Question

Q.2

Answer 1

Let the angles of a triangle be `(a-d), a and (a+d)`

Now, sum of the angles of a triangle=`180°`

`⇒ (a-d)+a+(a+d)=180°`

`⇒3a=180°`

`⇒a=60°`

Given that,

`(a+d)=2(a-d)`

`⇒60°+d=2(a-d)`

`⇒60°+d=120°-2d`

`⇒3d=60°`

`⇒d=20°`

`⇒a-d=60°-20°=40°,a=60° and a+d=60°+20°=80°`

Thus, the angle of a triangle are `40°,60° and 80°`