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MHTCET 20 Apr 2026 Shift 2 Paper

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Question : 112 of 150

Marks: +1, -0

The angles of

\triangle A B C

are in A.P. and

b: c=\sqrt{3}: \sqrt{2}

/ A=

30°
90°
105°
75°

Solution:

Concept:

Angles in arithmetic progression give one angle as

60^\circ

, and the sine rule connects side ratios to angle sines.

Explanation:

Let the angles be

A

,

B

,

C

in A.P.

So,

A + C = 2B

.

Using

A + B + C = 180^\circ

, we get

2B + B = 180^\circ

.

Thus,

3B = 180^\circ

, so

B = 60^\circ

.

By the sine rule,

\frac{b}{c} = \frac{\sin B}{\sin C}

.

Given

\frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}}

, we have

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

.

Since

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

,

\frac{\frac{\sqrt{3}}{2}}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}}

.

Cancelling

\sqrt{3}

, we get

\sin C = \frac{\sqrt{2}}{2}

.

Therefore,

C = 45^\circ

.

Now,

A = 180^\circ - B - C = 180^\circ - 60^\circ - 45^\circ

.

So,

A = 75^\circ

.

Answer:

A = 75^\circ

, which is Option D.

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