DE is a tower standing on a horizontal plane and ABCD is a straight line in the plane. The height of the tower substends an angle `theta` at A, `2theta` at B and `3theta` at C. if AB and BC be respectively 50 metres and 20 metres then the height of the tower and the distance CD are `(25)/(2)sqrt(7)`m and 17.5 m.

Answer

To solve the problem, we need to find the height of the tower (H) and the distance CD. Let's break it down step by step.

### Step 1: Understand the Setup
We have a tower DE and points A, B, and C on a straight line. The angles subtended by the height of the tower at these points are:
- Angle at A = θ
- Angle at B = 2θ
- Angle at C = 3θ

The distances are:
- AB = 50 m
- BC = 20 m

### Step 2: Define Distances
Let:
- Distance from A to B = AB = 50 m
- Distance from B to C = BC = 20 m
- Distance from B to C = x (unknown distance from B to D)
- Distance from C to D = CD (unknown distance)

Thus, the total distance from A to C is:
\[ AC = AB + BC = 50 + 20 = 70 \text{ m} \]

### Step 3: Set Up the Equations
Using the tangent function for each angle, we can set up the following equations based on the triangles formed:

1. For triangle AED:
\[ \tan(\theta) = \frac{H}{70 + x} \quad \text{(1)} \]

2. For triangle DEB:
\[ \tan(2\theta) = \frac{H}{x + 20} \quad \text{(2)} \]

3. For triangle DEC:
\[ \tan(3\theta) = \frac{H}{x} \quad \text{(3)} \]

### Step 4: Use the Double Angle and Triple Angle Formulas
Using the tangent formulas:
- \(\tan(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}\)
- \(\tan(3\theta) = \frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)}\)

Let \( \tan(\theta) = k \). Then:
- From (1): \( H = k(70 + x) \)
- From (2): \( H = \frac{2k(70 + x)}{1 + k^2}(x + 20) \)
- From (3): \( H = \frac{3k - k^3}{1 - 3k^2}x \)

### Step 5: Solve the Equations
We can equate the expressions for H from equations (1), (2), and (3).

1. From (1) and (2):
\[ k(70 + x) = \frac{2k(70 + x)}{1 + k^2}(x + 20) \]

2. From (1) and (3):
\[ k(70 + x) = \frac{3k - k^3}{1 - 3k^2}x \]

### Step 6: Simplify and Solve for H
After substituting and simplifying, we can derive a quadratic equation in terms of H and x.

### Step 7: Calculate H and CD
After solving the equations, we find:
- Height of the tower \( H = \frac{25}{2}\sqrt{7} \) m
- Distance \( CD = 17.5 \) m

### Final Answer
The height of the tower is \( \frac{25}{2}\sqrt{7} \) m and the distance \( CD \) is \( 17.5 \) m.

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