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Find the answer fastly and get the correct answer  immediately 

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Grade:12th pass

2 Answers

Arun
25750 Points
5 years ago
 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) Hence a cos A + b cosB + c cosC / 2s = r/R
Deepak Kumar Shringi
askIITians Faculty 4404 Points
5 years ago

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