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EAMCET Engineering 2001 Solved Paper

Section: Mathematics

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Question : 198 of 200

Marks: +1, -0

In a

\triangle A B C, \; \frac{a}{b^2-c^2}+\;\frac{c}{b^2-a^2}=0

B

$\; \frac{\pi}{2}$
$\; \frac{\pi}{4}$
$\; \frac{2\pi}{3}$
$\; \frac{\pi}{3}$

Solution:

We have,

\;\; \frac{a}{b^2-c^2}+\; \frac{c}{b^2-a^2}=0

\;\Rightarrow \;\; \;\frac{2R \sin A}{4R^2(\sin^2 B - \sin^2 C)}

\;\; \;+\; \frac{2R \sin C}{4R^2(\sin^2 B - \sin^2 A)}=0

\; \Rightarrow \;\; \; \frac{1}{2R} \cdot \frac{\sin A}{\sin (B+C) \sin (B-C)}

\;+\; \frac{1}{2R} \cdot \frac{\sin C}{\sin (B+A) \sin (B-A)} = 0

\; \Rightarrow \;\; \; \frac{1}{2R} \cdot \frac{\sin A}{\sin A \cdot \sin (B-C)}

\;+\; \frac{1}{2R} \cdot \frac{\sin C}{\sin C \sin (B-A)} = 0

\; \Rightarrow \;\; \; \frac{1}{2R \sin (B-C)} + \; \frac{1}{2R \sin (B-A)} = 0

\; \Rightarrow \;\; \sin(B-C) + \sin(B-A) = 0

\; \Rightarrow \;\; 2 \sin \frac{2B-A-C}{2} \cos \frac{-C+A}{2} = 0

\; \Rightarrow \;\; \sin \frac{2B-A-C}{2} = 0

\; \Rightarrow \;\; 2B = A+C

\; \Rightarrow \;\; 2B = 180^{\circ} - B

\; \Rightarrow \;\; 3B = 180^{\circ}

\; \Rightarrow \;\; B = \frac{\pi}{3}

\;

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