# let there be a traingle of side a,b,c and angles A,B,C then:(acosA+bcosB+ccosC)/(a+b+c)=?the answer must only be in the form of ex-radii and in-radii of the traingle

Sourabh Singh IIT Patna
6 years ago
Use properties of Triangles n Just substitute the values in order to get the answer

c2 = a2 + b2 – 2ab cos(C)
R = a/2 sin A and
r = 4R sin (A/2)sin (B/2)sin (C/2)
jagdish singh singh
173 Points
6 years ago
$\hspace{-0.70 cm} Using a=2R\sin A\;,b=2R\sin B\;,c=2R \sin C\\\\ So \frac{a\cos A+b\cos B+c\cos C}{a+b+c} = R\left(\frac{\sin 2A+\sin 2B+\sin 2C}{a+b+c}\right)\\\\ Using \sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C\\\\So We get =4R\left(\frac{\sin A\sin B\sin C}{a+b+c}\right) = \frac{abc}{2R^2(2s)}=\frac{4R\triangle}{4R^2s}\\\\ So we get = \frac{\triangle}{Rs} = \frac{r}{R}. Here we use r=\frac{\triangle}{s}.$