If in ABC 2 cos A - a -cos B - b -2 cos C - c - a - bc - b - ca - then - A is equal toA- 90-B- 45-C- 60-D- None of these

The correct option is A 90° We have 2cos A/a+cos B/b+2cos C/c=a/bc+b/ca or 2(b^2+c^2 a^2)/2bca+c^2+a^2 b^2/2bac+2(a^2+b^2 c^2)/2abc=a^2+b^2/abc ⇒ 2b^2+2c^2 2a^ ...

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Byju's Answer

Standard XII

Mathematics

Cosine Rule

If in ABC 2 c...

Question

If in $△ A B C \frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{c a} = t h e n ∠ A$ is equal to

90°

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45°

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60°

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None of these

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Solution

The correct option is A
90°

We have $\frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{c a}$
or $\frac{2 (b^{2} + c^{2} - a^{2})}{2 b c a} + \frac{c^{2} + a^{2} - b^{2}}{2 b a c} + \frac{2 (a^{2} + b^{2} - c^{2})}{2 a b c} = \frac{a^{2} + b^{2}}{a b c}$
$\Rightarrow 2 b^{2} + 2 c^{2} - 2 a^{2} + c^{2} + a^{2} - b^{2} + 2 a^{2} + 2 b - 2 c^{2} = 2 a^{2} + 2 b^{2}$
or $3 b^{2} + c^{2} + a^{2} = 2 a^{2} + 2 b^{2} o r b^{2} + c^{2} = a^{2}$
Hence $∠ A = 90^{o}$

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If in △ABC 2 cos Aa+cos Bb+2cos Cc=abc+bca= then ∠A is equal to

The correct option is A 90° We have 2cos Aa+cos Bb+2cos Cc=abc+bca or 2(b2+c2−a2)2bca+c2+a2−b22bac+2(a2+b2−c2)2abc=a2+b2abc ⇒2b2+2c2−2a2+c2+a2−b2+2a2+2b−2c2=2a2+2b2 or 3b2+c2+a2=2a2+2b2 or b2+c2=a2 Hence ∠A=90o

If in $△ A B C \frac{2 c o s A}{a} + \frac{c o s B}{b} + \frac{2 c o s C}{c} = \frac{a}{b c} + \frac{b}{c a} = t h e n ∠ A$ is equal to