Paragraph: Read the following information and answer: $A B C D$ is a trapezium such that $A B$ and $C D$ are parallel and $B C$ is perpendicular to them. Let $\angle A D B=\theta, \angle A B D=\alpha, B C=p$ and $C D=q$ and $C D=q$.
Question: Consider the following : 1. $A D \sin \theta=A B \sin \alpha$ 2. $B D \sin \theta=A B \sin (\theta+\alpha)$ Which of the above is/are correct?

• A

  1 only
• B

  2 only
• Correct answer
  C

  Both 1 and 2
• D

  Neither 1 nor 2

Solution

(1) Applying Sine ryke in $\triangle A B D$ \begin{aligned} & \frac{A D}{\sin \alpha}=\frac{A B}{\sin \theta} \\ & \Rightarrow A D \cdot \sin \theta=A B \cdot \sin \alpha \end{aligned} Hence, (1) is correct. (2) \begin{aligned} & \frac{A B}{\sin \theta}=\frac{B D}{\sin \left(180^{\circ}-(\alpha+\theta)\right)} \\ & \frac{A B}{\sin \theta}=\frac{B D}{\sin (\alpha+\theta)} \\ & A B \cdot \sin (\theta+\alpha)=B D \cdot \sin \theta \end{aligned} Hence, (2) is correct.

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