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 and when above B `30^(@)` as seen from A. the distance from A of the point at which it will touch the plane is_____

Answer

To solve the problem step by step, we will use trigonometric ratios and properties of similar triangles.

### Step 1: Understand the Problem
We have two points A and B on a horizontal plane that are 1000 feet apart. A balloon passes directly above these points at different angles of elevation as seen from A and B.

### Step 2: Draw the Diagram
Draw a horizontal line representing the plane with points A and B, where AB = 1000 ft. Above point A, the balloon is at an altitude where the angle of elevation from B is 60 degrees. Above point B, the angle of elevation from A is 30 degrees. Let the point where the balloon touches the ground be R.

### Step 3: Set Up the Triangles
1. When the balloon is above point A, let the height of the balloon be PA.
2. When the balloon is above point B, let the height of the balloon be QB.
3. The distance from A to the point where the balloon touches the ground is AR, and the distance from B to the point where the balloon touches the ground is BR.

### Step 4: Use Trigonometric Ratios
From point B, when the balloon is directly above A:
- The angle of elevation is 60 degrees.
- Using the tangent function:
\[ \tan(60^\circ) = \frac{PA}{AB} \]
\[ \sqrt{3} = \frac{PA}{1000} \]
Therefore,
\[ PA = 1000\sqrt{3} \quad \text{(Equation 1)} \]

From point A, when the balloon is directly above B:
- The angle of elevation is 30 degrees.
- Using the tangent function:
\[ \tan(30^\circ) = \frac{QB}{AB} \]
\[ \frac{1}{\sqrt{3}} = \frac{QB}{1000} \]
Therefore,
\[ QB = \frac{1000}{\sqrt{3}} \quad \text{(Equation 2)} \]

### Step 5: Set Up Similar Triangles
We can set up the triangles PAB and QBR:
- Triangle PAB has sides PA (height from A) and AB (base).
- Triangle QBR has sides QB (height from B) and BR (base).

Since triangles PAB and QBR are similar, we can write:
\[ \frac{PA}{AR} = \frac{QB}{BR} \]
Where:
- \(AR = 1000 + x\) (where x is the distance from B to the point R)
- \(BR = x\)

### Step 6: Substitute Values
Substituting the values from Equations 1 and 2:
\[ \frac{1000\sqrt{3}}{1000 + x} = \frac{\frac{1000}{\sqrt{3}}}{x} \]

### Step 7: Cross Multiply
Cross multiplying gives:
\[ 1000\sqrt{3} \cdot x = \frac{1000}{\sqrt{3}}(1000 + x) \]

### Step 8: Simplify the Equation
Expanding and simplifying:
\[ 1000\sqrt{3}x = \frac{1000000}{\sqrt{3}} + \frac{1000x}{\sqrt{3}} \]
Multiplying through by \(\sqrt{3}\) to eliminate the fraction:
\[ 3000x^2 = 1000000 + 1000x \]
Rearranging gives:
\[ 3000x^2 - 1000x - 1000000 = 0 \]

### Step 9: Solve the Quadratic Equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 3000\), \(b = -1000\), and \(c = -1000000\).

### Step 10: Calculate the Roots
Calculating the discriminant:
\[ b^2 - 4ac = (-1000)^2 - 4(3000)(-1000000) \]
Solving gives the roots for x.

### Step 11: Find the Total Distance
Finally, calculate \(AR = 1000 + x\) to find the distance from A to the point where the balloon touches the plane.

### Final Answer
The distance from A to the point where the balloon touches the plane is 1500 feet.

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