Question

In a triangle ABC, the tangent of half the difference of two angles is one third the tangent of half the sum of the angles, then the ratio of the sides opposite to these angles is:

• A. $2:1$
• B. $3:4$
• C. $1:3$
• D. $4:5$

Answer 1

Answer: (A)

$2:1$

Let $A,B$ and $C$ be the angles of a triangle.

Let $a,b$ and $c$ be the opposite sides of these angles respectively.

According to the question,

$\tan \left[\frac{1}{2}(A-B)\right]=\frac{1}{3}\tan \left[\frac{1}{2}(A+B)\right]$

According to Napiers Anology, we can write the above equation as

$\frac{a-b}{a+b}\cot \frac{C}{2}=\frac{1}{3}\tan [90-\frac{C}{2}]$

$\frac{a-b}{a+b}\cot \frac{C}{2}=\frac{1}{3}\cot \frac{C}{2}$

Cancelling $\cot \frac{C}{2}$ from the equation, weget

$\frac{a-b}{a+b}=\frac{1}{3}$

$3a-3b=a+b$

$2a=4b$

$a=2b$

$a:b=2:1$

Hence, option (A) is the correct answer