An aeroplane is vertically above the another plane flying at a height of 5000 feet from the ground. The angle of elevation of these two planes from a points on the ground are respectively `pi/3 and pi/4`. What is the vertical distance between these two planes ?

Options

A. `2500(sqrt(3)+1)` feet
B. `5000sqrt(3)` feet
C. `5000(sqrt(3)-1)` feet
D. `5000(sqrt(3)+1)` feet

Answer

To find the vertical distance between the two planes, we can follow these steps:

### Step 1: Understand the problem
We have two planes: Plane A (the higher plane) and Plane B (the lower plane). Plane B is flying at a height of 5000 feet from the ground, and we need to find the vertical distance between Plane A and Plane B. The angles of elevation from a point on the ground to Plane A and Plane B are given as \( \frac{\pi}{3} \) and \( \frac{\pi}{4} \) respectively.

### Step 2: Convert angles to degrees
Convert the angles from radians to degrees for easier calculations:
- \( \frac{\pi}{3} \) radians = 60 degrees
- \( \frac{\pi}{4} \) radians = 45 degrees

### Step 3: Set up the triangles
Let:
- \( H \) = height of Plane A above Plane B
- \( CD \) = horizontal distance from the point on the ground to the point directly below Plane A and Plane B

We can use the tangent function to relate the angles of elevation to the heights and distances.

### Step 4: Apply trigonometry to Plane A
For Plane A (angle of elevation = 60 degrees):
\[ \tan(60^\circ) = \frac{AC}{CD} \]
Where \( AC = H + 5000 \) (the height of Plane A) and \( CD \) is the horizontal distance.

Since \( \tan(60^\circ) = \sqrt{3} \):
\[ \sqrt{3} = \frac{H + 5000}{CD} \quad \text{(1)} \]

### Step 5: Apply trigonometry to Plane B
For Plane B (angle of elevation = 45 degrees):
\[ \tan(45^\circ) = \frac{BC}{CD} \]
Where \( BC = 5000 \) (the height of Plane B).

Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{5000}{CD} \quad \text{(2)} \]

### Step 6: Solve for CD from equation (2)
From equation (2):
\[ CD = 5000 \]

### Step 7: Substitute CD into equation (1)
Now substitute \( CD = 5000 \) into equation (1):
\[ \sqrt{3} = \frac{H + 5000}{5000} \]

### Step 8: Solve for H
Multiply both sides by 5000:
\[ 5000\sqrt{3} = H + 5000 \]
Rearranging gives:
\[ H = 5000\sqrt{3} - 5000 \]

### Step 9: Final expression for the vertical distance
Thus, the vertical distance between the two planes is:
\[ H = 5000(\sqrt{3} - 1) \text{ feet} \]

### Conclusion
The vertical distance between the two planes is \( 5000(\sqrt{3} - 1) \) feet.

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