Question

In a triangle ABC, write cos(B+C/2) in terms of angle A.

Plzzzzz help me.....

100 points...

Answer 1

Hey mate, here is your answer -:)

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$Since, A + B + C=180 \\ \: \: \: \: \: \: \: \: Hence, B + C = 180 - A \\ \\ cos( \frac{B + C}{2} ) = cos( \frac{180 - A}{2} ) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = cos(90 - \frac{A}{2} ) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = sin \frac{A}{2}$

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HOPE it helps to you!!

Answer 2

$<b > Answer :-$

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Given :-

ABC is a triangle.

To write :-

cos { ( B + C ) / 2 } in terms of angle A.

Salutation :-

As a triangle , angle A + B + C = 180°

Then , B + C = 180° - A

Now ,

cos { ( B + C ) / 2 }.        [ • Putting the value of ( B + C ) ]

= cos { ( 180° - A ) / 2 }

= cos { 180° / 2 - A / 2 }

= cos ( 90° - A / 2 )

= sin A / 2                [ ★ Required answer ]

[ • As we know , cos ( 90° - ∅ ) = sin∅ ]

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