Question

Question~

$\bold {If A, B \: and \: C \: are \: interior \: angles \: of \: a \: triangle \: ABC \: then \: show \: that \: \sin (\frac{B +C }{2} ) = \cos \: \frac{A}{2} }$
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Answer 1

Step-by-step explanation:

Since A+B+C=180 for interior angles of triangle ABC. then B+C=180-A.

Answer 2

♡Answer♡:-

Given △ABC

We know that sum of three angles of a triangle is 180

Hence ∠A+∠B+∠C=180°

or A+B+C=180°

B+C=180° −A

Multiply both sides by 1/2

1/2 (B+C)= 1/2 (180° −A)

1)2 (B+C)=90° − A/2 ...(1)

Now 1/2 (B+C)

Taking sine of this angle

sin( B+C/2) [B+C/2 =90° − A/2 ]

sin(90° − A/2 )

cos A/2 [sin(90°−θ)=cosθ]

Hence sin( B+C/2 )=cos A/2

proved

Itzgirl45 ❤️