Question

Two poles of equal height are placed opposite to each other on either side of a road 80 meters wide. From a point on the road between these two poles, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance from the point from the poles.

Hint:

To solve problems involving angles of depression or elevation, use trigonometric functions such as tangent, sine, and cosine.

Answer 1

Let the height of the poles be \( h \), and the distance from the point on the road to the first pole be \( x_1 \) and to the second pole be \( x_2 \).
Step 1: From the first pole:
\[ \tan(60^\circ) = \frac{h}{x_1} \] \[ \sqrt{3} = \frac{h}{x_1} \Rightarrow h = \sqrt{3} \cdot x_1 \]
Step 2: From the second pole:
\[ \tan(30^\circ) = \frac{h}{x_2} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{x_2} \Rightarrow h = \frac{x_2}{\sqrt{3}} \] Since the total distance between the poles is 80 meters: \[ x_1 + x_2 = 80 \] Now substitute the expression for \( h \) from both equations: \[ \sqrt{3} \cdot x_1 = \frac{x_2}{\sqrt{3}} \Rightarrow 3 \cdot x_1 = x_2 \] \[ x_1 + 3 \cdot x_1 = 80 \Rightarrow 4 \cdot x_1 = 80 \Rightarrow x_1 = 20 \] Now, substitute \( x_1 = 20 \) into \( h = \sqrt{3} \cdot x_1 \): \[ h = \sqrt{3} \cdot 20 \Rightarrow h = 20\sqrt{3} \Rightarrow h \approx 34.64 \, \text{meters} \] Thus, the height of the poles is approximately \( 34.64 \, \text{meters} \), and the distance from the point to the first pole is 20 meters, and to the second pole is 60 meters.