Question

Points $D$, and $E$ are taken on the side $BC$ of the $∆ABC$, such that $BD=DE=EC$. If $\angle BAD=x,\angle DAE=y$ and $\angle EAC=z$, then the value of $\frac{\sin (x+y)\sin (y+z)}{\sin x\sin z}$ is

• A. $1$
• B. $2$
• C. $4$
• D. None of these

Answer 1

The correct option is C

$4$

Explanation for the correct option :

Determining the value of $\frac{\sin (x+y)\sin (y+z)}{\sin x\sin z}$.

Applying Sine Rule in $∆\mathrm{ADC}$, we get :

$\frac{\sin (y+z)}{DC}=\frac{SinC}{AD}$ $\left(1\right)$

Again, applying Sine Rule in $∆ABD$, we have :

$\frac{\sin x}{BD}=\frac{SinB}{AD}$ $\left(2\right)$

Also, applying Sine Rule in $∆AEC$, we obtain :

$\frac{\sin z}{EC}=\frac{SinC}{AE}$ $\left(3\right)$

At last, applying the Sine rule in $∆ABE$, we get :

$\frac{\sin (x+y)}{BE}=\frac{SinB}{AE}$ $\left(4\right)$

Multiplying $\left(1\right)$ and $\left(4\right)$ and dividing it by the product of $\left(2\right)$ and $\left(3\right)$, we get

$\begin{array}{rcl}\frac{\sin \left(x+y\right)\sin \left(y+z\right)}{\left(\sin x\right)\left(\sin z\right)}& =& \frac{BE}{AE}\times \frac{DC}{AD}\times \frac{AD}{BD}\times \frac{AE}{EC}\\ & =& \frac{BE}{BD}\times \frac{DC}{EC}\\ & =& 2\times 2\\ & =& 4\end{array}$ $(\because BE=2BD\text{and}DC=2EC)$

Hence, option (C) is the correct answer.