Question

(a+b) sinc/2=c cos(a-b)/2


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Answer 1

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

Ask for details Follow Report by Ritamalhotra9554 06.03.2018

Answers

suraj397

suraj397 Ambitious

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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Comments Report

ujjwalksingh304 Helping Hand

Answer:

Step-by-step explanation:

Using sine formula ,

a/sinA =b/sinB =c/sinC =k(say)

,a =ksinA, b =ksinB, c =ksinC

LHS

(a+b)/c = (ksinA +ksinB)/ksinC

=k(2.sin(A+B)/2).cos(A-B)/2)

/(2.sinC/2.cosC/2)

= (sin(π/2-C/2).cos(A-B/2))

/(sinC/2.cosC/2)

=(cosC/2.cos(A-B/2))

/(sinC/2.cosC/2

=cos((A-B)/2)/sin(C/2)=RHS

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Answer 2

Step-by-step explanation:

Using sine formula ,

a/sinA =b/sinB =c/sinC =x(say)

,a =xsinA, b =xsinB, c =xsinC

LHS

(a+b)/c = (xsinA +xsinB)/xsinC

=x(2.sin(A+B)/2).cos(A-B)/2)

/(2.sinC/2.cosC/2)

= (sin(π/2-C/2).cos(A-B/2))

/(sinC/2.cosC/2)

=(cosC/2.cos(A-B/2))

/(sinC/2.cosC/2

=cos((A-B)/2)/sin(C/2)=RHS