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NIMCET 2021 Question Paper with solutions

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Question : 10 of 120

Marks: +1, -0

In triangle

A B C (a\cos)^2 \frac{C}{2}+c \cos^2 \frac{A}{2}=\frac{3b}{2}

then the sides of the triangle are in

A.P.
G.P.
H.P.
None

Solution:

(a\cos)^2 \frac{C}{2}+c \cos^2 \frac{A}{2}=\frac{3b}{2}

a \times \left( \sqrt{\frac{S(S-c)}{ab}} \right)^2 + c \times \left( \sqrt{\frac{S(S-a)}{bc}} \right)^2

=\frac{3b}{2}

a \times \frac{S(S-c)}{ab} + c \times \frac{S(S-a)}{bc} = \frac{3b}{2}

\frac{S(S-c)}{b} + \frac{s(S-a)}{b} = \frac{3b}{2}

S(S-c) + S(S-a) = \frac{3b^2}{2}

S\left[ S-c+S-a \right] = \frac{3b^2}{2}

2S(2S - c - a) = 3b^2

(a+b+c)(a+b+c-c-a) = 3b^2

Here

S = \frac{a+b+c}{2}

(semi parameter)

b(a+b+c) = 3b^2

a+b+c = 3b

a+c = 2b

here

a, b, c

are in A.P.

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