A pole of 50 meter high stands on a building 250 m high. To an observer at a height of 300 m, thebuilding and the pole subtend equal angles. The distance of the observer from the top of The pole

Options

A. `25sqrt(6)` m
B. 50 m
C. `25sqrt(3)` m
D. 25 m

Answer

To solve the problem step by step, we will use trigonometric principles related to heights and distances.

### Given:
- Height of the building (PQ) = 250 m
- Height of the pole (PR) = 50 m
- Height of the observer (O) = 300 m
- The angles subtended by the building and the pole at the observer are equal.

### Step 1: Define the scenario
Let:
- D = distance from the observer to the base of the pole (point R)
- The angle subtended by the pole at the observer is θ.
- The angle subtended by the building at the observer is also θ.

### Step 2: Set up the triangles
From the observer's point (O):
- In triangle PRO (for the pole):
\[ \tan(\theta) = \frac{\text{Height of pole}}{\text{Distance to pole}} = \frac{50}{D} \]

- In triangle QRO (for the building):
\[ \tan(\theta) = \frac{\text{Height of building}}{\text{Distance to building}} = \frac{300}{D} \]

### Step 3: Relate the angles
Since both angles are equal, we can express the tangent of double the angle for the building:
\[ \tan(2\theta) = \frac{300}{D} \]

Using the double angle formula for tangent:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]

### Step 4: Substitute tan(θ)
Substituting \(\tan(\theta) = \frac{50}{D}\) into the double angle formula:
\[ \tan(2\theta) = \frac{2 \cdot \frac{50}{D}}{1 - \left(\frac{50}{D}\right)^2} \]

### Step 5: Set up the equation
Setting the two expressions for \(\tan(2\theta)\) equal:
\[ \frac{2 \cdot \frac{50}{D}}{1 - \left(\frac{50}{D}\right)^2} = \frac{300}{D} \]

### Step 6: Cross-multiply
Cross-multiplying gives:
\[ 2 \cdot 50 = 300 \left(1 - \frac{2500}{D^2}\right) \]
\[ 100 = 300 - \frac{750000}{D^2} \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ \frac{750000}{D^2} = 300 - 100 \]
\[ \frac{750000}{D^2} = 200 \]

### Step 8: Solve for D^2
Cross-multiplying gives:
\[ 750000 = 200D^2 \]
\[ D^2 = \frac{750000}{200} = 3750 \]

### Step 9: Find D
Taking the square root:
\[ D = \sqrt{3750} = 25\sqrt{6} \]

### Final Answer:
The distance of the observer from the top of the pole is \(25\sqrt{6}\) meters.

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