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

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equilateral

none

isosceles

right angled

answer is B.

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Detailed Solution

$2 e^{i B} = e^{i A} + e^{i C}$

Equating real and imaginary parts,

$\begin{aligned} 2 \cos B = \cos A + \cos C \\ 2 \sin B = \sin A + \sin C \end{aligned}$

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

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

$\begin{array}{l} ∴ B = \frac{π}{2} - \frac{B}{2} or \frac{3 B}{2} = \frac{π}{2} \\ ∴ B = \frac{π}{3} = 60^{\circ} ∴ A + C = 120^{\circ} \\ ∴ 2 \cos 60^{\circ} = 2 \cos \frac{A + C}{2} \cos \frac{A - C}{2} by (I) \\ 1 = 2 \cos 60^{\circ} \cos \frac{A - C}{2} \\ ∴ \cos \frac{A - C}{2} = 1 \Rightarrow \frac{A - C}{2} = 0 \\ \Rightarrow A = C = 60^{\circ} ∴ A = B = C = 60^{\circ} \end{array}$