## Answer :

**Step-by-step explanation:**

Let's denote the height of pole AB as \( h \) meters and the distance AC between the tops of the poles as \( x \) meters.

From the problem statement, we can form two right triangles: △PBD and △PCD.

In △PBD:

- \( BP = 10 \) meters

- \( PD = 20 \) meters

In △PCD:

- \( CD = 20\sqrt{3} \) meters

- \( PD = 20 \) meters

Given that the angles of elevation to the tops of the poles from point P are complementary, we have:

\[ \tan^{-1}\left(\frac{h}{10}\right) + \tan^{-1}\left(\frac{h}{20\sqrt{3}}\right) = 90^\circ \]

Using the property of the tangent function that \( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \), we can rewrite the equation as:

\[ \tan^{-1}\left(\frac{\frac{h}{10} + \frac{h}{20\sqrt{3}}}{1 - \frac{h}{10} \cdot \frac{h}{20\sqrt{3}}}\right) = 90^\circ \]

Solving for \( h \):

\[ \frac{\frac{h}{10} + \frac{h}{20\sqrt{3}}}{1 - \frac{h}{10} \cdot \frac{h}{20\sqrt{3}}} = \tan(90^\circ) \]

\[ \frac{\frac{h}{10} + \frac{h}{20\sqrt{3}}}{1 - \frac{h^2}{200}} = \infty \]

Since the tangent of \(90^\circ\) is undefined, the denominator must equal zero:

\[ 1 - \frac{h^2}{200} = 0 \]

\[ h^2 = 200 \times 1 \]

\[ h = \sqrt{200} \]

\[ h = 10\sqrt{2} \]

So, the length of pole AB is \( 10\sqrt{2} \) meters.

Now, to find the distance AC, we'll use the Pythagorean theorem in △ABC:

\[ (10 + x)^2 = (10\sqrt{2})^2 + (20\sqrt{3})^2 \]

\[ (10 + x)^2 = 200 + 1200 \]

\[ (10 + x)^2 = 1400 \]

\[ 10 + x = \sqrt{1400} \]

\[ x = \sqrt{1400} - 10 \]

\[ x ≈ 20.98 \]

So, the distance AC between the tops of the poles is approximately \( 20.98 \) meters.