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MHTCET 25 Apr 2025 Shift 1 Paper

Question : 140 of 150

Marks: +1, -0

\frac{2 \cos A}{a}+\frac{\cos B}{b}+\frac{2 \cos C}{c}=\frac{a}{\mathrm{bc}}+\frac{b}{\mathrm{ca}}

\angle A=

Solution:

Concept:

Use the cosine rule to express each term of the given equation in terms of sides, then simplify.

Explanation:

\triangle ABC

, the cosine rule gives:

\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}

\cos B = \dfrac{a^2 + c^2 - b^2}{2ac}

\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}

Substitute these into the LHS:

\dfrac{2\cos A}{a} + \dfrac{\cos B}{b} + \dfrac{2\cos C}{c}

= \dfrac{b^2+c^2-a^2}{abc} + \dfrac{a^2+c^2-b^2}{2abc} + \dfrac{a^2+b^2-c^2}{abc}

= \dfrac{a^2 + 3b^2 + c^2}{2abc}

The RHS is:

\dfrac{a}{bc} + \dfrac{b}{ca} = \dfrac{a^2 + b^2}{abc} = \dfrac{2a^2 + 2b^2}{2abc}

Equating LHS and RHS:

\dfrac{a^2 + 3b^2 + c^2}{2abc} = \dfrac{2a^2 + 2b^2}{2abc}

\Rightarrow a^2 + 3b^2 + c^2 = 2a^2 + 2b^2

\Rightarrow a^2 = b^2 + c^2

Hence,

\triangle ABC

is right-angled at

A

Answer:

\angle A = \dfrac{\pi}{2}

Correct option: C.

\dfrac{\pi}{2}

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