$A$ and $B$ are two points on one bank of a straight river and $C, D$ are two other points on the other bank of river, if direction from $A$ to $B$ is same as that from $C$ to $D$ and $A B=a, \angle C A D=\alpha, \angle D A B=\beta, \angle C B A=\gamma$, then $C D$ is equal to

• A

  $\frac{a \sin \beta \cdot \sin \gamma}{\sin \alpha \cdot \sin (\alpha+\beta+\gamma)}$
• Correct answer
  B

  $\frac{a \sin \alpha \cdot \sin \gamma}{\sin \beta \cdot \sin (\alpha+\beta+\gamma)}$
• C

  $\frac{a \sin \alpha \cdot \sin \beta}{\sin \gamma \cdot \sin (\alpha+\beta+\gamma)}$
• D

  None of the above

Solution

Clearly, $\angle C D A=\beta, \angle D C B=\gamma, \angle A C B_0=\pi-(\alpha+\beta+\gamma)$ Using since rule in $\triangle C A B$, we get
$\frac{A B}{\sin (\pi-(\alpha+\beta+\gamma))}=\frac{C A}{\sin \gamma}$ $\Rightarrow C A=\frac{a \cdot \sin \gamma}{\sin (\alpha+\beta+\gamma)}$ Now, using sine rule in $\triangle C A D$, we get $\frac{C A}{\sin \beta}=\frac{C D}{\sin \alpha}$ $\Rightarrow C D=\frac{C A \cdot \sin \alpha}{\sin \beta}=\frac{a \sin \alpha \cdot \sin \gamma}{\sin \beta \cdot \sin (\alpha+\beta+\gamma)}$

Hence, the correct answer is option: 'B'.