Question

In a triangle $ABC$ with usual notation, the tangent of half the difference of two angles $A$ and $B$ is one-third the tangent of half the sum of these angles. If the ratio (in simplest form) of the side opposite to these angles is $a:b,$ then the value of $a+b$ is

Answer 1

Here, given that in $△ABC$
$\tan \left(\frac{A-B}{2}\right)=\frac{1}{3}\tan \left(\frac{A+B}{2}\right)\cdots \left(1\right)$
From Napier’s analogy,
$\tan \left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\cot \left(\frac{C}{2}\right)\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right),$ we get
$\frac{1}{3}\tan \left(\frac{A+B}{2}\right)=\frac{a-b}{a+b}\cot \left(\frac{C}{2}\right)$
$\Rightarrow \frac{1}{3}\cot \left(\frac{C}{2}\right)=\frac{a-b}{a+b}\cot \left(\frac{C}{2}\right)[\because A+B+C=\pi ]$
$\Rightarrow \frac{a-b}{a+b}=\frac{1}{3}$
$\Rightarrow 3a-3b=a+b$
$\Rightarrow 2a=4b$
$\Rightarrow \frac{a}{b}=\frac{2}{1}$
$\therefore a+b=3$