Question:

Step 1: Set up a coordinate system. Let the horizontal ground line AC be the x-axis. Let point A be at the origin $(0, 0)$. Let the distance between the poles be $d$. Then point C is at $(d, 0)$. Since AB and CD are vertical poles, B is at $(0, 15)$ and D is at $(d, 10)$. Step 2: Find the equation of line AD. Line AD passes through $A(0, 0)$ and $D(d, 10)$. The slope of AD is $m_{AD} = \frac{10 - 0}{d - 0} = \frac{10}{d}$. The equation of line AD is $y = \frac{10}{d}x$. Step 3: Find the equation of line BC. Line BC passes through $B(0, 15)$ and $C(d, 0)$. The slope of BC is $m_{BC} = \frac{0 - 15}{d - 0} = -\frac{15}{d}$. Using the point-slope form with $B(0, 15)$: $y - 15 = -\frac{15}{d}(x - 0)$. The equation of line BC is $y = -\frac{15}{d}x + 15$. Step 4: Find the intersection point P. The point P is the intersection of lines AD and BC. Equate the y-values: $$ \frac{10}{d}x = -\frac{15}{d}x + 15 $$ Multiply by $d$ to clear the denominators: $$ 10x = -15x + 15d $$ $$ 25x = 15d $$ $$ x = \frac{15d}{25} = \frac{3d}{5} $$ Step 5: Calculate the height of P. Substitute the x-coordinate of P into the equation of line AD: $$ y = \frac{10}{d}x = \frac{10}{d}\left(\frac{3d}{5}\right) $$ $$ y = \frac{30d}{5d} = 6 $$ The height of P above the line AC is 6 meters. Step 6: Alternative method using similar triangles. Let P be a point $(x, h)$ where $h$ is the height of P above AC. Let Q be the foot of the perpendicular from P to AC, so $PQ = h$. Consider $\triangle APQ$ and $\triangle ADC$. These triangles are similar. Thus, the ratio of their corresponding sides is equal: $$ \frac{PQ}{CD} = \frac{AQ}{AC} $$ $$ \frac{h}{10} = \frac{x}{d} \quad (*)$$ Consider $\triangle CPQ$ and $\triangle CAB$. These triangles are similar. Thus, the ratio of their corresponding sides is equal: $$ \frac{PQ}{AB} = \frac{CQ}{AC} $$ $$ \frac{h}{15} = \frac{d-x}{d} \quad (**)$$ From equation $(*)$, we have $x = \frac{hd}{10}$. Substitute this into equation $(**)$: $$ \frac{h}{15} = \frac{d - \frac{hd}{10}}{d} $$ $$ \frac{h}{15} = 1 - \frac{h}{10} $$ Rearrange the terms to solve for $h$: $$ \frac{h}{15} + \frac{h}{10} = 1 $$ Find a common denominator for the fractions: $$ \frac{2h}{30} + \frac{3h}{30} = 1 $$ $$ \frac{5h}{30} = 1 $$ $$ \frac{h}{6} = 1 $$ $$ h = 6 $$ Both methods consistently yield a height of 6 meters. The final answer is $\boxed{6}$.