(a+b) sinc/2=c cos(a-b)/2
❎❎No spam❎❎
Answers
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.
2.9
32 votes
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
please mark brainliest
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