b cos(C+0)+ccos(B-) =
Answers
Answered by
0
Answer:
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
2R
a
−a
2R
b
) [Since acosB+bcosA=c and using sine law of triangle]
=ccosθ.
Explanation:
I think it will help you
Answered by
0
Answer: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
2R
a
−a
2R
b
) [Since acosB+bcosA=c and using sine law of triangle]
=ccosθ.
Explanation:
iI HOPE ITS HELPFULL
plz mark me as brilliant
Similar questions