Question

In a triangle abc prove that for any angle theta b cos(A-theta) + a cos(B - theta) = c costheta

Answer 1

Step-by-step explanation:

Solution verified

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−2rb) [Since acosB+bcosA=c and using sinelaw of triangle]=ccosθ.

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