36. In the figure the poles AB and CD of different heights are standing vertically on a level ground. From a point P on the line joining the foots of the poles on the level ground, the angles of elevation to the tops of the poles are found to be complementary. The height of CD and the distance PD are 20√3 m and 20 m respectively. If BP is 10 m, then find the length of the pole AB and the distance AC between the tops of the poles.​

36 In the figure the poles AB and CD of different heights are standing vertically on a level ground From a point P on the line joining the foots of the poles on class=

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.

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