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If the bisector an angle A of the triangle ABC makes an angle $θ$ with BC , then sin $θ$ =

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$\cos \frac{B - C}{2}$

$\cos \frac{B + C}{2}$

$\cos (B + \frac{A}{2})$

$\cos (C + \frac{A}{2})$

answer is A.

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Detailed Solution

$A + B + C = π \Rightarrow \frac{A}{2} = \frac{π}{2} - \frac{B}{2} - \frac{C}{2} In Δ^{le} ADC$

$\Rightarrow θ + C + \frac{A}{2} = π \Rightarrow θ + C + \frac{π}{2} - \frac{B}{2} - \frac{C}{2} = π \Rightarrow θ = \frac{π}{2} + (\frac{B - C}{2}) \Rightarrow \sin θ = \cos (\frac{B - C}{2})$

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