Question

Given that $e^{iA},e^{iB},e^{iC}$ are in $A.P.$, where $A,B,C$ are the angles of a triangle then the triangle is

• A. isosceles
• B. equilateral
• C. right angled
• D. none

Answer 1

Answer: (B)

equilateral

Ans. $\left(a\right)$, $\left(b\right)$.
$2e^{iB}=e^{iA}+e^{iC}$
Equating real and imaginary parts,
$2\cos B=\cos A+\cos C$,
$2\cos B=\cos A+\cos C$
$\therefore \tan B=\frac{2\sin \frac{A+C}{2}\cos \frac{A-C}{2}}{\cos \frac{A+C}{2}\cos \frac{A-C}{2}}=\tan \frac{A+C}{2}$
or $\tan B=\tan \frac{A+C}{2}\Rightarrow \tan B=\cot \frac{B}{2}$
$\therefore B=\frac{\pi }{2}-\frac{B}{2}$ or $\frac{3B}{2}=\frac{\pi }{2}$
$\therefore B=\frac{\pi }{3}={60}^o\therefore A+C={120}^o$
$\therefore 2\cos {60}^o=2\cos \frac{A+C}{2}\cos \frac{A-C}{2}$ by $\left(I\right)$
$1=2\cos {60}^o\cos \frac{A-C}{2}=\cos \frac{A-C}{2}$
$\Rightarrow$ $\frac{A-C}{2}=0$ $\Rightarrow$ $A=C$
$\therefore$ Isosceles, since $A=C$ and $B={60}^o$, $\Delta$ is equilateral also.