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

Asked by Abhisek | 1 year ago |  46

##### Solution :-

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

To find the value of $$\dfrac{(B+C)}{2}$$, simplify the equation (1)

⇒ B + C = 180° – A

⇒ $$\dfrac{(B+C)}{2}$$ =$$\dfrac{(180°-A)}{2}$$

⇒ $$\dfrac{(B+C)}{2}$$ = (90°-$$\dfrac{A}{2}$$)

Now, multiply both sides by sin functions, we get

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

Since sin (90°-$$\dfrac{A}{2}$$) = $$\dfrac{cosA}{2}$$, the above equation is equal to

sin $$\dfrac{(B+C)}{2}$$$$\dfrac{cosA}{2}$$

Hence proved.

Answered by Pragya Singh | 1 year ago

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