Question

The angles of a triangle are in A.P. and ratio of the number of degrees in the least to the number of radians in the greatest is $60:\pi$. Then find the smallest angle in degrees.

Answer 1

Let the angles of the triangle be $(a-d{)}^0,a^0$ and $(a+d{)}^0$.
Then $(a-d)+a+(a+d)=180$
$\Rightarrow 3a=180$
$\Rightarrow a=60$
Thus, the angles are $(60-d{)}^0,{60}^0$ and $(60+d{)}^0$
Number of degrees in the least angle $=(60-d)$

and number of radius in the greatest angle $=(60+d)\times \frac{\pi }{180}$
$=\frac{\pi }{180}(60+d)$
According to question,
$\frac{(60-d)}{\frac{\pi }{180}(60+d)}=\frac{60}{\pi }$
$\Rightarrow (60-d)=\frac{1}{3}(60+d)$
$\Rightarrow 180-3d=60+d$
$\therefore d=30$
Hence, the angles of the triangle are $(60-30{)}^0,{60}^0,(60+30{)}^0$ i,e., ${30}^0,{60}^0,{90}^0$.