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.
As the angle of the triangle are in A.P.
Let, $A=a-d,B=a\text{and}C=a+d$
Sum of angles of a triangle is ${180}^{\circ }$, so
$a-d+a+a+d={180}^{\circ }$
$\Rightarrow a={60}^{\circ }$

And in radians, we know that
$1^{\circ }={\left(\frac{\pi }{180}\right)}^c$

$a={60}^{\circ }=60\left(\frac{\pi }{180}\right)=\frac{\pi }{3}$ radians

Given: $\frac{\text{No. of degrees in least angle}}{\text{No. of degrees in mean angle}}=\frac{1}{120}$

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

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

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

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

​​​​​​​$\Rightarrow d=\frac{119}{2}\left({60}^{\circ }\right)=\frac{{119}^{\circ }}{2}$

​​​​​​​$\Rightarrow d=\frac{119}{2}\left(\frac{\pi }{180}\right)=\frac{119\pi }{360}$ radians

$\therefore B=a={\left(\frac{\pi }{3}\right)}^c$

$A=a-d={(\frac{\pi }{3}-\frac{119\pi }{360})}^c={\left(\frac{\pi }{360}\right)}^c$

$C=a+d={(\frac{\pi }{3}+\frac{119\pi }{360})}^c={\left(\frac{239\pi }{360}\right)}^c$

$\therefore$ Angles in radians are $\frac{\pi }{360},\frac{\pi }{3},\frac{239\pi }{360}$.