right)=\frac{1}{3}$
$\Rightarrow 3a-3b=a+b$
$\Rightarrow 2a=4b\Rightarrow \frac{b}{a}=\frac{1}{2}$
Thus the ratio of the sides opposite to the angles is $b:a=1:2$