ABCD is a trapezium such that AB and CD are parallel and `BC bot CD`. If `angleADB = theta, BC = p and CD = q`, then AB is equal to

Options

A. `((p^(2) + q^(2) sin theta)/(p cos theta + q sin theta)`
B. `((p^(2) + q^(2)) cos theta)/(p cos theta + q sin thet)`
C. `(p^(2) + q^(2))/(p^(2) cos theta + q^(2) sin theta)`
D. `((p^(2) + q^(2)) sin theta)/((p cos theta + q sin theta))`

Explanation

To solve the problem, we need to find the length of side AB in trapezium ABCD, given that AB is parallel to CD, BC is perpendicular to CD, and we have the angles and lengths defined.

### Step-by-Step Solution:

1. **Understanding the Trapezium**:
- We have trapezium ABCD where AB || CD.
- BC is perpendicular to CD, which means angle BCD = 90°.

2. **Identifying Angles**:
- Given angle ADB = θ.
- Let angle CDB = α. Since AB || CD, angle CDB = angle ADB = α.

3. **Using Right Triangle Properties**:
- In triangle BCD, we can apply the Pythagorean theorem:
\[ DB = \sqrt{BC^2 + CD^2} = \sqrt{p^2 + q^2} \]
- Here, BC = p and CD = q.

4. **Finding Cosine and Sine of Angle α**:
- From triangle CDB:
- Cosine of angle α:
\[ \cos \alpha = \frac{CD}{DB} = \frac{q}{\sqrt{p^2 + q^2}} \]
- Sine of angle α:
\[ \sin \alpha = \frac{BC}{DB} = \frac{p}{\sqrt{p^2 + q^2}} \]

5. **Applying the Sine Rule in Triangle ADB**:
- According to the sine rule:
\[ \frac{AB}{\sin \theta} = \frac{DB}{\sin(\pi - \theta - \alpha)} \]
- Since \(\sin(\pi - x) = \sin x\), we have:
\[ \frac{AB}{\sin \theta} = \frac{\sqrt{p^2 + q^2}}{\sin(\theta + \alpha)} \]

6. **Expanding \(\sin(\theta + \alpha)\)**:
- Using the sine addition formula:
\[ \sin(\theta + \alpha) = \sin \theta \cos \alpha + \cos \theta \sin \alpha \]
- Substitute the values of \(\cos \alpha\) and \(\sin \alpha\):
\[ \sin(\theta + \alpha) = \sin \theta \left(\frac{q}{\sqrt{p^2 + q^2}}\right) + \cos \theta \left(\frac{p}{\sqrt{p^2 + q^2}}\right) \]

7. **Substituting Back**:
- Now substituting this back into the sine rule equation:
\[ AB = \frac{\sqrt{p^2 + q^2} \cdot \sin \theta}{\sin \theta \cdot \frac{q}{\sqrt{p^2 + q^2}} + \cos \theta \cdot \frac{p}{\sqrt{p^2 + q^2}}} \]
- Simplifying gives:
\[ AB = \frac{(p^2 + q^2) \cdot \sin \theta}{q \cdot \sin \theta + p \cdot \cos \theta} \]

### Final Answer:
Thus, the length of AB is given by:
\[ AB = \frac{p^2 + q^2 \cdot \sin \theta}{p \cdot \cos \theta + q \cdot \sin \theta} \]