Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If the median $AD$ of a triangle $ABC$ makes an angle $\alpha$ with $AB$,

then $sin(A-\alpha )$ is equal to?

• A. $\frac{bsin\alpha }{c}$
• B. $\frac{csin\alpha }{b}$
• C. $\frac{b}{csin\alpha }$
• D. $\frac{c}{bsin\alpha }$

Answer 1

The correct option is B $\frac{csin\alpha }{b}$

Since, $\angle D={90}^o,CD=DB=\frac{a}{2},AD\left(common\right)$

According to SAS congruency Rule, ${△}_{CAD}$ is congruent to ${△}_{BAD}$

Now, $CD=DB\Rightarrow bsin(A-\alpha )=csin\alpha \Rightarrow sin(A-\alpha )=\frac{csin\alpha }{b}$

Therefore, Correct Answer is $B$