The angles of depression of two points A and B on a horizontal plane such that AB= 200 from the top P of a tower PQ of height 100 are `45 - theta` and `45 + theta`. If the line AB Passes through Q the foot of the tower, then angle `theta` is equal to

Options

A. `45^(@)`
B. `30^(@)`
C. `22.5^(@)`
D. `15^(@)`

Answer

To solve the problem step by step, we will follow the geometric relationships and trigonometric identities involved in the angles of depression from the top of the tower to points A and B.

### Step 1: Diagram and Setup
Let's visualize the problem. We have a tower PQ with height 100 meters. The top of the tower is point P, and the foot of the tower is point Q. Points A and B are on the horizontal plane, with a distance of 200 meters between them. The angles of depression from point P to points A and B are \(45 - \theta\) and \(45 + \theta\) respectively.

### Step 2: Identify the Triangles
From point P, we can draw two right triangles:
1. Triangle PQA for angle \(45 - \theta\)
2. Triangle PBQ for angle \(45 + \theta\)

### Step 3: Apply the Tangent Function
Using the definition of tangent in right triangles:
- For triangle PQA:
\[ \tan(45 - \theta) = \frac{AQ}{PQ} = \frac{x}{100} \]
where \(AQ = x\).

- For triangle PBQ:
\[ \tan(45 + \theta) = \frac{BQ}{PQ} = \frac{200 + x}{100} \]

### Step 4: Use Trigonometric Identities
Using the tangent subtraction and addition formulas:
- For \( \tan(45 - \theta) \):
\[ \tan(45 - \theta) = \frac{1 - \tan(\theta)}{1 + \tan(\theta)} \]
Thus,
\[ \frac{x}{100} = \frac{1 - \tan(\theta)}{1 + \tan(\theta)} \]

- For \( \tan(45 + \theta) \):
\[ \tan(45 + \theta) = \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \]
Thus,
\[ \frac{200 + x}{100} = \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \]

### Step 5: Solve the Equations
Now we have two equations:
1. \( x = 100 \cdot \frac{1 - \tan(\theta)}{1 + \tan(\theta)} \)
2. \( 200 + x = 100 \cdot \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \)

Substituting the value of \(x\) from the first equation into the second:
\[ 200 + 100 \cdot \frac{1 - \tan(\theta)}{1 + \tan(\theta)} = 100 \cdot \frac{1 + \tan(\theta)}{1 - \tan(\theta)} \]

### Step 6: Cross Multiply and Simplify
Cross-multiplying gives:
\[ (200 + 100 \cdot \frac{1 - \tan(\theta)}{1 + \tan(\theta)}) \cdot (1 - \tan(\theta)) = 100 \cdot (1 + \tan(\theta)) \cdot (1 + \tan(\theta)) \]

### Step 7: Rearranging and Solving for \(\tan(\theta)\)
After simplifying and rearranging the above equation, we will get a quadratic equation in terms of \(\tan(\theta)\):
\[ 2\tan^2(\theta) + 4\tan(\theta) - 2 = 0 \]

### Step 8: Use the Quadratic Formula
Using the quadratic formula:
\[ \tan(\theta) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 2\), \(b = 4\), and \(c = -2\):
\[ \tan(\theta) = \frac{-4 \pm \sqrt{16 + 16}}{4} = \frac{-4 \pm 4\sqrt{2}}{4} = -1 \pm \sqrt{2} \]

### Step 9: Determine Valid Solution
Since \(\tan(\theta)\) must be positive:
\[ \tan(\theta) = \sqrt{2} - 1 \]
Thus, \(\theta = \tan^{-1}(\sqrt{2} - 1)\).

### Step 10: Calculate \(\theta\)
Calculating \(\theta\) gives:
\[ \theta \approx 22.5^\circ \]

### Final Answer
Thus, the angle \(\theta\) is equal to \(22.5^\circ\).

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