Question

In a triangle $ABC,$ If two sides are $a=6,b=3$ and $\cos (A-B)=\frac{4}{5}$, then the area of the triangle is:

• A. $9$ Sq. units
• B. $3$ Sq. units
• C. $18$ Sq. units
• D. $\frac{9}{2}$ Sq. units

Answer 1

The correct option is A $9$ Sq. units

Given: $a=6,b=3$
$\cos (A-B)=\frac{4}{5}$
$\Rightarrow \frac{1-{\tan }^2\left(\frac{A-B}{2}\right)}{1+{\tan }^2\left(\frac{A-B}{2}\right)}=\frac{4}{5}$
$\Rightarrow 5-5{\tan }^2\left(\frac{A-B}{2}\right)=4+4{\tan }^2\left(\frac{A-B}{2}\right)$
$\Rightarrow 9{\tan }^2\left(\frac{A-B}{2}\right)=1$
$\Rightarrow \tan \left(\frac{A-B}{2}\right)=\pm \frac{1}{3}$
$\Rightarrow \frac{a-b}{a+b}\cot \frac{C}{2}=\pm \frac{1}{3}$
$\Rightarrow \frac{6-3}{6+3}\cot \frac{C}{2}=\pm \frac{1}{3}$
$\Rightarrow \cot \frac{C}{2}=\pm 1$
$\Rightarrow C={90}^{\circ }$
$\Delta =\frac{1}{2}ab\sin C$
$\Delta =\frac{1}{2}\cdot 6\cdot 3\cdot \sin {90}^0=9\text{sq. unit}$