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

1 Answer

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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