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Question
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
Options
\[\sin \frac{A}{2}\]
\[\cos \frac{A}{2}\]
\[- \sin \frac{A}{2}\]
\[- \cos \frac{A}{2}\]
Solution
We know that in triangle `ABC`
`A+B+C=180°`
⇒ `B+C=180°-A`
⇒` (B+C)/2=(90°)/2-A/2`
⇒ `sin ((B+C)/2)=sin (90°-A/2)`
`"since" sin (90°-A)=cos A`
So
`sin ((B+C)/2)= cos A`
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