Problem 26 - Algebra 2 and Trigonometry

Question

An airplane travels at an altitude of 6 miles. At a point on the ground, the measure of the angle of elevation to the airplane is \(\theta .\) Find the distance to the plane from the point on the ground when: $$ \csc \theta=2 \quad \text { b. } \sec \theta=\frac{5}{3} \quad \text { c. } \cot =\sqrt{3} $$

Step-by-Step Solution

Answer

(a) 12 miles, (b) 10 miles, (c) 12 miles.

Step 1: Understanding the Relationship

The problem involves trigonometric relationships in a right triangle. Given the angle of elevation \(\theta\), the altitude (6 miles) serves as one side of the right triangle, while the hypotenuse represents the distance \(d\) to the airplane.

Step 2: Solve for Part (a) using \(\csc \theta\)

Given \(\csc \theta = 2\), we know \(\csc \theta = \frac{1}{\sin \theta}\). Therefore, \(\sin \theta = \frac{1}{2}\). Since \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\), the relationship is \(\frac{6}{d} = \frac{1}{2}\). Solving for \(d\):\[d = 6 \times 2 = 12 \, \text{miles}\]

Step 3: Solve for Part (b) using \(\sec \theta\)

Given \(\sec \theta = \frac{5}{3}\), we know \(\sec \theta = \frac{1}{\cos \theta}\). Therefore, \(\cos \theta = \frac{3}{5}\). Since \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\) and the adjacent side is 6, the equation becomes \(\frac{6}{d} = \frac{3}{5}\). Solving for \(d\):\[d = 6 \times \frac{5}{3} = 10 \, \text{miles}\]

Step 4: Solve for Part (c) using \(\cot \theta\)

Given \(\cot \theta = \sqrt{3}\), we know \(\cot \theta = \frac{1}{\tan \theta}\). Therefore, \(\tan \theta = \frac{1}{\sqrt{3}}\). Since \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\) and the opposite side (altitude) is 6, \(\frac{6}{\text{adjacent}} = \frac{1}{\sqrt{3}}\). Solving for the adjacent side:\[\text{adjacent} = 6 \times \sqrt{3}\]Finally, calculate \(d\) using the Pythagorean theorem:\[d = \sqrt{6^2 + (6 \sqrt{3})^2} = \sqrt{36 + 108} = \sqrt{144} = 12 \, \text{miles}\]

Key Concepts

Angle of Elevation

The angle of elevation is a fundamental concept when dealing with right triangles, especially in problems involving real-world applications like flight paths. It is defined as the angle between the line of sight from an observer to an object above and the horizontal plane of the observer. For this problem, imagine standing on the ground looking up at an airplane in the sky.

• The horizontal line is like the ground you are standing on.
• Your line of sight extends from the point on the ground up to the plane.
• The angle that forms between this line of sight and the horizontal is the angle of elevation heta.

This angle tells us how steep or shallow the view is from the ground to the airplane. It becomes crucial when calculating distances in trigonometric problems, as it allows us to apply trigonometric functions accurately. In this case, the angle of elevation helps in finding the plane's distance by forming a right triangle, with one side being the height and the other side being the hypotenuse or the direct line to the airplane.

Cosecant Function

The cosecant function is one of the reciprocal trigonometric functions related to the sine function. It's written as \[ \csc \theta = \frac{1}{\sin \theta} \] This function is particularly useful when we know the hypotenuse and need to understand the angle's relationship in a right triangle. In our example, the problem involves the cosecant function since \( \csc \theta = 2 \). To solve this, recognize that since \( \csc \theta = \frac{1}{\sin \theta} \), it follows that \( \sin \theta = \frac{1}{2} \).

• The sine function here represents \( \frac{\text{opposite}}{\text{hypotenuse}} \).
• The opposite side is usually the vertical height of 6 miles (where the airplane is flying).

By setting up the equation \( \frac{6}{d} = \frac{1}{2} \), you can find the hypotenuse \( d \) or the distance directly to the airplane.The rationale behind using \( \csc \) is it transforms easily for calculations already given in terms of sine, helping us quickly determine necessary measures.

Secant Function

The secant function is another reciprocal trigonometric function, related to the cosine. It's denoted as \[ \sec \theta = \frac{1}{\cos \theta} \] This function is vital for problems where the adjacent side's relation to the hypotenuse in a right triangle is needed. In our scenario, knowing that \( \sec \theta = \frac{5}{3} \) allows us to find \( \cos \theta = \frac{3}{5} \).This tells us that the adjacent side (which is 6 miles across the ground) to the hypotenuse (distance to the airplane) has this ratio.

• \( \cos \theta \) represents \( \frac{\text{adjacent}}{\text{hypotenuse}} \).

Putting the pieces together helps us apply the formula: \( \frac{6}{d} = \frac{3}{5} \). By resolving this equation, we compute the hypotenuse, giving us the sought distance directly to the plane from our position on the ground.

Cotangent Function

The cotangent function is the reciprocal of the tangent function, and it is expressed as \[ \cot \theta = \frac{1}{\tan \theta} \] In this problem, the given condition \( \cot \theta = \sqrt{3} \) means \( \tan \theta = \frac{1}{\sqrt{3}} \). The tangent relates the opposite to the adjacent side of the triangle:

• \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \).

Here, the altitude where the plane is flying (6 miles) is the opposite side. Therefore, you set up the equation: \( \frac{6}{\text{adjacent}} = \frac{1}{\sqrt{3}} \) which lets you calculate the adjacent side (ground distance).Ultimately, employing the Pythagorean theorem helps find the triangle's hypotenuse, finalizing the calculation of distance from the initial point to the plane. The cotangent, though not as commonly used as sine or cosine, equally offers insight into angular relationships within right triangles.