A) 1/r
B) r/R
C) R/r
D) 1/R
Correct Answer: B
Solution :
\[\frac{a\cos A+b\cos B+c\cos C}{a+b+c}\] = \[\frac{k\,[\sin A\cos A+\sin B\cos B+\sin C\cos C]}{k\,(\sin A+\sin B+\sin C]}\] = \[\frac{1}{2}\,\frac{(\sin 2A+\sin 2B+\sin 2C)}{(\sin A+\sin B+\sin C)}\] = \[\frac{1}{2}\,\left[ \frac{2\sin (A+B)\,\cos (A-B)+2\sin C\cos C\,}{2\sin \left( \frac{A+B}{2} \right)\,\cos \left( \frac{A-B}{2} \right)+2\sin \frac{C}{2}\cos \frac{C}{2}} \right]\] = \[\frac{1}{2}\left[ \frac{\sin C\{\cos \,(A-B)-\cos \,(A+B\})}{\cos \frac{C}{2}\left\{ \cos \left( \frac{A-B}{2} \right)+\cos \left( \frac{A+B}{2} \right) \right\}} \right]\] = \[\frac{1}{2}\,\left[ \frac{\sin C(2\sin A\sin B)}{\cos \frac{C}{2}\left( 2\cos \frac{A}{2}\cos \frac{B}{2} \right)} \right]\] = \[\frac{1}{2}\left[ \frac{2\sin \frac{A}{2}\cos \frac{A}{2}\,.\,2\sin \frac{B}{2}\cos \frac{B}{2}.2\sin \frac{C}{2}\cos \frac{C}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}} \right]\] =\[4\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\]=\[\frac{r}{R}\],\[\left[ \because \,\,r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \right]\].
You need to login to perform this action.
You will be redirected in
3 sec
