The angles of a triangle are in AP, and the greatest angle is double the least.

Question:

The angles of a triangle are in AP, and the greatest angle is double the least. Find all the angles in degrees and radians. 

Solution:

 Let a - d, a, a + d be the three angles of the triangle that form AP.

Given that the greatest angle is double the least. Now, $a+d=2(a-d) 2 a-2 d=a+d a$ $=3 d \ldots . .(1)$ Now by angle sum property, $(a-d)+a+(a+d)=180^{\circ} 3 a=180^{\circ} a=60^{\circ}$

$\ldots \ldots$ (2) From (1) and (2), 3d $=60^{\circ} \mathrm{d}=20^{\circ}$ Now, the angles are, $\mathrm{a}-\mathrm{d}=60^{\circ}-20^{\circ}=40^{\circ} \mathrm{a}=$ $60^{\circ} \mathrm{a}+\mathrm{d}=60^{\circ}+20^{\circ}=80^{\circ}$

Therefore the required angles are 40° 60° 80°

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