Question

a+b/c=cos[A-B/2]\sinC/2

Answer 1

Answer:

Brainly.in

What is your question?

Secondary School Math 13+7 pts

For a triangle ABC, prove that a+b/c=Cos(A-B/2)/SinC/2

Report by Ritamalhotra9554 06.03.2018

Answers

suraj397

Suraj397Ambitious

The sum of all angles in a triangle is 180°

Therefore,

A + B + C = 180°

=> A + B = 180° - C

=> (A + B)/2 = (180° - C)/2 = 90° - C/2

In the above question,

LHS = cos((A + B)/2)

substituting (A + B)/2 with 90°- C/2

= cos(90° - C/2)

cos(90° - θ) = sinθ

Therefore,

= sin(C/2) which is also = RHS

LHS = RHS

Hence proved.

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Answer 2

Answer:

A+B+C=180

A+B=180-C

(A+B)/2=(180-C)/2=90-C/2

cos(90-C/2)

cos(90-Q)=sinQ

sin(c/2)

LHS=RHS

Step-by-step explanation: