Question

Prove that: acosA + bcosB + ccosC=∆/R
where a,b,c are the sides of a triangle and A,B,C are it's angles and ∆ is the area of the triangle ABC and R is the circumradius.

Answer 1

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 Hope it helps All the best.