Question

please help!

A+B+C=180°, cos B + cos C ??

a. 2 cos (A/2) cos {(B-C)/2}
b. 2 sin (A/2) cos {(B-C)/2}
c. 2 cos (C/2) cos {(B-C)/2}
d. 2 cos (B/2) cos (C/2)
e. 2 sin (B/2) sin (C/2)
(choose one answer that is correct and please give the explanation)

Answer 1

its option B

Answer 2

Here its explanation :-
\[\huge \bf Given: A+B+C=180^o\]
so
\[\huge \bf B+C=180-A\]
\[\huge \bf \frac{B+C}{2}=\frac{180}{2}-\frac{A}{2}\]

Remember this formula :-
\[\huge \bf Cos~B+Cos~C=2\cos \frac{B+C}{2}\cos \frac{B-C}{2}\]

here,
\[\huge \bf \frac{B+C}{2}=90^o-\frac{A}{2}\]
therefore,we get
\[\huge \bf 2\cos (90-\frac{A}{2})\cos \frac{B-C}{2}\]
we know that \(\large \bf \cos(90- \theta)=\sin \theta\)

here in this case ,we get
\[\huge \bf 2\sin \frac{A}{2}\cos \frac{B-C}{2}\]

Answer 3

therefore,option B is correct

Answer 4

*remember this formula:-
\[\large \bf \cos~B+\cos~C=2\cos \frac{B+C}{2}\cos \frac{B-C}{2}\]

Answer 5

@gabriellalyn understood?

Answer 6

yea i got it. thank you!! @mayankdevnani

Answer 7

welcome :)