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# If A, B and C are interior angles of a $$∆ \ ABC$$, then show that $$sin( \frac{B+C}{2}) \ = \ cos \frac{A}{2}$$

Answer :

Given, A, B and C are the interior angles of a $$∆ \ ABC$$

We know that, sum of interior angles of a triangle is 180°

$$\because A + B + C \ = \ 180°$$

=> $$\frac{A+B+C}{2} \ = \ 90°$$

=> $$\ \frac{B+C}{2} \ = \ 90° - \frac{A}{2}$$

=> $$sin( \frac{B+C}{2} ) \ = \ sin( 90° - \frac{A}{2} )$$

=> $$sin( \frac{B+C}{2} ) \ = \ cos \frac{A}{2}$$    [ ∵ $$sin(90° - \theta) \ = \ cos\theta$$ ]