Question

The angle 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 $60:\pi$.Find the angle of the triangle in degrees ?

Answer 1

Let, the angles of the triangle are,

${(a-d)}^o,a^o$ and ${(a+d)}^o$.

Then, $a-d+a+a+d={180}^o\Rightarrow a={60}^o$

So, the angles are ${(60-d)}^o,{60}^o,{(60+d)}^o$.

Here, ${(60-d)}^o$ is the least angle and ${(60+d)}^o$ is the greatest angle.

Now, greater angle$={(60+d)}^o={\left\{\right(60+d\left)\frac{\pi }{180}\right\}}^c$

Also, $\frac{numberofdegreesintheleastangle}{numberofradiansinthegreatestangle}=\frac{60}{\pi }\Rightarrow \frac{60-d}{{\left\{\right(60+d\left)\frac{\pi }{180}\right\}}^c}=\frac{60}{\pi }\Rightarrow \frac{180(60-d)}{\pi (60+d)}=\frac{60}{\pi }\Rightarrow 4d=120\Rightarrow d=30$

Hence, the angles are ${(60-30)}^o,{60}^o,{(60+30)}^o$ i.e., ${30}^o,{60}^0,{90}^o.$