ABCDEF is a regular polygon. Two poles at C and D are standing vertically and subtend angles of elevation 30° and 60° at A respectively. What is the ratio of the height of the pole at C to that of the pole at D?
1 : 2√3
Analyzing the Regular Polygon and Observation Points
The problem describes a regular polygon named ABCDEF. Poles are situated vertically at vertices C and D. An observer is located at vertex A. The angles of elevation from point A to the top of the poles are given: \(30^\circ\) for the pole at C and \(60^\circ\) for the pole at D. The objective is to determine the ratio of the height of the pole at C (\(h_C\)) to the height of the pole at D (\(h_D\)).
We will use the properties of a regular hexagon (assuming ABCDEF represents the vertices in order) and basic trigonometric ratios.
Calculating Distances from Observation Point A
Let the side length of the regular hexagon be denoted by '\(s\)'. We need to find the distances from point A to points C and D on the ground.
- Distance AC: This is the distance between vertex A and vertex C. In a regular hexagon, the distance between vertices separated by one vertex (like A and C) forms the shorter diagonal. Consider the triangle formed by vertices A, B, and C. Sides AB and BC have length '\(s\)', and the angle \(\angle ABC\) is an interior angle of a regular hexagon. The measure of an interior angle of a regular n-gon is given by \(\frac{(n-2) \times 180^\circ}{n}\). For a hexagon (\(n=6\)), the interior angle is \(\frac{(6-2) \times 180^\circ}{6} = 120^\circ\). Using the Law of Cosines on triangle ABC to find AC: \[ AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC) \] \[ AC^2 = s^2 + s^2 - 2(s)(s)\cos(120^\circ) \] Since \(\cos(120^\circ) = -\frac{1}{2}\): \[ AC^2 = 2s^2 - 2s^2(-\frac{1}{2}) = 2s^2 + s^2 = 3s^2 \] \[ AC = \sqrt{3s^2} = s\sqrt{3} \] So, the distance \(d_{AC} = s\sqrt{3}\).
- Distance AD: This is the distance between vertex A and vertex D. In a regular hexagon, AD is a longer diagonal that passes through the center of the hexagon. The length of the longest diagonal is twice the side length. \[ AD = 2s \] So, the distance \(d_{AD} = 2s\).
Calculating the Heights of the Poles
We use the tangent function (\(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\)) for the right-angled triangles formed by each pole, the ground, and the line of sight from A.
- Height of the pole at C (\(h_C\)): The angle of elevation is \(30^\circ\), and the distance \(d_{AC} = s\sqrt{3}\). \[ \tan(30^\circ) = \frac{h_C}{d_{AC}} \] \[ h_C = d_{AC} \tan(30^\circ) = (s\sqrt{3}) \times \frac{1}{\sqrt{3}} \] \[ h_C = s \]
- Height of the pole at D (\(h_D\)): The angle of elevation is \(60^\circ\), and the distance \(d_{AD} = 2s\). \[ \tan(60^\circ) = \frac{h_D}{d_{AD}} \] \[ h_D = d_{AD} \tan(60^\circ) = (2s) \times \sqrt{3} \] \[ h_D = 2s\sqrt{3} \]
Finding the Ratio of Pole Heights
We need to find the ratio \(h_C : h_D\).
- Substitute the expressions for \(h_C\) and \(h_D\): \[ \frac{h_C}{h_D} = \frac{s}{2s\sqrt{3}} \]
- Simplify the fraction by canceling the common factor '\(s\)': \[ \frac{h_C}{h_D} = \frac{1}{2\sqrt{3}} \]
Thus, the ratio of the height of the pole at C to the height of the pole at D is \(1 : 2\sqrt{3}\).
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