Trigonometric Equations Question 245
Question: $ \frac{a\cos A+b\cos B+c\cos C}{a+b+c}= $
[Orissa JEE 2004]
Options:
A) 1/r
B) r/R
C) R/r
D) 1/R
Show Answer
Answer:
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},[ \frac{2\sin (A+B),\cos (A-B)+2\sin C\cos C,}{2\sin ( \frac{A+B}{2} ),\cos ( \frac{A-B}{2} )+2\sin \frac{C}{2}\cos \frac{C}{2}} ] $ = $ \frac{1}{2}[ \frac{\sin C{\cos ,(A-B)-\cos ,(A+B})}{\cos \frac{C}{2}{ \cos ( \frac{A-B}{2} )+\cos ( \frac{A+B}{2} ) }} ] $ = $ \frac{1}{2},[ \frac{\sin C(2\sin A\sin B)}{\cos \frac{C}{2}( 2\cos \frac{A}{2}\cos \frac{B}{2} )} ] $ = $ \frac{1}{2}[ \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}} ] $ = $ 4\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2} $ = $ \frac{r}{R} $ , $ [ \because r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} ] $ .
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