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If in a triangle, ABC

p \cos ^{2}(\frac{C}{2})+r \cos ^{2}(\frac{A}{2})=\frac{3q}{2}, then the sides p, q and r

  • Option 1)

    are in AP

  • Option 2)

    are in GP

  • Option 3)

    are in HP

  • Option 4)

    satisfy p+q=r

 

Answers (1)

best_answer

As we discussed in concept

Double Angle Formula -

Double angle formula

- wherein

These are formulae for double angles.

 

 Pcos^{2}(\frac{C}{2})+rcos^{2}(\frac{A}{2})=\frac{3q}{2}

=> P[\frac{1+cosC}{2}]+r[\frac{1+cosA}{2}]=\frac{3q}{2}

\therefore (P+r)+(PcosC+rcosA)=3q

\therefore P+r+q=3q                                                                          [\because Since\:\:acosC+CcosA=b]

\therefore P+r=2q                                                                                 \therefore\:P,q,r\:in\:AP


Option 1)

are in AP

This option is correct.

Option 2)

are in GP

This option is incorrect.

Option 3)

are in HP

This option is incorrect.

Option 4)

satisfy p+q=r

This option is incorrect.

Posted by

prateek

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