Question

If A, B and C are interior angles of a triangles ABC, then show

that Sin B + C/2 = cos (A/2) ​

Answer 1

To prove :

sin ( A + B / 2 ) = cos C / 2

Solution :

A + B + C = 180° ( sum of angles of a ∆le = 180° )

A + B = 180° - C

divide both side by 2

A + B / 2 = 180° - C / 2

A + B / 2 = 180° / 2 - C / 2

A + B / 2 = 90° - C / 2

==> tan ( A + B / 2 ) = tan ( 90° - C / 2 )

tan ( A + B / 2 ) = cot ( C / 2 )

( bcoz ,, tan ( 90 - A) = cot A )

Hope it's helpful ..... Please mark it as Brainliest ...... ❤

Answer 2

Sin[(B+C)/2]

Since A+B+C=180 for interior angles of triangle ABC.

then B+C=180-A.

NOW   Sin [(180-A)/2]

              =Sin[90-(A/2)]                      since Sin(90-A)=CosA

               =Cos(A/2)

hope it will help u......✌