In triangle $A B C$ , the length of the altitude drawn from $A$ to $B C$ is

\frac{a sin B sin C}{sin A}

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\frac{b sin C sin A}{sin B}

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\frac{c sin A sin B}{sin C}

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\frac{a cos B sin C}{sin A}

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Solution

The correct option is A $\frac{a sin B sin C}{sin A}$
$a = C c o s B + b c o s C$
$s i n B = \frac{l}{C}$
$C = C s i n B$
$\frac{a}{s i n A} = \frac{C}{s i n C}$
$C = \frac{a s i n C}{s i n A}$
$C = \frac{a s i n B s i n C}{s i n A}$ .

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a (\sin B - \sin C) + (\sin C - \sin A) + c (\sin A - \sin B) = 0

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