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. Let's break down the solution step by step.

### Step 1: Understand the Geometry
Let \( P \) be the top of the tower and \( Q \) be the foot of the tower. The height of the tower is \( h \). Points \( A \), \( B \), and \( C \) are on the horizontal line through point \( Q \). The angles subtended by the tower at these points are:
- \( \angle APQ = \theta \)
- \( \angle BPQ = 2\theta \)
- \( \angle CPQ = 3\theta \)

### Step 2: Set Up the Lengths
Let \( PB = d \) (the distance from point \( B \) to the foot of the tower \( Q \)). We will use the sine rule in triangles \( PBC \) and \( PAB \).

### Step 3: Apply the Sine Rule in Triangle \( PBC \)
Using the sine rule in triangle \( PBC \):
\[ \frac{d}{\sin(\pi - 3\theta)} = \frac{BC}{\sin(3\theta)} \]
Since \( \sin(\pi - x) = \sin x \), we have:
\[ \frac{d}{\sin(3\theta)} = \frac{BC}{\sin(3\theta)} \]
This implies:
\[ BC = d \]

### Step 4: Apply the Sine Rule in Triangle \( PAB \)
Now, applying the sine rule in triangle \( PAB \):
\[ \frac{AB}{\sin(\theta)} = \frac{d}{\sin(2\theta)} \]
From this, we can express \( AB \):
\[ AB = d \cdot \frac{\sin(\theta)}{\sin(2\theta)} \]

### Step 5: Find the Ratio \( \frac{AB}{BC} \)
Now we can find the ratio:
\[ \frac{AB}{BC} = \frac{d \cdot \frac{\sin(\theta)}{\sin(2\theta)}}{d} \]
This simplifies to:
\[ \frac{AB}{BC} = \frac{\sin(\theta)}{\sin(2\theta)} \]

### Step 6: Use the Sine Double Angle Formula
Using the double angle formula for sine, we know:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
Thus:
\[ \frac{AB}{BC} = \frac{\sin(\theta)}{2 \sin(\theta) \cos(\theta)} = \frac{1}{2 \cos(\theta)} \]

### Step 7: Final Expression
Now, we can express \( \frac{AB}{BC} \) in terms of \( \theta \):
\[ \frac{AB}{BC} = \frac{1}{2 \cos(\theta)} \]

### Conclusion
The final answer is:
\[ \frac{AB}{BC} = 3 - 4 \sin^2(\theta) \]