Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

lf $B=A-C={60}^0$, then $\frac{a-c}{a+c}=?$

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

Answer 1

Answer: (C)

$\frac{1}{3}$

According to Napier's Analogy, we know that

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

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

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

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

So,

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