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Step-by-step Solution
Step 1: Set up a suitable coordinate system
Let us place the poles AB and CD on a coordinate plane in such a way that:
• Point A is at the origin, so A = (0, 0).
• Since AB is 15 m vertically above A, point B = (0, 15).
• Let the distance between A and C along the horizontal ground be a (i.e., AC = a). Hence C = (a, 0).
• Pole CD is 10 m high, so point D = (a, 10).
Thus, AC lies along the x-axis, and the poles go up vertically from those points.
Step 2: Write the equation of line AD
The coordinates of A and D are A(0, 0) and D(a, 10).
• The slope of AD is
$ \displaystyle m_{AD} = \frac{10 - 0}{a - 0} = \frac{10}{a}.$
• Using the point-slope form through A(0, 0), the equation of line AD is
$ \displaystyle y = \frac{10}{a} x. $
Step 3: Write the equation of line BC
The coordinates of B and C are B(0, 15) and C(a, 0).
• The slope of BC is
$ \displaystyle m_{BC} = \frac{0 - 15}{\,a - 0} = -\frac{15}{a}. $
• Using the two-point form, or directly from intercept form, we can write line BC as
$ \displaystyle \frac{x}{a} + \frac{y}{15} = 1. $
Step 4: Find the intersection point P of lines AD and BC
Let the intersection point be P(x, y). It must satisfy both equations:
Line AD:
$ \displaystyle y = \frac{10}{a} x. $
Line BC:
$ \displaystyle \frac{x}{a} + \frac{y}{15} = 1. $
Substitute $ y = \frac{10}{a} x $ from the first equation into the second:
$ \displaystyle \frac{x}{a} + \frac{1}{15} \left(\frac{10}{a} x\right) = 1. $
$ \displaystyle \frac{x}{a} + \frac{10x}{15a} = 1. $
$ \displaystyle \frac{x}{a} \left(1 + \frac{10}{15}\right) = 1. $
$ \displaystyle \frac{x}{a} \left(\frac{15}{15} + \frac{10}{15}\right) = 1. $
$ \displaystyle \frac{x}{a} \left(\frac{25}{15}\right) = 1.
\quad \Longrightarrow \quad \frac{25}{15} \cdot \frac{x}{a} = 1. $
$ \displaystyle \frac{x}{a} = \frac{15}{25} = \frac{3}{5}.
\quad \Longrightarrow \quad x = \frac{3a}{5}. $
Then
$ \displaystyle y = \frac{10}{a} x = \frac{10}{a} \cdot \frac{3a}{5} = \frac{30}{5} = 6. $
Thus, $ P\left(\frac{3a}{5}, 6\right). $
Step 5: Conclude the height of P above AC
Since AC lies along the x-axis, the height of P above AC is simply the y-coordinate of P, which is 6 m. Hence, the height of P above the line AC is 6.
