Solving the Angle of Elevation Problem: Pole and Tower Distance

This problem involves trigonometry, specifically the concept of the angle of elevation, applied to find the distance between two upright structures: a pole and a tower.

Let's represent the scenario with points:

• Let the base of the pole be A and its top be B. The height of the pole is AB = 17.75 m.
• Let the base of the tower be C and its top be D. The height of the tower is CD = 57.75 m.
• The pole and the tower are upright, meaning they are perpendicular to the ground (AC).
• The distance we need to find is the distance between the foot of the pole and the foot of the tower, which is AC.

Setting up the Geometry

Imagine a horizontal line drawn from the top of the pole (B) parallel to the ground (AC). Let this line intersect the tower (CD) at point E.

• Since BE is parallel to AC, ABEC forms a rectangle.
• Therefore, BE = AC (the distance we need to find), and CE = AB = 17.75 m.
• The total height of the tower is CD = 57.75 m.
• The segment ED is the part of the tower above the horizontal line from B. Its length is ED = CD - CE.
• ED = 57.75 m - 17.75 m = 40 m.

Now consider the right-angled triangle BDE. The angle of elevation from the top of the pole (B) to the top of the tower (D) is given as 60°. This angle is ∠DBE = 60°.

In the right triangle BDE:

• The side opposite to the angle ∠DBE (60°) is ED = 40 m.
• The side adjacent to the angle ∠DBE (60°) is BE, which is equal to AC.

Applying Trigonometry to Find the Distance

In a right-angled triangle, the tangent of an angle is defined as the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. The trigonometric relation is:

\(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\)

In triangle BDE, with respect to the angle ∠DBE = 60°:

• Opposite side = ED = 40 m
• Adjacent side = BE

So, we can write the equation:

\(\tan(60^\circ) = \frac{ED}{BE}\)

We know that the value of \(\tan(60^\circ)\) is \(\sqrt{3}\).

Substituting the known values into the equation:

\(\sqrt{3} = \frac{40}{BE}\)

Calculating the Distance Between the Pole and the Tower

Now, we solve for BE:

\(BE = \frac{40}{\sqrt{3}}\)

To rationalize the denominator (remove the square root from the bottom), we multiply both the numerator and the denominator by \(\sqrt{3}\):

\(BE = \frac{40}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

\(BE = \frac{40\sqrt{3}}{3}\)

Since BE is equal to AC, the distance between the foot of the pole and the foot of the tower is \(\frac{40\sqrt{3}}{3}\) m.

Comparing this result with the given options, we find that our calculated distance matches one of the options.

Revision Table: Key Concepts

Additional Information: Trigonometric Values

It is useful to remember the trigonometric values for common angles like 0°, 30°, 45°, 60°, and 90°. For this problem, the value of \(\tan(60^\circ)\) was crucial.

• \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\)
• \(\cos(60^\circ) = \frac{1}{2}\)
• \(\tan(60^\circ) = \sqrt{3}\)

Understanding how to set up the geometric diagram and identify the right triangle is the first step in solving such angle of elevation or depression problems.