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Grade: 11
`        proove that:acosA + bcosB + ccosC = 8(area of triangle)2/(abc)`
one year ago

Answers : (1)

Arun
22540 Points
```							Dear Rahul Let ... Δ = area of ΔABC. Then ... bc sin A = ca sin B = ab sin C = 2Δ ... (1) Also ... bc sin A = 2Δ . . . . . sin A = 2Δ / bc. Similarly, ..... sin B = 2Δ / ca ... and ... sin C = 2Δ / ab. ... (2) ......................................... ... LHS . ▬▬▬ = a cos A + b cos B + c cos C, ... where ... a/sin A = ... = k = (k/2) [ 2 sin A cos A + 2 sin B cos B + 2 sin C cos C ] = (k/2) [ ( sin 2A + sin 2B ) + sin 2C ] = (k/2) [ 2 sin (A+B)· cos(A-B) + sin 2C ] = (k/2) [ 2 sin C. cos(A-B) + 2 sin C cos C ] = k sin C [ cos(A-B) + cos C ] ... here .. cos C = cos [ π - (A+B) ] = - cos (A+B) = k sin C [ cos(A-B) - cos(A+B) ] = k sin C [ 2 sin A sin B ] = 2 ( k sin A ) sin B sin C = 2 a sin B sin C = Middle Side ▬▬▬▬▬▬▬ = 2a [ 2Δ / ca ] [ 2Δ / ab ] ...... from (2) = 8 Δ² / ( abc ) = RHS   RegardsArun (askIITians forum expert)
```
one year ago
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