In a triangle abc prove that for any angle theta b cos(A-theta) + a cos(B - theta) = c costheta
Answers
Answered by
0
PLZ MARK ME AS BRAINIST
Step-by-step explanation:
bcos(A−θ)+acos(B+θ)
=b(cosA.cosθ+sinA.sinθ)+a(cosB.cosθ−sinB.sinθ)
=cosθ(bcosA+acosB)+sinθ(bsinA−asinB)
=cosθ(c)+sinθ(b 2Ra−a 2Rb) [SinceacosB+bcosA=c and using sine law of triangle]
=ccosθ.
Similar questions