A pole divided into 3:7 by a mark on it , lower part is small , this pole subtends equal angle with a point on ground at a distance 18m from pole , find the height of pole.

Answer

To solve the problem, we need to find the height of a pole that is divided into two parts in the ratio of 3:7, where the lower part is smaller. The pole subtends equal angles with a point on the ground that is 18 meters away from the pole.

### Step-by-Step Solution:

1. **Define the Variables:**
Let the height of the lower part of the pole be \(3x\) and the height of the upper part be \(7x\). Therefore, the total height \(H\) of the pole is:
\[ H = 3x + 7x = 10x \]

2. **Understand the Geometry:**
We have a point \(O\) on the ground that is 18 meters away from the base of the pole. The angles subtended by the lower part (point B) and the upper part (point C) of the pole at point O are equal. Let this angle be \(\alpha\).

3. **Using Trigonometry:**
From point O, we can form two right triangles:
- Triangle AOB (lower part)
- Triangle AOC (upper part)

For triangle AOB:
\[ \tan(\alpha) = \frac{3x}{18} \]
So, we can write:
\[ \tan(\alpha) = \frac{x}{6} \quad \text{(1)} \]

For triangle AOC:
\[ \tan(2\alpha) = \frac{10x}{18} \]
Thus, we have:
\[ \tan(2\alpha) = \frac{5x}{9} \quad \text{(2)} \]

4. **Using the Double Angle Formula:**
The double angle formula for tangent is:
\[ \tan(2\alpha) = \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)} \]
Substituting equation (1) into this formula:
\[ \tan(2\alpha) = \frac{2 \cdot \frac{x}{6}}{1 - \left(\frac{x}{6}\right)^2} \]
This simplifies to:
\[ \tan(2\alpha) = \frac{\frac{2x}{6}}{1 - \frac{x^2}{36}} = \frac{\frac{x}{3}}{1 - \frac{x^2}{36}} \]

5. **Setting the Equations Equal:**
Now we equate the two expressions for \(\tan(2\alpha)\):
\[ \frac{5x}{9} = \frac{\frac{x}{3}}{1 - \frac{x^2}{36}} \]

6. **Cross-Multiplying:**
Cross-multiplying gives:
\[ 5x(1 - \frac{x^2}{36}) = \frac{9x}{3} \]
Simplifying further:
\[ 5x - \frac{5x^3}{36} = 3x \]
Rearranging leads to:
\[ 2x - \frac{5x^3}{36} = 0 \]
Factoring out \(x\):
\[ x(2 - \frac{5x^2}{36}) = 0 \]
Thus, we have:
\[ 2 - \frac{5x^2}{36} = 0 \]

7. **Solving for \(x\):**
Rearranging gives:
\[ \frac{5x^2}{36} = 2 \implies 5x^2 = 72 \implies x^2 = \frac{72}{5} \implies x = \sqrt{\frac{72}{5}} = \frac{6\sqrt{2}}{\sqrt{5}} = \frac{6\sqrt{10}}{5} \]

8. **Finding the Height of the Pole:**
Now substituting \(x\) back into the height equation:
\[ H = 10x = 10 \cdot \frac{6\sqrt{10}}{5} = 12\sqrt{10} \]

### Final Answer:
The height of the pole is \(12\sqrt{10}\) meters.