A balloon moving in a straight line passes vertically above two points A and B on a horizontal plane 10ft apart. When above A the balloon has an angle of elevation of `60^(@)` as seen from B. When above B it has an angle of elevation of `45^(@)` as seen from A. The distance of B from the point C where it will touch the plane is

Options

A. `5(sqrt3+1)" ft"`
B. 15 ft
C. `5(3+sqrt3)ft`
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Draw the Diagram
Draw a horizontal line representing the ground. Mark points A and B, which are 10 ft apart. Draw vertical lines from points A and B to represent the position of the balloon above these points.

### Step 2: Identify Angles of Elevation
From point B, the angle of elevation to the balloon when it is above point A is \(60^\circ\). From point A, the angle of elevation to the balloon when it is above point B is \(45^\circ\).

### Step 3: Set Up the Triangles
Let:
- \(P\) be the position of the balloon when it is above point A.
- \(Q\) be the position of the balloon when it is above point B.
- \(C\) be the point where the balloon will touch the ground.

### Step 4: Use Trigonometry to Find Heights
Using the triangle formed by points A, B, and P:
- From point B to point P, we have:
\[ \tan(60^\circ) = \frac{h_A}{AB} \]
where \(h_A\) is the height of the balloon above point A and \(AB = 10\) ft. Thus,
\[ h_A = AB \cdot \tan(60^\circ) = 10 \cdot \sqrt{3} = 10\sqrt{3} \text{ ft} \]

Using the triangle formed by points A, B, and Q:
- From point A to point Q, we have:
\[ \tan(45^\circ) = \frac{h_B}{BA} \]
where \(h_B\) is the height of the balloon above point B. Thus,
\[ h_B = BA \cdot \tan(45^\circ) = 10 \cdot 1 = 10 \text{ ft} \]

### Step 5: Set Up the Similar Triangles
Using the similarity of triangles:
\[ \frac{h_A}{AC} = \frac{h_B}{BC} \]
Let \(BC = x\) and \(AC = 10 + x\). Substituting the heights:
\[ \frac{10\sqrt{3}}{10 + x} = \frac{10}{x} \]

### Step 6: Cross-Multiply and Solve for x
Cross-multiplying gives:
\[ 10\sqrt{3} \cdot x = 10(10 + x) \]
Expanding:
\[ 10\sqrt{3}x = 100 + 10x \]
Rearranging terms:
\[ 10\sqrt{3}x - 10x = 100 \]
Factoring out \(x\):
\[ x(10\sqrt{3} - 10) = 100 \]
Thus,
\[ x = \frac{100}{10(\sqrt{3} - 1)} = \frac{10}{\sqrt{3} - 1} \]

### Step 7: Rationalize the Denominator
To rationalize:
\[ x = \frac{10(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{10(\sqrt{3} + 1)}{3 - 1} = \frac{10(\sqrt{3} + 1)}{2} = 5(\sqrt{3} + 1) \]

### Final Answer
The distance from point B to point C where the balloon will touch the ground is:
\[ \boxed{5(\sqrt{3} + 1)} \text{ ft} \]