Question

If $A,B,C$ are the angles of a triangle and $e^{iA},e^{iB},e^{iC}$ are in $A.P.,$ then the triangle must be

• A. right angle
• B. isosceles triangle
• C. equilateral
• D. None of these

Answer 1

The correct option is C equilateral

We have, $e^{iA},e^{iB},e^{iC}$ are in $A.P.$

$\Rightarrow 2e^{iB}=e^{iA}+e^{iC}$

$\Rightarrow 2(\cos B+i\sin B)=(\cos A+\cos C)+i(\sin A+\sin B)$

$\Rightarrow 2\cos B=\cos A+\cos C$and, $2\sin B=\sin A+\sin C$

$\Rightarrow 2\cos B=2\cos \frac{A+C}{2}\cos \frac{A-C}{2}$ ...(1)

and, $2\sin B=2\sin \frac{A+C}{2}\cos \frac{A-C}{2}$ ...(2)

Dividing (1) by (2), we get

$\cot B=\cot \left(\frac{A+C}{2}\right)=\tan \frac{B}{2}\Rightarrow \cos \frac{3B}{2}=0$

$\Rightarrow \frac{3B}{2}=\frac{\pi }{2}$ or $B=\frac{\pi }{3}\Rightarrow A+C=\frac{2\pi }{3}.$

Putting this value in (1), we get

$2\cos \frac{\pi }{3}=2\cos \frac{\pi }{3}\cos \left(\frac{A-C}{2}\right)\Rightarrow \cos \left(\frac{A-C}{2}\right)=1$

$\Rightarrow \left(\frac{A-C}{2}\right)=0$ or $A=C\Rightarrow A=B=C=\frac{\pi }{3}.$