Question

The angles of a triangle are in A.P. If the greatest angle is twice the least, find the greatest angle.

Answer 1

Let three angles be $(a-d{)}^{\circ },a^{\circ }and(a+d{)}^{\circ }$
Then, $(a-d{)}^{\circ }+a^{\circ }+(a+d{)}^{\circ }={180}^{\circ }$
$\Rightarrow (3a{)}^{\circ }={180}^{\circ }\Rightarrow a={60}^{\circ }$
Given, greatest angle is twice of least angle
$\therefore (a+d{)}^{\circ }=2(a-d{)}^{\circ }$
$\Rightarrow a=3d\Rightarrow {60}^{\circ }=3d\Rightarrow d={20}^{\circ }$
Hence, required angles are $({60}^{\circ }-{20}^{\circ }),{60}^{\circ }$ and $({60}^{\circ }+{20}^{\circ })$, i.e. ${40}^{\circ },{60}^{\circ }$ and ${80}^{\circ }$.
$\therefore$ The greatest angle is 80${}^{\circ }$