The angles of a triangle are in A.P. and the | vedclass

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Question

(A) Let the angles of the triangle be $(A-d)$, $A$, and $(A+d)$.Since the sum of angles in a triangle is $180^\circ$, we have $(A-d) + A + (A+d) = 180

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(A) Let the angles of the triangle be $(A-d)$, $A$, and $(A+d)$.Since the sum of angles in a triangle is $180^\circ$, we have $(A-d) + A + (A+d) = 180^\circ$, which gives $3A = 180^\circ$, so $A = 60^\circ$.The angles are $(60^\circ-d)$, $60^\circ$, and $(60^\circ+d)$.The greatest angle is $(60^\circ+d)$ and the least angle is $(60^\circ-d)$.Given that the greatest angle is double the least angle: $60^\circ+d = 2(60^\circ-d)$.$60^\circ+d = 120^\circ - 2d \implies 3d = 60^\circ \implies d = 20^\circ$.The angles are $60^\circ-20^\circ = 40^\circ$, $60^\circ$, and $60^\circ+20^\circ = 80^\circ$.The third angle is $60^\circ$.Therefore, $\sin(60^\circ) = \frac{\sqrt{3}}{2}$.

The angles of a triangle are in $A.P.$ and the greatest angle is double the least angle. Find the sine of the third angle.

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