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If ABC.be triangle then acosBcosC+bcosCcosA+ccosAcosB=abc/4R^2 prove it

If ABC.be triangle then acosBcosC+bcosCcosA+ccosAcosB=abc/4R^2 prove it 

Grade:12th pass

1 Answers

Aditya Varshney
20 Points
4 years ago
 
=cosAcosBcosC[a/cosA + b/cosB + c/cosC]
 
using Sine rule:
=cosAcosBcosC(2R)[tanA + tanB + tanC]
{R= circumradius }
{tanA + tanB + tanC = tanAtanBtanC}
=cosAcosBcosC(2R)sinAsinBsinC/cosAcosBcosC
=2RsinAsinBsinC
 
again using the sine rule
= 2Rabc/2R*2R*2R 
= abc/4R2 
 
where a,b,c = sides of triangle

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