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} {C}/{2}+c \cos ^{2} {A}/{2}={3 b}/{2}

then the sides of the triangle are in

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

Solution:

{a\cos}^{2} {C}/{2}+c \cos ^{2} {A}/{2}={3 b}/{2}

a \×(\√{{S(S-c)}/{a b}})^{2}+c \×(\√{{S(S-a)}/{b c}})^{2}

={3 b}/{2}

a \×{S(S-c)}/{a b}+c \×{S(S-a)}/{b c}={3 b}/{2}

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

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

S[S-c+S-a]={3 b^{2}}/{2}

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

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

Here

{S}={{a}+{b}+{c}}/{2}

(semi parameter)

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

(a+b+c)=3 b

a+c=2 b

here

{a}, {b}, {c}

are in A.P.

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