Question

Point $D,E$ are taken on the side $BC$ of a triangle $\vartriangle ABC$ such that $BD=DE=EC$. If $\angle BAD=x,\angle DAE=y,\angle EAC=z$, then the value of $\dfrac{\sin \left( x+y \right)\sin \left( y+z \right)}{\sin x\sin z}=$
A. $1$
B. $2$
C. $4$
D. None of these.

Answer 1

Hint: We will first draw the diagram of the triangle $\vartriangle ABC$ using given data. Then we will apply Law of sines in each of the triangles $\vartriangle ABD$, $\vartriangle ADC$, $\vartriangle ABE$ and $\vartriangle AEC$ and form equations for each of the triangles.
We will then use each of the equations and perform some arithmetic operations to form this equation $\dfrac{\sin \left( x+y \right)\sin \left( y+z \right)}{\sin x\sin z}$ and then find its value.

Formula Used:

Complete step by step solution: We are given a triangle $\vartriangle ABC$ in which point $D,E$ are taken on the side of $BC$ in such a way that $BD=DE=EC$. We are given the value of angles $\angle BAD=x,\angle DAE=y,\angle EAC=z,$ and we need to find the value of $\dfrac{\sin \left( x+y \right)\sin \left( y+z \right)}{\sin x\sin z}$.
First we will draw the diagram of $\vartriangle ABC$,

Now from the diagram using law of sines we will determine,
In triangle $\vartriangle ABD$,
$\dfrac{a}{3\sin x}=\dfrac{AD}{\sin B}$……. (i)
In triangle $\vartriangle ABE$,
$\dfrac{2a}{3\sin (x+y)}=\dfrac{AE}{\sin B}$…………(ii)
In triangle $\vartriangle ADC$,
$\dfrac{2a}{3\sin (y+z)}=\dfrac{AD}{\sin C}$………………. (iii)
In triangle $\vartriangle AEC$,
$\dfrac{2a}{3\sin z}=\dfrac{AE}{\sin C}$………………. (iv)

We will divide the equation (i) by equation (ii).
$\left( \dfrac{\dfrac{a}{3\sin x}}{\dfrac{2a}{3\sin (x+y)}} \right)=\left( \dfrac{\dfrac{AD}{\sin B}}{\dfrac{AE}{\sin B}} \right)$
The resultant equation will be,
$\dfrac{\sin (x+y)}{2\sin x}=\dfrac{AD}{AE}$……………(v)
We will now divide equation (iv) by equation (iii).
$\left( \dfrac{\dfrac{2a}{3\sin z}}{\dfrac{2a}{3\sin (y+z)}} \right)=\left( \dfrac{\dfrac{AE}{\sin C}}{\dfrac{AD}{\sin C}} \right)$
The resultant equation will be,
$\dfrac{\sin (y+z)}{2\sin z}=\dfrac{AE}{AD}$………………(vi)
Now we will multiply the equations (v) and equation (vi).
$\begin{align}
  & \dfrac{\sin (x+y)}{2\sin x}\times \dfrac{\sin (y+z)}{2\sin z}=\dfrac{AE}{AD}\times \dfrac{AD}{AE} \\
 & \dfrac{\sin (x+y)\sin (y+z)}{4\sin x\sin z}=1 \\
 & \dfrac{\sin (x+y)\sin (y+z)}{\sin x\sin z}=4
\end{align}$

The triangle $\vartriangle ABC$ on which when point $D,E$ are taken on the side $BC$ of a triangle $\vartriangle ABC$ such that $BD=DE=EC$ and angles $\angle BAD=x,\angle DAE=y,\angle EAC=z,$ then the value of $\dfrac{\sin \left( x+y \right)\sin \left( y+z \right)}{\sin x\sin z}$ is $\dfrac{\sin \left( x+y \right)\sin \left( y+z \right)}{\sin x\sin z}=4$. Hence the correct option is (C).

Note: In solving this question we should keep in mind which operations must be performed in order to form this equation $\dfrac{\sin \left( x+y \right)\sin \left( y+z \right)}{\sin x\sin z}$.