Problem 26
Question
Two boats leave a dock to cross a river that is 80 meters wide. The first boat travels to a point that is 100 meters downstream from a point directly opposite the starting point, and the second boat travels to a point that is 200 meters downstream from a point directly opposite the starting point. a. Let \(x\) be the measure of the angle between the river's edge and the path of the first boat and \(y\) be the measure of the angle between the river's edge and the path of the second boat. Find \(\tan x\) and \(\tan y .\) b. Find the tangent of the measure of the angle between the paths of the boats.
Step-by-Step Solution
Verified Answer
\(\tan x = 0.8\), \(\tan y = 0.4\), angle between paths: \(\approx 0.303\).
1Step 1: Understand the problem
We need to find the tangent of angles for two boats as they cross a river. These angles are formed by the path of each boat and the perpendicular line from the starting point to their respective destinations on the other bank. The first boat lands 100 meters downstream and the second 200 meters, both starting from the same 80 meters width across the river.
2Step 2: Identify the right triangles
For each boat, we form a right triangle. The width of the river is the opposite side (80 meters) for both triangles. The base, or adjacent side to the angle with the river's edge for the first boat is 100 meters, and for the second boat, it is 200 meters.
3Step 3: Calculate \(\tan x\) for the first boat
Using the definition of tangent in a right triangle, \(\tan x = \frac{\text{opposite}}{\text{adjacent}}\). For the first boat, this is \(\tan x = \frac{80}{100} = 0.8\).
4Step 4: Calculate \(\tan y\) for the second boat
Similarly, for the second boat, \(\tan y = \frac{\text{opposite}}{\text{adjacent}}= \frac{80}{200} = 0.4\).
5Step 5: Calculate the tangent of the difference of angles \(x\) and \(y\)
The tangent of the angle between two paths is given by the formula \(\tan(x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}\). Substitute the values \(\tan x = 0.8\) and \(\tan y = 0.4\): \[\tan(x-y) = \frac{0.8 - 0.4}{1 + 0.8 \times 0.4} = \frac{0.4}{1 + 0.32} = \frac{0.4}{1.32} \approx 0.303\].
Key Concepts
Right TriangleTangent FunctionAngle CalculationMath Problem Solving
Right Triangle
In trigonometry, the right triangle is a fundamental shape. It consists of two legs that form a right angle (90 degrees) and a hypotenuse, which is the side opposite the right angle. As we explore problems involving right triangles, it's important to understand the roles of each side:
- The hypotenuse is the longest side of a right triangle.
- The opposite side is the side opposite the angle of interest.
- The adjacent side is the side next to the angle of interest, but not the hypotenuse.
Tangent Function
The tangent function is one of the primary functions in trigonometry, playing a crucial role in solving right triangle problems. It relates the measures of angles to the lengths of the opposite and adjacent sides of a right triangle. For an angle \( \theta \) in a right triangle:
- The tangent is the ratio of the length of the opposite side to the length of the adjacent side.
- Mathematically, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).
Angle Calculation
When faced with right triangle problems, calculating the angle often involves finding the tangent and using inverse trigonometric functions for precise angle measures. The exercise shows how we compute the tangents of angles \( x \) and \( y \):
- For the first boat, \( \tan x = \frac{80}{100} = 0.8 \).
- For the second boat, \( \tan y = \frac{80}{200} = 0.4 \).
Math Problem Solving
Effective math problem solving in trigonometry requires clear understanding, breaking the problem into manageable steps, and applying mathematical principles correctly. Each step should:
- Clarify the problem, e.g., identifying right triangles as in the boat paths.
- Utilize known formulas, like the tangent function for angle calculations.
- Consider all given data, such as the river’s width and downstream distances.
Other exercises in this chapter
Problem 25
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 b
View solution Problem 25
In \(3-26,\) prove that each equation is an identity. $$ \frac{\csc \theta}{\sin \theta}-\cot ^{2} \theta=1 $$
View solution Problem 26
In \(3-26,\) prove that each equation is an identity. $$ \frac{\cos \theta}{\sec \theta}+\frac{\sin \theta}{\csc \theta}=1 $$
View solution Problem 27
For what values of \(\theta\) is the identity \(\frac{\cos \theta}{\sec \theta}+\frac{\sin \theta}{\csc \theta}=1\) undefined?
View solution