Question

The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians.

Answer 1

Let A, B and C are the angles of triangle ABC

According to the question,

A, B and C are in A.P

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

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

[By angle sum property]

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

$\Rightarrow 3a={180}^{\circ }$

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

Also,

greatest angle is 5 times the least

$\therefore$ a + d = 5(a-d)

$\Rightarrow$ 4a = 6d

$\Rightarrow$ $d=\frac{2}{3}a$

$\Rightarrow d=\frac{2}{3}\times 60={40}^{\circ }$ . . . (ii)

$\therefore$ A = a - d = ${20}^{\circ }$

B = a = ${60}^{\circ }$

C = a + d = ${100}^{\circ }$

$\because 1^{\circ }=\left(\frac{\pi }{{180}^{\circ }}\right)$ radians

$\therefore A=20\times \frac{\pi }{180}=\frac{\pi }{9}$

$B=60\times \frac{\pi }{180}=\frac{\pi }{3}$

$C=100\times \frac{\pi }{180}=\frac{5\pi }{9}$

Thus,

$A=\frac{\pi }{9},B=\frac{\pi }{3},C=\frac{5\pi }{9}$