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

From the identity of an A.P
$2e^{iB}=e^{iA}+e^{iC}$
$=(cosA+cosC)+i(sinA+sinC)$
$=2\left(cosB\right)+2sinBi$
Hence
$2cosB=cosC+cosA=2cos\left(\frac{A+C}{2}\right)cos\left(\frac{A-C}{2}\right)$
$2sinB=sinA+sinC=2sin\left(\frac{A+C}{2}\right)cos\left(\frac{A-C}{2}\right)$
$tanB=\frac{sin\left(\frac{A+C}{2}\right)}{cos\left(\frac{A+C}{2}\right)}$
$=tan\left(\frac{A+C}{2}\right)$
$=tan\left(\frac{\pi -B}{2}\right)$
$=tan(\frac{\pi }{2}-\frac{B}{2})$
Hence
$B=\frac{\pi }{2}-\frac{B}{2}$
$3B=\pi$
$B={60}^0$
Hence
$A+C={120}^0$
Now
$2cos{60}^0=2cos\left(\frac{{120}^0}{2}\right)cos\left(\frac{A-C}{2}\right)$
$1=1.cos\left(\frac{A-C}{2}\right)$
Hence
$A-C=0$
$A=C$
Therefore
$A=B=C={60}^0$
This proves that it is a equilateral triangle.