Lf the median AD of a triangle ABC makes an angle - with AB- then SinA- is equal to

The correct option is B csinα/b sin #160; αa/2=sin #160; acsin(A α )a/2=sin(π a)b sin #160; αsin(A α )=b c sin(A α )=c #160; sin α ...

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Standard XII

Mathematics

In Radius

lf the median...

Question

lf the median $A D$ of a triangle $A B C$ makes an angle $α$ with $A B$ , then $S i n (A - α)$ is equal to

\frac{b sin α}{c}

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\frac{c sin α}{b}

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\frac{b}{c sin α}

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\frac{c}{b sin α}

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Solution

The correct option is B $\frac{c sin α}{b}$
$\frac{\frac{s i n α}{a / 2} = \frac{s i n a}{c}}{\frac{s i n (A - α)}{a / 2} = \frac{s i n (π - a)}{b}}$
$\frac{s i n α}{s i n (A - α)} = \frac{b}{c}$
$s i n (A - α) = \frac{c s i n α}{b}$ .

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lf the median AD of a triangle ABC makes an angle α with AB, then Sin(A−α) is equal to

The correct option is B csinαbsinαa/2=sinacsin(A−α)a/2=sin(π−a)bsinαsin(A−α)=bcsin(A−α)=csinαb.

lf the median $A D$ of a triangle $A B C$ makes an angle $α$ with $A B$ , then $S i n (A - α)$ is equal to

The correct option is B $\frac{c sin α}{b}$
$\frac{\frac{s i n α}{a / 2} = \frac{s i n a}{c}}{\frac{s i n (A - α)}{a / 2} = \frac{s i n (π - a)}{b}}$
$\frac{s i n α}{s i n (A - α)} = \frac{b}{c}$
$s i n (A - α) = \frac{c s i n α}{b}$ .