If A , B and C are interior angles of a triangle ABC , then show that sin left(frac{B+C}{2}right)=co

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8. Introduction to Trigonometry

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If $A , B$ and $C$ are interior angles of a triangle $ABC ,$ then show that

$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$

Option A

Option B

Option C

Option D

Solution

We know that for a triangle $ABC$

$\angle A+\angle B+\angle C=180^{\circ}$

$\angle B+\angle C=180^{\circ}-\angle A$

$\frac{\angle B+\angle C}{2}=90^{\circ}-\frac{\angle A}{2}$

$\sin \left(\frac{B+C}{2}\right)=\sin \left(90^{\circ}-\frac{A}{2}\right)$

$=\cos \left(\frac{ A }{2}\right)$

Standard 10

Mathematics

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