Question

The angles of a triangle are in AP and the number of degrees in the least is to the number of radians in the greatest as $60:\pi$. Find the angles in degrees.

Answer 1

Let the angles be $a-d$, $a$, $a+d$ (as in A.P.)

$(a-d)+a+(a+d)={180}^o$

$\Rightarrow a={60}^o$

$\frac{a-d}{a+d}=\frac{{60}^o}{\pi rad}=\frac{60}{180}=\frac{1}{3}$

$\Rightarrow 3a-3d=a+d$

$\Rightarrow a=2d$

$\Rightarrow d={30}^o$

$\therefore$ Angles are $30,60,90$.