Problem 25 - Algebra 2 and Trigonometry

Question

A tower that is 20 feet tall stands at the edge of a 30 -foot cliff. From a point on level ground that is 20 feet from a point directly below the tower at the base of the cliff, the measure of the angle of elevation of the top of the tower is \(x\) and the measure of the angle of elevation of the foot of the tower is \(y .\) a. Find the exact value of \(\tan (x-y),\) the tangent of the angle between the lines of sight to the foot and top of the tower. b. Find to the nearest degree the measure of the angle between the lines of sight to the foot and the top of the tower.

Step-by-Step Solution

Answer

\( \tan(x-y) = \frac{1}{4.75} \), angle \( x-y \approx 12^\circ \).

Step 1: Calculate the angle of elevation to the base of the cliff (y)

To find the angle of elevation to the base of the cliff, use \( \tan(y) = \frac{30}{20} \). This simplifies to \( \tan(y) = 1.5 \). Taking the inverse tangent: \( y = \tan^{-1}(1.5) \).

Step 2: Calculate the angle of elevation to the top of the tower (x)

The top of the tower is 50 feet above the level ground (20 + 30 feet). Use \( \tan(x) = \frac{50}{20} \), which simplifies to \( \tan(x) = 2.5 \). Taking the inverse tangent: \( x = \tan^{-1}(2.5) \).

Step 3: Use angle difference identity for tangent

The identity for tangent of an angle difference is given by \( \tan(x-y) = \frac{\tan(x) - \tan(y)}{1 + \tan(x)\tan(y)} \). Substitute \( \tan(x) = 2.5 \) and \( \tan(y) = 1.5 \) to get:\[ \tan(x-y) = \frac{2.5 - 1.5}{1 + 2.5 \times 1.5} = \frac{1}{4.75} \].

Step 4: Find the measure of the angle (x-y)

Now calculate the angle \( x-y \) by taking the inverse tangent of the result from Step 3: \( x-y = \tan^{-1}\left(\frac{1}{4.75}\right) \). This gives \( x-y \approx 11.93^\circ \). The nearest degree is \( 12^\circ \).

Key Concepts

Tangent Function

The tangent function is a fundamental concept in trigonometry. It relates the angle of a right triangle to the ratio of the length of the opposite side and the adjacent side. In mathematical terms, if we have an angle \( \theta \) in a right triangle, the tangent of this angle is expressed as:\[\tan(\theta) = \frac{\text{opposite side length}}{\text{adjacent side length}}.\]This ratio helps us understand how steep a line is from a given angle. In problems involving the angle of elevation, as in the original exercise, tangent becomes particularly useful because it directly links the height and the distance on the ground. For example, in the exercise, the tangent function helps find angles of elevation to different parts of the tower and cliff structure.

Inverse Tangent

The inverse tangent, or arctangent, is a function that reverses what the tangent function does. While the tangent gives us the ratio of two sides of a triangle, the inverse tangent allows us to find the angle when this ratio is known. It is represented as \( \tan^{-1}(x) \) or \( \text{arctan}(x) \).To solve real-world problems like those dealing with elevation angles, inverse tangent is essential. It helps convert a known tangent value back into an angle measure. In the original exercise, after calculating the tangent values for specific angles, the inverse tangent is used to find the actual angles of elevation \( x \) and \( y \).

• For the angle to the base of the cliff: \( y = \tan^{-1}(1.5) \).
• For the angle to the top of the tower: \( x = \tan^{-1}(2.5) \).

The key idea here is that the inverse tangent converts back from ratio values to the angle in degrees, solving the equation for an angle expression.

Angle Difference Identity

The angle difference identity for tangent is a crucial formula in trigonometry, especially when determining angles between two given angles. This identity states:\[\tan(x-y) = \frac{\tan(x) - \tan(y)}{1 + \tan(x)\tan(y)}.\]It allows you to calculate the tangent of the difference of two angles, utilizing the tangents of the individual angles \( x \) and \( y \). In the exercise, this identity is used to compute \( \tan(x-y) \) by substituting the calculated tangent values:

• \( \tan(x) = 2.5 \)
• \( \tan(y) = 1.5 \)

The result from applying the identity is \( \tan(x-y) = \frac{1}{4.75} \), which simplifies the process of determining the angle between the lines of sight in the exercise.

Trigonometry Problem Solving

Trigonometry problem solving often involves understanding how to use functions like tangent and identities such as the angle difference identity to find unknown values or angles. In real-world scenarios like the one presented in the exercise, these skills are vital for making accurate calculations. To tackle such problems, follow these steps:

• Identify the right triangles in the scenario and determine what needs to be solved - such as angles or side lengths.
• Use the tangent function to relate angles to side ratios, and the inverse tangent to find angle measures from these ratios.
• Apply trigonometric identities, like the angle difference identity, to calculate more complex angles.
• Convert the angles to degree form when necessary for interpretation.

The exercise demonstrated how these trigonometric concepts and problem-solving techniques can be used to determine the angle of elevation between two lines of sight, providing both the exact and approximate measures needed.