Question

If A, B, C are the interior angles of a ΔABC, show that:
(i)sinB+C/2=cosA/2 (ii)cosB+C/2=sinA/2

Answer 1

Showing that:

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

Given : A, B, C are the interior angles of a ΔABC.

Therefore sum of all angles is equal to 180°

i.e A + B + C = 180°

Now dividing both side by 2,

A+B+C / 2 = 180 / 2

B+C / 2 + A / 2 = 90

B+C / 2 = 90 - A/2

Substitute this value in (i)

sin (B+C)/2 = sin 90 - A/2

{ sin (90-theta) = cos theta }

= cos A/2

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

B+C/2 = 90 - A/2

cos (B+C)/2 = cos 90 - A/2

{ cos (90 - theta) = sin theta }

= sin A/2