Question

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

lf $A=3B$, then $\frac{a-b}{2b}$ is equal to

• A. $\sin \frac{A}{2}$
• B. $\sin \frac{C}{2}$
• C. $\cos \frac{B}{2}$
• D. $\cos \frac{C}{2}$

Answer 1

The correct option is B $\sin \frac{C}{2}$

Given- In a triangle ABC,

$A=3B$

Since, $A+B+C=\pi$ ------------$(\text{Sum of all angles of a triange is}$ $\pi )$

$\Rightarrow 2B=\frac{\pi }{2}-\frac{C}{2}$

$\text{Formula used-}$

$sinX-sinY=2\times sin\left(\frac{X-Y}{2}\right)\times cos\left(\frac{X+Y}{2}\right)$

By $\text{sine formula}$,

$b=\left(2R\right)\times \left(sinB\right)$

$a=\left(2R\right)\times \left(sinA\right)$

Therefore,

$\frac{a-b}{2b}=\frac{\left(2R\right)\times \left(sinA\right)-\left(2R\right)\times \left(sinB\right)}{2\times \left(2R\right)\times \left(sinB\right)}$

$=\frac{sinA-sinB}{2\times sinB}$

$=\frac{2\times sin\left(\frac{A-B}{2}\right)\times cos\left(\frac{A+B}{2}\right)}{2\times sinB}$

$=\frac{sin\left(\frac{3B-B}{2}\right)\times cos\left(\frac{3B+B}{2}\right)}{sinB}$

$=cos\left(2B\right)$

Putting value of $2B$, we get

$\frac{a-b}{2b}=sin\left(\frac{C}{2}\right)$

Hence, option (B) is correct.