A balloon moving in a straight line passes vertically above two points A and B on a horizontal plane 1000 ft. apart, when above A it has an altitude of `60^(@)` as seen from B when above B it has an altitude of `45^(@)` as seen from A. the distance of A from the point at which it will touch the plane is _____

Answer

To solve the problem, we will use trigonometric relationships in right triangles formed by the balloon's position and the points A and B on the ground.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two points A and B on the ground, which are 1000 ft apart. The balloon is directly above point A at an angle of elevation of \(60^\circ\) as seen from B, and directly above point B at an angle of elevation of \(45^\circ\) as seen from A. We need to find the distance from point A to the point where the balloon will touch the ground.

2. **Setting Up the Triangles**:
- Let \(P\) be the position of the balloon when it is directly above point A.
- Let \(Q\) be the position of the balloon when it is directly above point B.
- Let \(R\) be the point where the balloon will touch the ground.

3. **Using Triangle APB**:
In triangle \(APB\):
- The angle of elevation from B to P is \(60^\circ\).
- The distance \(AB = 1000\) ft.
- Using the tangent function:
\[ \tan(60^\circ) = \frac{PA}{AB} \]
\[ \sqrt{3} = \frac{PA}{1000} \]
\[ PA = 1000\sqrt{3} \text{ ft} \]

4. **Using Triangle BQA**:
In triangle \(BQA\):
- The angle of elevation from A to Q is \(45^\circ\).
- Using the tangent function:
\[ \tan(45^\circ) = \frac{QB}{AB} \]
\[ 1 = \frac{QB}{1000} \]
\[ QB = 1000 \text{ ft} \]

5. **Finding the Distance AR**:
Now, we have two triangles \(APR\) and \(BQR\). Since the balloon moves in a straight line, triangles \(APR\) and \(BQR\) are similar.
- Let \(AR = x\) and \(BR = 1000 + x\).
- From triangle \(APR\):
\[ \frac{PA}{AR} = \frac{QB}{BR} \]
Substituting the known values:
\[ \frac{1000\sqrt{3}}{x} = \frac{1000}{1000 + x} \]

6. **Cross-Multiplying**:
\[ 1000\sqrt{3}(1000 + x) = 1000x \]
\[ 1000\sqrt{3} \cdot 1000 + 1000\sqrt{3}x = 1000x \]
Rearranging gives:
\[ 1000\sqrt{3} \cdot 1000 = 1000x - 1000\sqrt{3}x \]
\[ 1000\sqrt{3} \cdot 1000 = 1000x(1 - \sqrt{3}) \]
\[ x = \frac{1000\sqrt{3} \cdot 1000}{1000(1 - \sqrt{3})} \]
Simplifying:
\[ x = \frac{1000\sqrt{3}}{1 - \sqrt{3}} \]

7. **Rationalizing the Denominator**:
Multiply numerator and denominator by \(1 + \sqrt{3}\):
\[ x = \frac{1000\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \]
\[ = \frac{1000\sqrt{3}(1 + \sqrt{3})}{1 - 3} = \frac{1000\sqrt{3}(1 + \sqrt{3})}{-2} \]
\[ = -500\sqrt{3}(1 + \sqrt{3}) \]

8. **Final Calculation**:
The distance \(AR\) is:
\[ AR = 1000 + x = 1000 - 500\sqrt{3}(1 + \sqrt{3}) \]

### Final Answer:
The distance of A from the point at which the balloon will touch the plane is approximately \(500(3 + \sqrt{3})\) ft.