Question

in ∆ABC ,sin (B+C)/2 = ?​

Answer 1

Answer:

Pls mark as brain

Step-by-step explanation:

Hi friend,

We know that the sum of the angles of a triangle is 180°

Therefore,

angle A + Angle B + Angle C = 180°

Angle B + Angle C = 180°-A

Then,

Angle B + Angle C/2 = 90°-A/2

Therefore,

Sin(B+C/2) = Sin(90°-A/2)

=> Sin(B+C/2) = Cos A/2 [ Sin(90°-theta) = cos theta]

Hence,  

LHS = RHS

Sin (B+C/2) = Cos A/2

Answer 2

As ∠A = 90°

so , ∠B + ∠C = 90°

so ,

$\sf \sin( \frac{B + C}{2} ) \degree \\ \\ = \sf \: \sin( \frac{90 \degree}{2} ) = \sin45 \degree \\ \\ \\ \sf = \frac{1}{ \sqrt{2} } \\ \\ \\ \rm so \\ \\ \\ \\ \\ \sf\sin( \frac{B + C}{2} ) = \frac{1}{ \sqrt{2} }$