Question

$ABC$ is a triangle in which $a=6,b=3$ and $\cos$ ($A-B$) $=\frac{4}{5},AD$ is the median through $A,\angle BAD=\theta ,CL$ is perpendicular to $AD$. Then, $△ABC$ is

• A. isosceles triangle.
• B. right angled triangle.
• C. obtuse angled triangle
• D. none of the above

Answer 1

Answer: (B)

right angled triangle.

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

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

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

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

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

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

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

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