Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.
lf $C=A-B={60}^0$, then $\frac{a-b}{a+b}=?$

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{1}{3}$

According to Napiers Analogy, we know that

$\tan \left[\frac{1}{2}(A-B)\right]=\frac{a-b}{a+b}\cot \left[\frac{C}{2}\right]$

Putting the values of$A-B=C={60}^0$ from the question in the above equation, we get

$\tan \left[{30}^0\right]=\frac{a-b}{a+b}\cot \left[{30}^0\right]$

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

So,

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