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Question:
If the angles of a triangle are in A.P., then the measures of one of the angles in radians is
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{3}$
(c) $\frac{\pi}{2}$
(d) $\frac{2 \pi}{3}$
Solution:
(b) $\frac{\pi}{3}$
Let the angles of the triangle be $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$.
Thus, we have:
$a-d+a+a+d=180$
$\Rightarrow 3 a=180$
$\Rightarrow a=60$
Hence, the angles are $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$, i.e., $(60-d)^{\circ}, 60^{\circ}$ and $(60+d)^{\circ}$. $60^{\circ}$ is the only angle which is independent of $d$.
$\therefore$ One of the angles of the triangle (in radians) $=\left(60 \times \frac{\pi}{180}\right)=\frac{\pi}{3}$
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