Calculating Distance Between Poles Using Trigonometry

This problem involves finding the horizontal distance between two vertical poles of different heights, given the angle that the line connecting their tops makes with the horizontal. We can solve this using basic trigonometry, specifically the tangent function.

Understanding the Geometry

Imagine the two poles standing vertically. Let the shorter pole have height \(h_1\) and the taller pole have height \(h_2\). Let the distance between the poles be \(d\). A line joining the tops of the poles forms the hypotenuse of a right-angled triangle. The vertical side of this triangle is the difference in height between the two poles, and the horizontal side is the distance \(d\) between the poles.

• Height of the shorter pole, \(h_1 = 10 \text{ m}\).
• Height of the taller pole, \(h_2 = 20 \text{ m}\).
• The angle the line joining the tops makes with the horizontal, \(\theta = 15^\circ\).
• We need to find the distance between the poles, \(d\).

Setting up the Trigonometric Relationship

The difference in height between the poles forms the side opposite the angle \(\theta\) in our right-angled triangle. This difference in height is:

Difference in height \( \Delta h = h_2 - h_1 = 20 \text{ m} - 10 \text{ m} = 10 \text{ m} \).

The distance between the poles, \(d\), is the side adjacent to the angle \(\theta\).

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

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

In our case:

$$ \tan(15^\circ) = \frac{\Delta h}{d} $$

We know \(\Delta h = 10 \text{ m}\) and \(\theta = 15^\circ\). We want to find \(d\).

$$ \tan(15^\circ) = \frac{10}{d} $$

Calculating the Distance

To find \(d\), we rearrange the equation:

$$ d = \frac{10}{\tan(15^\circ)} $$

The value of \(\tan(15^\circ)\) can be calculated or found from trigonometric tables. A common way to calculate \(\tan(15^\circ)\) is using the tangent subtraction formula: \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \).

Using \(A = 45^\circ\) and \(B = 30^\circ\):

$$ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} $$

We know \(\tan 45^\circ = 1\) and \(\tan 30^\circ = \frac{1}{\sqrt{3}}\).

$$ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} $$

To simplify, multiply the numerator and denominator by \( \sqrt{3}-1 \):

$$ \tan(15^\circ) = \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} $$

Using the approximate value \(\sqrt{3} \approx 1.732\):

$$ \tan(15^\circ) \approx 2 - 1.732 = 0.268 $$

Now, substitute this value back into the equation for \(d\):

$$ d = \frac{10}{\tan(15^\circ)} \approx \frac{10}{0.268} $$

$$ d \approx 37.31 \text{ m} $$

Rounding to one decimal place, the distance between the poles is approximately 37.3 m.

Summary of Steps

1. Identify the heights of the two poles.
2. Calculate the difference in height between the poles.
3. Recognize that the difference in height, the distance between poles, and the line joining their tops form a right-angled triangle.
4. Use the tangent function relating the angle with the horizontal to the opposite side (difference in height) and the adjacent side (distance between poles).
5. Solve the equation for the distance between the poles.
6. Calculate or use the value of \(\tan(15^\circ)\).
7. Compute the final distance.

Comparing with Options

Our calculated distance is approximately 37.3 m. Let's compare this with the given options:

• 36.3 m
• 37.3 m
• 38.3 m
• 39.3 m

The calculated value is closest to 37.3 m.

Revision Table: Pole Distance Calculation

Additional Information: Trigonometric Values for Distance Problems

Understanding standard trigonometric values can be helpful for solving geometry and distance problems. For less common angles like \(15^\circ\) or \(75^\circ\), you can derive their values using sum/difference formulas or half-angle formulas if a calculator isn't allowed.

• \( \sin(15^\circ) = \frac{\sqrt{6}-\sqrt{2}}{4} \)
• \( \cos(15^\circ) = \frac{\sqrt{6}+\sqrt{2}}{4} \)
• \( \tan(15^\circ) = \frac{\sin(15^\circ)}{\cos(15^\circ)} = \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} = 2 - \sqrt{3} \)

Using \(2 - \sqrt{3}\) gives a more accurate result before approximation than using a pre-rounded decimal value of \(\tan(15^\circ)\). For example, \(10 / (2 - \sqrt{3}) = 10(2 + \sqrt{3}) / ((2 - \sqrt{3})(2 + \sqrt{3})) = 10(2 + \sqrt{3}) / (4 - 3) = 10(2 + \sqrt{3})\). Using \(\sqrt{3} \approx 1.73205\), \(10(2 + 1.73205) = 10(3.73205) = 37.3205\), which is very close to 37.3 m.

These calculations show that the distance between the poles relies directly on the difference in pole heights and the angle with the horizontal, using the tangent function as the key trigonometric relationship.