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m - n Theorem

In Δ ABC, the...

Question

In $Δ A B C$ , the median $A D$ divides $∠ B A C$ such that $∠ B A D : ∠ C A D = 2 : 1.$ Then $cos (\frac{A}{3})$ is equal to

\frac{sin B}{2 sin C}

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\frac{sin C}{2 sin B}

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\frac{2 sin B}{sin C}

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\frac{sin B}{sin C}

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Solution

The correct option is A $\frac{sin B}{2 sin C}$
Let, $∠ C D A = α$ .
Now by $m - n$ theorem
$(\frac{a}{2} + \frac{a}{2}) cot α = \frac{a}{2} \cdot cot \frac{2 A}{3} - \frac{a}{2} \cdot cot \frac{A}{3}$
$\Rightarrow 2 cot (B + \frac{2 A}{3}) = cot \frac{2 A}{3} - cot \frac{A}{3}$
$\Rightarrow cot (B + \frac{2 A}{3}) + cot \frac{A}{3} = cot \frac{2 A}{3} - cot (B + \frac{2 A}{3})$
$\Rightarrow \frac{cos (B + \frac{2 A}{3}) \times sin \frac{A}{3} + sin (B + \frac{2 A}{3}) \times cos \frac{A}{3}}{sin (B + \frac{2 A}{3}) \times sin \frac{A}{3}} = \frac{sin (B + \frac{2 A}{3}) \times cos \frac{2 A}{3} - cos (B + \frac{2 A}{3}) \times sin \frac{2 A}{3}}{sin (B + \frac{2 A}{3}) \times sin \frac{2 A}{3}}$
$\Rightarrow \frac{sin (B + \frac{3 A}{3})}{sin \frac{A}{3}} = \frac{sin (B)}{sin \frac{2 A}{3}}$
$\Rightarrow sin (B + A) = \frac{sin (B)}{2 sin \frac{A}{3} cos \frac{A}{3}} \times sin \frac{A}{3}$
$\Rightarrow sin (π - C) = \frac{sin (B)}{2 cos \frac{A}{3}}$
$\Rightarrow cos (\frac{A}{3}) = \frac{sin B}{2 sin C}$

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