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: (A)

Isosceles

Answer 2

Sol. 2e^iB = e^iA + e^iC

Equating real and imaginary parts

2 cos B = cos A + cos C, 2 sin B = sin A + sin C

\ tan B = $\frac{2\sin\frac{A + C}{2}\cos\frac{A - C}{2}}{2\cos\frac{A + C}{2}\cos\frac{A - C}{2}} = \tan\frac{A + C}{2}$

or tan B = tan $\frac{A + C}{2}$ Ž tan B = cot $\frac{B}{2}$

\ B = $\frac{\pi}{2}$– $\frac{B}{2}$ or $\frac{3B}{2}$ = $\frac{\pi}{2}$

\ B = $\frac{\pi}{3}$ = 60°\ A + C = 120°

\ 2 cos 60° = 2 cos $\frac{A + C}{2}$ cos $\frac{A - C}{2}$by (1)

1 = 2 cos 60° cos $\frac{A - C}{2}$

\ cos $\frac{A - C}{2}$ = 1 Ž $\frac{A - C}{2}$ = 0 Ž A = C.