Question

The angles of a triangle are in A.P. and the number of degrees in the least angle is to the number of degrees in the mean angle as 1 : 120. Find the angles in radians.

Answer 1

Let A, B and C be the angles of triangle ABC.

We are given that A, B and C are in A.P.

$\therefore$ Let A = a - d, B = a and C = a + d

According to the question,

A + B + C = ${180}^{\circ }$

[By angle sum property]

$\therefore$ a - d + a + a + d = ${180}^{\circ }$

$\Rightarrow 3a={180}^{\circ }\Rightarrow a={60}^{\circ }$ . . . (i)

Again,

$\frac{\text{least angle}}{\text{mean angle}}=\frac{1}{{120}^{\circ }}$

$\Rightarrow \frac{a-d}{a}=\frac{1}{120}$ $\Rightarrow 119a=120d$

$\Rightarrow d=\frac{119a}{120}$

$\Rightarrow d=\frac{119}{120}\times {60}^{\circ }$

$={\left(\frac{119}{2}\right)}^{\circ }$

$=\frac{119}{2}\times \frac{\pi }{180}=\frac{119\pi }{360}\text{radians}$

Now,

$1^{\circ }=\frac{\pi }{180}$ radians

$\therefore B=a={60}^{\circ }=\frac{\pi }{3}$ radians

$A=a-d=\frac{\pi }{3}-\frac{119\pi }{360}=\frac{\pi }{360}$ radians

$C=a+d=\frac{\pi }{3}+\frac{119\pi }{360}=\frac{239\pi }{360}$ radians