Question

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

If $a=6,b=3$ and $\cos (A-B)=\frac{4}{5}$

Then which of the following is correct?

• A. $C=\frac{\pi }{4}$
• B. $A=\sin \frac{2}{\sqrt{5}}$
• C. $Area\left(\Delta ABC\right)=9$
• D. none of these

Answer 1

The correct option is C $Area\left(\Delta ABC\right)=9$

$\cos (A-B)=\frac{4}{5}$

Hence, $\sin (A-B)=\frac{3}{5}$.

Now $\cos \left(\frac{A-B}{2}\right)=\sqrt{\frac{1+\cos (A-B)}{2}}$
$=\frac{3}{\sqrt{10}}$.

And similarly , $\sin \left(\frac{A-B}{2}\right)=\frac{1}{\sqrt{10}}$.

Hence, $\tan \left(\frac{A-B}{2}\right)=\frac{1}{3}$.

$=\frac{a-b}{a+b}.\cot \frac{C}{2}$
$\frac{1}{3}=\frac{6-3}{6+3}.\cot \frac{C}{2}$
$\cot \frac{C}{2}=1$
$\frac{C}{2}=\frac{\pi }{4}$
$C=\frac{\pi }{2}$.

Hence, area of triangle ABC
$=\frac{6\times 3}{2}$
$=9$