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UPSC NDA II 2015 Mathematics Paper

Question Numbers: 111-112

For the next two (02) items that follow:

Consider a triangle ABC satisfying

2{\rm{a}}{\sin ^2}\left( {\frac{{\rm{C}}}{2}} \right) + 2{\rm{c}}{\sin ^2}\left( {\frac{{\rm{A}}}{2}} \right) = 2{\rm{a}} + 2{\rm{c}} - 3{\rm{b}}

Question : 112 of 120

Marks: +1, -0

sin A, sin B, sin C are in

Solution:

Concept:

Using the sine rule and the given condition that sides

a, b, c

are in arithmetic progression (AP).

Explanation:

From the previous question, it is known that

a, b, c

are in AP.

So,

a + c = 2b

Apply the sine rule:

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k

(constant).

Thus,

a = k \sin A

b = k \sin B

c = k \sin C

Substitute into

a + c = 2b

k \sin A + k \sin C = 2k \sin B

Divide both sides by

k

\sin A + \sin C = 2 \sin B

This condition means

\sin A

\sin B

\sin C

are in AP.

Answer:

Option B: A.P.

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