Question

For any angle x, show that acosx=bcos(C+X)+c cos(B-x) [Properties of triangle]

Answer 1

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.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

SINE LAW OF TRIANGLES

$\displaystyle \: \frac{a}{ \sin A } = \frac{b}{ \sin B} = \frac{c}{ \sin C} = {2R}$

2.

$b \cos C + c \: \cos B = a$

TO PROVE

For any angle x

$a \cos x = b \cos \: (C + x) + c \cos (B - x)$

PROOF

RHS

$= b \cos \: (C + x) + c \cos (B - x)$

$= b (\cos C \cos x -\sin C \sin x ) + c \: ( \cos B \cos x + \sin B \sin x)$

$= \cos x(b \cos C + c \: \cos B )- \sin x ( b \sin C - c \: \sin B)$

$\displaystyle \: = \cos x \times a- \sin x ( b \times \frac{c}{2R} - c \: \times \frac{b}{2R} )$

$\displaystyle \: = a\cos x - \sin x \times ( \frac{bc}{2R} - \frac{bc}{2R} )$

$\displaystyle \: = a\cos x - \sin x \times 0$

$\displaystyle \: = a\cos x$

= LHS

Hence proved