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

Answers : (1)

Arun
15912 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 
 
 
Regards
Arun (askIITians forum expert)
11 months ago
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