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in a triangle ABC prove that: (acosA+bcosB+ccosC)=r/R(a+b+c)

in a triangle ABC prove that: (acosA+bcosB+ccosC)=r/R(a+b+c)

Grade:12

1 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
7 years ago
Dear Student,

Here is the proof:-
= a cos A + b cos B + c cos C, ... where ... a/sin A = ... = 2R

= R [ 2 sin A cos A + 2 sin B cos B + 2 sin C cos C ]

= R [ ( sin 2A + sin 2B ) + sin 2C ]

= R [ 2 sin (A+B)· cos(A-B) + sin 2C ]

= R [ 2 sin C. cos(A-B) + 2 sin C cos C ]

= R sin C [ cos(A-B) + cos C ] ... here .. cos C = cos [ π - (A+B) ] = - cos (A+B)

= R sin C [ cos(A-B) - cos(A+B) ]

= R sin C [ 2 sin A sin B ]

= 2 ( 2R sin A ) sin B sin C

= 2 a sin B sin C

=2a [ 2Δ / ca ] [ 2Δ / ab ] ...... from (2)

= 8 Δ² / ( abc )
= r/R(a+b+c)


as r =Δ/s and R = abc/4Δ

Thanks

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