Question

$ABC$ is a triangle in which $a=6,b=3$ and $\cos$ ($A-B$) $=\frac{4}{5}$. Area of the triangle $ABC$ is

• A. $9$ sq. units
• B. $12$ sq. units
• C. $18$ sq. units
• D. $36$ sq. units

Answer 1

The correct option is A $9$ sq. units

$\frac{\cos (A-B)}{1}=\frac{1-{\tan }^2\left(\frac{A-B}{2}\right)}{1+{\tan }^2\left(\frac{A-B}{2}\right)}$

${\tan }^2\left(\frac{A-B}{2}\right)=\frac{1-\cos (A-B)}{1+\cos \left(\frac{A+B}{2}\right)}$

${\tan }^2\left(\frac{A-B}{2}\right)=\frac{1}{9}$

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

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

$\frac{1}{3}=\left(\frac{6-3}{6+3}\right)\cot \left(\frac{C}{2}\right)$

$\cot \frac{C}{2}=1$

$\frac{C}{2}=\frac{\pi }{4}$

$C=\frac{\pi }{2}$

Area of $△ABC=\frac{1}{2}ab=\frac{1}{2}\times 6\times 3=9$ sq units.