Trigonometric Equations Question 351
Question: If $ x,,y,,z $ are perpendicular drawn $ a,,b $ and $ c $ , then the value of $ \frac{bx}{c}+\frac{cy}{a}+\frac{az}{b} $ will be
[UPSEAT 1999]
Options:
A) $ \frac{a^{2}+b^{2}+c^{2}}{2R} $
B) $ \frac{a^{2}+b^{2}+c^{2}}{R} $
C) $ \frac{a^{2}+b^{2}+c^{2}}{4R} $
D) $ \frac{2(a^{2}+b^{2}+c^{2})}{R} $
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Answer:
Correct Answer: A
Solution:
- Let area of triangle be $ \Delta $ , then according to question, $ \Delta =\frac{1}{2}ax=\frac{1}{2}by=\frac{1}{2}cz $
$ \therefore $ $ ,\frac{bx}{c}+\frac{cy}{a}+\frac{az}{b}=\frac{b}{c}( \frac{2\Delta }{a} )+\frac{c}{a}( \frac{2\Delta }{b} )+\frac{a}{b}( \frac{2\Delta }{c} ) $ $ =\frac{2\Delta (b^{2}+c^{2}+a^{2})}{abc}=\frac{2(a^{2}+b^{2}+c^{2})}{abc}\text{. }\frac{abc}{4R}=\frac{a^{2}+b^{2}+c^{2}}{2R} $ .
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