A tower subtends angle `theta, 2theta, 3theta` at the three points A,B and C respectively lying on a horizontal line through the foot of the tower. Then AB/BC is equal to

Options

A. `3-4sin^(2)theta`
B. `(sin 3theta)/(sin 2theta)`
C. `(sin 2theta)/(sin theta)`
D. `(sin 3theta)/(cos theta)`

Answer

To solve the problem, we need to find the ratio \( \frac{AB}{BC} \) given that a tower subtends angles \( \theta \), \( 2\theta \), and \( 3\theta \) at points A, B, and C respectively.

### Step-by-Step Solution:

1. **Understanding the Angles**:
- Let the tower be represented by point E and the foot of the tower be point D.
- The angles subtended at points A, B, and C are \( \angle AEB = \theta \), \( \angle BEC = 2\theta \), and \( \angle CED = 3\theta \).

2. **Using Exterior Angle Property**:
- In triangle AEB, the exterior angle \( \angle AEB = 2\theta \) can be expressed as:
\[ 2\theta = \theta + \angle AEB \]
Hence, \( \angle AEB = \theta \).

- Similarly, in triangle BEC:
\[ 3\theta = 2\theta + \angle BEC \]
Thus, \( \angle BEC = \theta \).

3. **Identifying Angles**:
- Now we have:
- \( \angle AEB = \theta \)
- \( \angle BEC = \theta \)

4. **Using Angle Bisector Theorem**:
- Since \( \angle AEB \) and \( \angle BEC \) are equal, line segment BE acts as an angle bisector of \( \angle AEC \).
- According to the angle bisector theorem:
\[ \frac{AE}{EC} = \frac{AB}{BC} \]

5. **Finding Lengths AE and EC**:
- Let the height of the tower be \( h \).
- In triangle AED:
\[ AE = \frac{h}{\sin \theta} \]
- In triangle CED:
\[ EC = \frac{h}{\sin 3\theta} \]

6. **Substituting Values**:
- Now substitute \( AE \) and \( EC \) into the angle bisector theorem:
\[ \frac{AE}{EC} = \frac{AB}{BC} \implies \frac{\frac{h}{\sin \theta}}{\frac{h}{\sin 3\theta}} = \frac{AB}{BC} \]
- The height \( h \) cancels out:
\[ \frac{\sin 3\theta}{\sin \theta} = \frac{AB}{BC} \]

7. **Using Sine Addition Formula**:
- We can express \( \sin 3\theta \) using the sine addition formula:
\[ \sin 3\theta = 3\sin \theta - 4\sin^3 \theta \]
- Thus, we can rewrite:
\[ \frac{AB}{BC} = \frac{3\sin \theta - 4\sin^3 \theta}{\sin \theta} = 3 - 4\sin^2 \theta \]

8. **Final Result**:
- Therefore, we have:
\[ \frac{AB}{BC} = 3 - 4\sin^2 \theta \]