From the top of a 120 m hight tower a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as `60^(@)` and `45^(@)`. Find the distance between the cars.

Answer

To solve the problem step by step, we will use the properties of right triangles and trigonometric ratios.

### Step 1: Understand the scenario
We have a tower of height 120 m. The man at the top of the tower observes two cars on opposite sides of the tower with angles of depression of 60° and 45°.

### Step 2: Draw a diagram
Draw a vertical line representing the tower (AM) of height 120 m. Mark the top of the tower as point A, the bottom as point M, and the positions of the two cars as points B and C on either side of the tower.

### Step 3: Identify the angles of depression
From point A (top of the tower), the angle of depression to car B is 45°, and to car C is 60°.

### Step 4: Apply trigonometric ratios
1. **For car B (angle of depression = 45°):**
- In triangle ABM, we have:
\[ \tan(45°) = \frac{AM}{MB} \]
- Here, \(AM = 120\) m and \(\tan(45°) = 1\).
- Therefore:
\[ MB = \frac{AM}{\tan(45°)} = \frac{120}{1} = 120 \text{ m} \]

2. **For car C (angle of depression = 60°):**
- In triangle ACM, we have:
\[ \tan(60°) = \frac{AM}{MC} \]
- Here, \(AM = 120\) m and \(\tan(60°) = \sqrt{3}\).
- Therefore:
\[ MC = \frac{AM}{\tan(60°)} = \frac{120}{\sqrt{3}} = 40\sqrt{3} \text{ m} \]

### Step 5: Calculate the distance between the cars
The distance between the two cars (BC) can be calculated as:
\[ BC = BM + MC \]
Substituting the values we found:
\[ BC = 120 + 40\sqrt{3} \text{ m} \]

### Step 6: Final expression
To express the distance in a simplified form, we can factor out 40:
\[ BC = 40(3 + \sqrt{3}) \text{ m} \]

### Conclusion
The distance between the two cars is:
\[ BC = 120 + 40\sqrt{3} \text{ m} \quad \text{or} \quad BC = 40(3 + \sqrt{3}) \text{ m} \]