In $△$ ABC,the median AD divides $∠$ BAC such that $∠$ BAD: $∠$ CAD=2:1. Then $cos (A / 3)$ 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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none of these

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Solution

The correct option is A $\frac{sin B}{2 sin C}$
Consider the below given triangle.
Applying sine rule to triangle $B A D$ we get
$\frac{a}{2 s i n \frac{2 A}{3}} = \frac{A D}{s i n B}$ .
Or
$\frac{a . s i n B}{2 s i n \frac{2 A}{3}} = A D$ .
And
Applying sine rule to triangle $C A D$
$\frac{a . s i n C}{2 s i n \frac{A}{3}} = A D$
Hence
$\frac{a . s i n C}{2 s i n \frac{A}{3}} = \frac{a . s i n B}{2 s i n \frac{2 A}{3}}$
Or
$s i n C . s i n \frac{2 A}{3} = s i n B . s i n \frac{A}{3}$ .
Or
$s i n C [2 s i n \frac{A}{3} . c o s \frac{A}{3}] = s i n B . s i n \frac{A}{3}$
Or
$2 s i n C . c o s \frac{A}{3} = s i n B$
Or
$c o s \frac{A}{3} = \frac{s i n B}{2 s i n C}$ .

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Similar questions

In $Δ A B C,$ the median AD divides $∠ B A C$ such that $∠ B A D : ∠ C A D = 2 : 1.$ then $c o s \frac{A}{3}$ is equal to

△ A B C,

∠ B A C

∠ B A C : ∠ C A D = 2 : 1.

cos \frac{A}{3}

\frac{sin 2 A + sin 2 B + sin 2 C}{sin A + sin B + sin C} = 8 sin (\frac{A}{2}) sin (\frac{B}{2}) sin (\frac{C}{2})

A D

A B C

∠ B A C

1 : 2,

\frac{sin B}{sin C}

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