Question

The angles of a triangle are in A.P. If the smallest angle is ${50}^o$, find the other two angles.

Answer 1

It is given that the smallest angle is ${50}^0$.

Let the difference be $d$. Then other angle are ${50}^0+d$ and ${50}^0+2d$.

Since, in a triangle, the sum of three interior angles is ${180}^0$, therefore,

${50}^0+({50}^0+d)+({50}^0+2d)={180}^0\Rightarrow {150}^0+3d={180}^0\Rightarrow 3d={180}^0-{150}^0\Rightarrow 3d={30}^0\Rightarrow d=\frac{{30}^0}{3}={10}^0$

Now, the other angles are:

${50}^0+d={50}^0+{10}^0={60}^0$ and

${50}^0+2d={50}^0+(2\times {10}^0)={50}^0+{20}^0={70}^0$

Hence, the other two angles are ${60}^0$ and ${70}^0$.