From the top of a cliff 90 m high, the angles of depression of the top and bottom of a tower are observed to be `30^(@)` and `60^(@)` respectively. The height of the tower is :

Options

A. 45 m
B. 60 m
C. 75 m
D. 30 m

Answer

To solve the problem, we will break it down step by step.

### Step 1: Understand the Problem
We have a cliff that is 90 m high. From the top of this cliff, the angles of depression to the top and bottom of a tower are given as 30° and 60°, respectively. We need to find the height of the tower.

### Step 2: Set Up the Diagram
1. Let the height of the cliff be \( AB = 90 \) m.
2. Let the bottom of the tower be point \( C \) and the top of the tower be point \( D \).
3. Let the horizontal distance from the base of the cliff to the tower be \( x \).
4. The height of the tower \( CD = h \).

### Step 3: Analyze the Angles of Depression
- The angle of depression to the bottom of the tower \( C \) is 60°.
- The angle of depression to the top of the tower \( D \) is 30°.

### Step 4: Use Triangle Properties
1. For angle \( C \) (60°):
- In triangle \( ABC \), where \( AB = 90 \) m and \( BC = x \):
\[ \tan(60°) = \frac{AB}{BC} = \frac{90}{x} \]
- Since \( \tan(60°) = \sqrt{3} \):
\[ \sqrt{3} = \frac{90}{x} \implies x = \frac{90}{\sqrt{3}} = 30\sqrt{3} \text{ m} \]

2. For angle \( D \) (30°):
- In triangle \( AED \), where \( AE = 90 - h \):
\[ \tan(30°) = \frac{AE}{ED} = \frac{90 - h}{x} \]
- Since \( \tan(30°) = \frac{1}{\sqrt{3}} \):
\[ \frac{1}{\sqrt{3}} = \frac{90 - h}{30\sqrt{3}} \implies 30\sqrt{3} = \sqrt{3}(90 - h) \]
- Simplifying gives:
\[ 30 = 90 - h \implies h = 90 - 30 = 60 \text{ m} \]

### Conclusion
The height of the tower \( h \) is **60 meters**.

---