A pilot in an aeroplane at an altitude of 200 metre observes two points lying on either side of a river .If the angles of depression of the two points be `45^(@)and60^(@)`, then the width of the river is

Options

A. `(200+(200)/(sqrt(3)))`metre
B. `(200-(200)/(sqrt(3)))metre`
C. `400sqrt(3)` metre
D. `((400)/(sqrt(3)))metre`

Answer

To solve the problem, we need to determine the width of the river based on the angles of depression observed by the pilot at an altitude of 200 meters.

### Step-by-Step Solution:

1. **Understanding the Angles of Depression**:
- The angles of depression from the pilot to points P and Q on either side of the river are given as 45° and 60°, respectively.

2. **Drawing the Diagram**:
- Let's denote the altitude of the airplane as point O, which is 200 meters above the ground.
- Let point P be the point on one side of the river and point Q be the point on the other side.
- The angles of depression from point O to points P and Q create two right triangles: triangle OAP (for angle 45°) and triangle OBQ (for angle 60°).

3. **Using Trigonometry for Triangle OAP**:
- In triangle OAP, where the angle of depression is 45°, we know:
\[ \tan(45°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{200}{d_1} \]
where \(d_1\) is the horizontal distance from point O to point P.
- Since \(\tan(45°) = 1\):
\[ 1 = \frac{200}{d_1} \implies d_1 = 200 \text{ meters} \]

4. **Using Trigonometry for Triangle OBQ**:
- In triangle OBQ, where the angle of depression is 60°, we have:
\[ \tan(60°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{200}{d_2} \]
where \(d_2\) is the horizontal distance from point O to point Q.
- Since \(\tan(60°) = \sqrt{3}\):
\[ \sqrt{3} = \frac{200}{d_2} \implies d_2 = \frac{200}{\sqrt{3}} \text{ meters} \]

5. **Calculating the Width of the River**:
- The total width of the river (W) is the sum of distances \(d_1\) and \(d_2\):
\[ W = d_1 + d_2 = 200 + \frac{200}{\sqrt{3}} \]

6. **Simplifying the Expression**:
- To express the width in a single fraction:
\[ W = 200 + \frac{200}{\sqrt{3}} = 200 \left(1 + \frac{1}{\sqrt{3}}\right) \]
- To combine the terms:
\[ W = 200 \left(\frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}}\right) = 200 \left(\frac{\sqrt{3} + 1}{\sqrt{3}}\right) \]

### Final Answer:
The width of the river is:
\[ W = \frac{200(\sqrt{3}+1)}{\sqrt{3}} \text{ meters} \]