The angles of a triangle are in AP. The great | vedclass

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(B) Let the angles of the triangle be $a-d, a, a+d$ in $AP$.The sum of the angles of a triangle is $180^{\circ}$.$(a-d) + a + (a+d) = 180^{\circ}$$3a

Vedclass · Accepted Answer

$80, 60, 40$

Vedclass · Answer

(B) Let the angles of the triangle be $a-d, a, a+d$ in $AP$.The sum of the angles of a triangle is $180^{\circ}$.$(a-d) + a + (a+d) = 180^{\circ}$$3a = 180^{\circ} \Rightarrow a = 60^{\circ}$.The angles are $60^{\circ}-d, 60^{\circ}, 60^{\circ}+d$.Given that the greatest angle is twice the least angle:$60^{\circ}+d = 2(60^{\circ}-d)$$60^{\circ}+d = 120^{\circ}-2d$$3d = 60^{\circ} \Rightarrow d = 20^{\circ}$.Substituting $d = 20^{\circ}$ into the angles:Least angle: $60^{\circ}-20^{\circ} = 40^{\circ}$.Middle angle: $60^{\circ}$.Greatest angle: $60^{\circ}+20^{\circ} = 80^{\circ}$.Thus,the angles are $80^{\circ}, 60^{\circ}, 40^{\circ}$.

The angles of a triangle are in $AP$. The greatest angle is twice the least. Find all the angles of the triangle (in $^\circ$).

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