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KCET 2026 Mathematics Solved Paper

Question : 7 of 60

Marks: +1, -0

Solution:

Concept:

If three angles are in A.P., they can be written as

a-d

a

a+d

. Their sum is

180^\circ

Explanation:

Let the angles be

a-d

a

a+d

Since the sum of angles of a triangle is

180^\circ

(a-d)+a+(a+d)=180^\circ

3a=180^\circ \implies a=60^\circ

The greatest angle is

a+d

and the least is

a-d

Given greatest

=2 \times

least,

a+d=2(a-d)

Substitute

a=60^\circ

60^\circ+d=2(60^\circ-d)

60^\circ+d=120^\circ-2d \implies 3d=60^\circ \implies d=20^\circ

Thus the angles are

40^\circ

60^\circ

80^\circ

The third angle (the one left after taking greatest and least) is

60^\circ

\sin 60^\circ=\frac{\sqrt{3}}{2}

Answer:

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

— Option A.

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