Question

If the median $AD$ of a triangle $ABC$ makes an angle $\theta$ with side $AB$, then $\sin (A-\theta )$ is equal to:

• A. $\frac{b}{c}\sin \theta$
• B. $\frac{c}{b}\sin \theta$
• C. $\frac{c}{b}\cos \theta$
• D. None of these

Answer 1

The correct option is B $\frac{c}{b}\sin \theta$


In $\Delta ABD,\frac{BD}{\sin \theta }=\frac{AD}{\sin B}$
$\Rightarrow AD=BD\cdot \frac{\sin B}{\sin \theta }\cdots \left(i\right)$
In $\Delta ACD,\frac{CD}{\sin (A-\theta )}=\frac{AD}{\sin C}$
$\Rightarrow AD=CD\cdot \frac{\sin C}{\sin (A-\theta )}\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$ we have: $BD\cdot \frac{\sin B}{\sin \theta }=CD\cdot \frac{\sin C}{\sin (A-\theta )}$
$\Rightarrow \frac{\sin B}{\sin C}=\frac{\sin \theta }{\sin (A-\theta )}$
$\Rightarrow \frac{b}{c}=\frac{\sin \theta }{\sin (A-\theta )}$
$\Rightarrow \sin (A-\theta )=\frac{c}{b}\sin \theta$