Question

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

• A. $\frac{b}{c}\text{cosec}\theta$
• B. $\frac{b}{c}\sin \theta$
• C. $\frac{c}{b}\sin \theta$
• D. $\frac{c}{b}\text{cosec}\theta$

Answer 1

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

Let $\angle ADC=\alpha$

Using Sine law in $\Delta ADC$ we get;

$\frac{\frac{a}{2}}{\sin (A-\theta )}=\frac{b}{\sin \alpha }$ (1)

Using Sine law in $\Delta ABD$ we get;

$\frac{\frac{a}{2}}{\sin \theta }=\frac{c}{\sin (\pi -\alpha )}⟹\frac{\frac{a}{2}}{\sin \theta }=\frac{c}{\sin \alpha }$

$⟹\sin \alpha =\frac{c}{\frac{a}{2}}\sin \theta$

Putting the value of $\sin \alpha$ in equation 1; we get

$\frac{\left(\frac{a}{2}\right)}{\sin (A-\theta )}=\frac{b\left(\frac{a}{2}\right)}{c\sin \theta }$

$⟹\sin (A-\theta )=\frac{c}{b}\sin \theta$

Hence; $\sin (A-\theta )=\frac{c}{b}\sin \theta$