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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

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

Grade:12

2 Answers

Sourabh Singh IIT Patna
askIITians Faculty 2104 Points
8 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
8 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 $r=\frac{\triangle}{s}.$

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