From the peak of a hill which is 300 m high ,the angle of depression of two sides of a bridge lying on a ground are `45^(@)and30^(@)`(both ends of the bridge are on the same side of the hill) .Then the length of the bridge is

Options

A. `300(sqrt(3)-1)m`
B. `300(sqrt(3)+1)`m
C. `300sqrt(3)`m
D. `(300)/(sqrt(3))`m

Answer

To solve the problem, we will use trigonometric principles involving angles of depression and the properties of right triangles.

### Step-by-Step Solution:

1. **Understand the Problem**:
We have a hill that is 300 m high. From the top of the hill, we observe two points on a bridge below, with angles of depression of 45° and 30°.

2. **Draw a Diagram**:
Visualize the situation. Let:
- Point A be the top of the hill (300 m high).
- Point B be the point on the ground directly below A.
- Point C be one end of the bridge where the angle of depression is 45°.
- Point D be the other end of the bridge where the angle of depression is 30°.

3. **Identify the Right Triangles**:
- Triangle ABC (for angle of depression 45°)
- Triangle ABD (for angle of depression 30°)

4. **Using the Angle of Depression**:
- The angle of depression from A to C is 45°. Therefore, the angle of elevation from C to A is also 45°.
- The angle of depression from A to D is 30°. Therefore, the angle of elevation from D to A is also 30°.

5. **Calculate Distances**:
- For triangle ABC (with angle 45°):
\[ \tan(45°) = \frac{AB}{BC} \]
Since \(\tan(45°) = 1\):
\[ 1 = \frac{300}{BC} \implies BC = 300 \text{ m} \]

- For triangle ABD (with angle 30°):
\[ \tan(30°) = \frac{AB}{BD} \]
Since \(\tan(30°) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{300}{BD} \implies BD = 300\sqrt{3} \text{ m} \]

6. **Find the Length of the Bridge**:
The length of the bridge \(CD\) is the difference between \(BD\) and \(BC\):
\[ CD = BD - BC = 300\sqrt{3} - 300 \]
Factoring out 300:
\[ CD = 300(\sqrt{3} - 1) \text{ m} \]

### Final Answer:
The length of the bridge is \(300(\sqrt{3} - 1)\) meters.

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