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`        in a triangle ABC prove that: (acosA+bcosB+ccosC)=r/R(a+b+c)`
3 years ago Nishant Vora
IIT Patna
2467 Points
```							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/4ΔThanks
```
3 years ago
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