A tower is observed from two stations A and B where B is East of A at a distance 100 metres. The tower is due North of A and due North-West of B. the angles of elevation of the tower from A and B are complementary, the height of the tower is_____.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry
- We have two points A and B. Point B is 100 meters east of point A.
- The tower is located north of point A and northwest of point B.
- The angles of elevation from A and B to the top of the tower are complementary.

### Step 2: Define the Angles
- Let the angle of elevation from A to the tower be θ.
- Then, the angle of elevation from B to the tower will be 90° - θ (since they are complementary).

### Step 3: Set Up the Triangles
- From point A, we can form a right triangle where:
- Height of the tower = PQ = H
- Distance from A to the base of the tower = AP
- We can use the tangent function:
\[ \tan(θ) = \frac{H}{AP} \quad \Rightarrow \quad H = AP \cdot \tan(θ) \quad \text{(1)} \]

- From point B, we can form another right triangle where:
- Distance from B to the base of the tower = BP
- Using the angle of elevation from B:
\[ \tan(90° - θ) = \cot(θ) = \frac{H}{BP} \quad \Rightarrow \quad H = BP \cdot \cot(θ) \quad \text{(2)} \]

### Step 4: Relate AP and BP
- Since B is 100 meters east of A, we can use the Pythagorean theorem to relate AP and BP:
\[ AB^2 = AP^2 + BP^2 \quad \Rightarrow \quad 100^2 = AP^2 + BP^2 \quad \Rightarrow \quad AP^2 + BP^2 = 10000 \quad \text{(3)} \]

### Step 5: Substitute Equations (1) and (2) into (3)
- From equations (1) and (2), we can express H in terms of AP and BP:
\[ AP \cdot \tan(θ) = BP \cdot \cot(θ) \]
- Rearranging gives:
\[ AP \cdot \tan(θ) = BP \cdot \frac{1}{\tan(θ)} \quad \Rightarrow \quad AP \cdot \tan^2(θ) = BP \]

### Step 6: Substitute BP in terms of AP into (3)
- Substitute BP into equation (3):
\[ AP^2 + (AP \cdot \tan^2(θ))^2 = 10000 \]
- Let \( x = AP \):
\[ x^2 + x^2 \tan^4(θ) = 10000 \]
- Factor out \( x^2 \):
\[ x^2(1 + \tan^4(θ)) = 10000 \]
- Thus, we have:
\[ x^2 = \frac{10000}{1 + \tan^4(θ)} \quad \Rightarrow \quad AP = \sqrt{\frac{10000}{1 + \tan^4(θ)}} \]

### Step 7: Calculate H
- Substitute back into equation (1) to find H:
\[ H = AP \cdot \tan(θ) = \sqrt{\frac{10000}{1 + \tan^4(θ)}} \cdot \tan(θ) \]
- Since \( \tan(θ) \) and \( \cot(θ) \) are related, we can simplify this further.

### Step 8: Solve for H
- Since the angles are complementary, we can use the property that \( \tan(θ) \cdot \cot(θ) = 1 \) to find the height:
\[ H^2 = AP \cdot BP \]
- Substitute \( AP = 100 \) and \( BP = 100 \sqrt{2} \):
\[ H^2 = 100 \cdot 100\sqrt{2} = 10000\sqrt{2} \]
- Finally, taking the square root gives:
\[ H = 100 \sqrt[4]{2} \]

### Final Answer
The height of the tower is \( H = 100 \sqrt[4]{2} \) meters.

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