An aeroplane is flying over two houses which are at a distance of 300 m from each other. If at some moment from the aeroplane the angle of depression of two houses on the same side are `45^(@)" and "60^(@)`, at what height the plane is flying ?

Answer

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

### Step 1: Draw the Diagram
First, we need to visualize the problem. Draw a horizontal line representing the ground where the two houses are located. Mark the two houses as points A1 and A2, with a distance of 300 m between them. Above this line, draw a point P representing the aeroplane flying at a certain height (h). Draw lines from P to A1 and A2, making angles of depression of 60° and 45°, respectively.

### Step 2: Identify the Triangles
From the point P, we can identify two right triangles:
1. Triangle PA1Q, where angle PA1Q = 60°.
2. Triangle PA2Q, where angle PA2Q = 45°.

### Step 3: Set Up the Equations
Let PQ be the horizontal distance from the point directly below the aeroplane (point Q) to the house A1. Therefore, the distance from Q to A2 is (300 - PQ).

Using the tangent function for both angles of depression:
- For angle 60° in triangle PA1Q:
\[ \tan(60°) = \frac{h}{PQ} \]
We know that \(\tan(60°) = \sqrt{3}\), so:
\[ \sqrt{3} = \frac{h}{PQ} \quad \Rightarrow \quad h = PQ \cdot \sqrt{3} \quad \text{(1)} \]

- For angle 45° in triangle PA2Q:
\[ \tan(45°) = \frac{h}{300 - PQ} \]
We know that \(\tan(45°) = 1\), so:
\[ 1 = \frac{h}{300 - PQ} \quad \Rightarrow \quad h = 300 - PQ \quad \text{(2)} \]

### Step 4: Equate the Two Expressions for h
From equations (1) and (2), we have:
\[ PQ \cdot \sqrt{3} = 300 - PQ \]

### Step 5: Solve for PQ
Rearranging the equation gives:
\[ PQ \cdot \sqrt{3} + PQ = 300 \]
\[ PQ(\sqrt{3} + 1) = 300 \]
\[ PQ = \frac{300}{\sqrt{3} + 1} \]

### Step 6: Rationalize the Denominator
To simplify \(PQ\), we can multiply the numerator and denominator by \(\sqrt{3} - 1\):
\[ PQ = \frac{300(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{300(\sqrt{3} - 1)}{3 - 1} = \frac{300(\sqrt{3} - 1)}{2} = 150(\sqrt{3} - 1) \]

### Step 7: Substitute PQ back to find h
Now substitute \(PQ\) back into equation (1) to find h:
\[ h = PQ \cdot \sqrt{3} = 150(\sqrt{3} - 1) \cdot \sqrt{3} \]
\[ h = 150(3 - \sqrt{3}) = 450 - 150\sqrt{3} \]

### Step 8: Calculate the Approximate Value of h
Using \(\sqrt{3} \approx 1.732\):
\[ h \approx 450 - 150 \cdot 1.732 \approx 450 - 259.8 \approx 190.2 \text{ m} \]

### Final Answer
The height at which the plane is flying is approximately **190.2 meters**.

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