Q.In triangle ABC , what is the value of sin⁡\(\cfrac{(B+C)}{2} \) ? (a) sin⁡\(\cfrac{A}{2}\) (b) sinA (c) cosA (d) cos⁡ \(\cfrac{A}{2}\)
Answer: D
A+B+C=180°
or, (B+C)=180°-A
or, \(\cfrac{B+C}{2}=\cfrac{180°-A}{2}\)
or, \(\cfrac{B+C}{2}\)=90°-\(\cfrac{A}{2}\)
∴sin\(( \cfrac{B+C}{2})\)=sin⁡\(⁡(90°-\cfrac{A}{2})\)=cos⁡\(\cfrac{A}{2}\)
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