A wireless pole 25 metres high is fixed on a top of a verandah of a house which is 15 metres high. At a point R on the ground, directly opposite, the wireless pole and verandah subtend equal angles. The distance of R from the verandah is____

Answer

To solve the problem step by step, we will analyze the situation involving the wireless pole and the verandah, and use trigonometric principles to find the distance from point R to the verandah.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The height of the verandah (H1) = 15 meters.
- The height of the wireless pole (H2) = 25 meters.
- The total height of the pole from the ground = H1 + H2 = 15 + 25 = 40 meters.

2. **Identifying the Angles**:
- Let the angle subtended by the top of the verandah at point R be θ.
- Let the angle subtended by the top of the wireless pole at point R be 2θ (since the pole is higher).

3. **Using Tangent Function**:
- From point R, the tangent of angle θ can be expressed in terms of the height of the verandah and the distance from R to the base of the verandah (let's call this distance X):
\[ \tan(θ) = \frac{H1}{X} = \frac{15}{X} \]
- For the angle 2θ, we can express it as:
\[ \tan(2θ) = \frac{H2 + H1}{X} = \frac{40}{X} \]

4. **Using the Double Angle Formula**:
- The double angle formula for tangent is:
\[ \tan(2θ) = \frac{2\tan(θ)}{1 - \tan^2(θ)} \]
- Substituting the value of \(\tan(θ)\):
\[ \tan(2θ) = \frac{2 \cdot \frac{15}{X}}{1 - \left(\frac{15}{X}\right)^2} \]

5. **Setting Up the Equation**:
- Equating the two expressions for \(\tan(2θ)\):
\[ \frac{2 \cdot \frac{15}{X}}{1 - \left(\frac{15}{X}\right)^2} = \frac{40}{X} \]
- Cross-multiplying gives:
\[ 2 \cdot 15 = 40 \left(1 - \frac{225}{X^2}\right) \]
- Simplifying this:
\[ 30 = 40 - \frac{9000}{X^2} \]
\[ \frac{9000}{X^2} = 10 \]
\[ X^2 = 900 \]
\[ X = 30 \text{ meters} \]

### Final Answer:
The distance of point R from the verandah is **30 meters**.