Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, $a=2b$ and $|A-B|=\frac{\pi }{3}.$

The measure of $\angle C$ is?

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. none of these

Answer 1

The correct option is B $\frac{\pi }{3}$

We have, $\sin A=2\sin B$
$\sin A-2\sin B=0$ ...(i)
${\sin }^{2}A=4{\sin }^{2}B$
$1-{\cos }^{2}A=4-4{\cos }^{2}B$
$4{\cos }^{2}B-co{s}^{2}A=3$
And
$\sin (A-B)=\frac{\sqrt{3}}{2}$.
$\sin A\cos B-\cos A\sin B=\frac{\sqrt{3}}{2}$
Now, $\sin B(2\cos B-\cos A)=\frac{\sqrt{3}}{2}$.
$\frac{2\cos B+\cos A}{\sin B}=\frac{2.3}{\sqrt{3}}$
$2\cot B+\frac{\cos A}{\sin B}=2\sqrt{3}$.
Considering $A={90}^0$ we get
$2\cot B=2\sqrt{3}$
$B={30}^0$.
Hence,
$C={90}^0-{30}^0$
$={60}^0$
Thus the above triangle is right angled and $\angle C={60}^0$ or $\frac{\pi }{3}$