Problem 11 - Algebra 2

Question

Tom wants to build a rope bridge between his tree house and Roy’s tree house. Suppose Tom’s tree house is directly behind Roy’s tree house. At a distance of 20 meters to the left of Tom’s tree house, an angle of \(52^{\circ}\) is measured between the two tree houses. Find the length of the rope bridge.

Step-by-Step Solution

Answer

The length of the rope bridge is approximately 25.40 meters.

Step 1: Identify the Problem Type

This is a trigonometry problem where we need to find the length of the side opposite the known angle in a right triangle.

Step 2: Draw the Triangle

Visualize a right triangle with the right angle at Roy's tree house, one side along the horizontal line 20 meters from Tom's tree house, and the hypotenuse as the rope bridge. The angle of 52° is between the horizontal side and the hypotenuse.

Step 3: Apply Trigonometric Ratio

Use the tangent function, as the tangent of an angle in a right triangle is the opposite side over the adjacent side, to find the opposite side length, i.e., the length of the rope bridge. Calculate using: \( \tan(52^{\circ}) = \frac{\text{Length of Rope Bridge}}{20} \)

Step 4: Calculate the Length

Solve the equation from Step 3 to find the length of the rope bridge:\(\text{Length of Rope Bridge} = 20 \times \tan(52^{\circ})\).

Key Concepts

Right Triangle

Understanding right triangles is essential in trigonometry. A right triangle is a triangle that has one angle measuring exactly 90 degrees. This 90-degree angle is also known as a right angle. Such a triangle has two sides forming the right angle called the legs and another side that is opposite to the right angle known as the hypotenuse. In trigonometry, right triangles are central because they allow us to define trigonometric functions based on their angles and sides.

Two characteristics of right triangles are:

• The sum of all interior angles is 180 degrees.
• The Pythagorean Theorem, which states that the square of the hypotenuse equals the sum of the squares of the other two sides \(c^2 = a^2 + b^2\).

In our specific exercise, visualizing the layout helps: the rope bridge acts as the hypotenuse, and one leg runs the 20 meters horizontally between Tom’s and Roy’s tree house. The challenge here focuses on finding the length of the rope bridge using this setup.

Tangent Function

Trigonometric functions are relationships between the angles and sides of right triangles. The tangent function is one of these, and it's particularly useful when you know the angle and one side of a right triangle.

The formula for tangent in a right triangle is given by:

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

This tells us that the tangent of an angle \(\theta\) is the ratio of the length of the opposite side to the length of the adjacent side. In our problem, we've got an angle of 52° and an adjacent side of 20 meters. To find the length of the rope bridge, we apply the formula:

\( \tan(52^{\circ}) = \frac{\text{Length of Rope Bridge}}{20} \).

By multiplying both sides by 20, we rearrange this to find the length of the rope bridge.

Angle Calculation

Knowing how to calculate angles is crucial in solving trigonometry problems. When we talk about angle calculation here, we are using the known angle to find an unknown side of the triangle using trigonometric functions.

Given an angle of \(52^{\circ}\) between the base of 20 meters and the intended course of the rope bridge, we rely on this angle to use the tangent function efficiently. This step involves:

• Recognizing the type of trigonometric function suitable for our data; in this case, it's tangent because we have an angle and need to find an opposite side.
• Applying the tangent formula: \( \tan(52^{\circ}) = \frac{\text{Length of Rope Bridge}}{20} \).

Afterwards, solving for the length leverages a calculator to determine \(20 \times \tan(52^{\circ})\), giving us the exact measure of the rope bridge between the tree houses. This coupling of geometric and algebraic principles exemplifies real-world trigonometry application.