From a tower 125 metres high the angle of depression of two objects ,which are in horizontal line through the base of the tower , are `45^(@)and30^(@)` and they are on the same side of the tower The distance (in metres) between the objects is

Options

A. `125sqrt(3)`
B. `125(sqrt(3)-1)`
C. `125//(sqrt(3)-1)`
D. `125(sqrt(3)+1)`

Answer

To solve the problem, we will use trigonometric ratios based on the angles of depression from the top of the tower to the two objects on the ground.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We have a tower of height 125 meters. The angle of depression to the first object is \(45^\circ\) and to the second object is \(30^\circ\). We need to find the distance between these two objects.

2. **Drawing the Diagram:**
Let's denote:
- The height of the tower as \(h = 125\) meters.
- The distance from the base of the tower to the first object (at \(45^\circ\)) as \(d_1\).
- The distance from the base of the tower to the second object (at \(30^\circ\)) as \(d_2\).

3. **Finding \(d_1\):**
For the first object, using the angle of depression of \(45^\circ\):
\[ \tan(45^\circ) = \frac{h}{d_1} \]
Since \(\tan(45^\circ) = 1\):
\[ 1 = \frac{125}{d_1} \implies d_1 = 125 \text{ meters} \]

4. **Finding \(d_2\):**
For the second object, using the angle of depression of \(30^\circ\):
\[ \tan(30^\circ) = \frac{h}{d_2} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{125}{d_2} \implies d_2 = 125\sqrt{3} \text{ meters} \]

5. **Calculating the Distance Between the Objects:**
The distance between the two objects is:
\[ \text{Distance} = d_2 - d_1 = 125\sqrt{3} - 125 \]
Factoring out \(125\):
\[ \text{Distance} = 125(\sqrt{3} - 1) \text{ meters} \]

### Final Answer:
The distance between the two objects is \(125(\sqrt{3} - 1)\) meters.

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