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Chapter 1: Problem 14

A hill that has a 12.0\(\%\) grade is one that rises 12.0 \(\mathrm{m}\) vertically for every 100.0 \(\mathrm{m}\) of distance in the horizontal direction. At what angle is such a hill inclined above the horizontal?

Short Answer

Expert verified
The hill is inclined at approximately 6.84 degrees above the horizontal.

Step by step solution

01

Understand the Problem

The problem gives us the grade of the hill, which is 12.0%. This means that for every 100 meters of horizontal distance, the hill rises 12 meters vertically. We are tasked with finding the angle of inclination above the horizontal.
02

Identify the Right Triangle

We can visualize the hill as forming a right triangle, where the vertical rise is 12 m, the horizontal run is 100 m, and the angle we need to find is the angle of inclination, \( \theta \), above the horizontal.
03

Use the Tangent Function

The angle of inclination \( \theta \) can be calculated using the tangent function in trigonometry, where \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). Here, the opposite side is the vertical rise (12 m), and the adjacent side is the horizontal distance (100 m).
04

Calculate \( \tan(\theta) \)

Calculate \( \tan(\theta) \) using the values: \( \tan(\theta) = \frac{12}{100} = 0.12 \). This value represents the tangent of the angle of inclination.
05

Find the Angle

To find \( \theta \), we need to find the angle whose tangent is 0.12. Use the inverse tangent function (also known as arctan or \( \tan^{-1} \)): \( \theta = \tan^{-1}(0.12) \).
06

Use a Calculator

Using a calculator, compute \( \theta = \tan^{-1}(0.12) \). This yields an angle of approximately 6.84 degrees.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angle of Inclination
When we talk about the angle of inclination, we're basically looking at how steep a slope is compared to the horizontal plane. It's like looking at a road and asking, "How steep is this part of the road?" To determine this, we consider how much the hill rises vertically compared to how much horizontal distance it covers. In our example, for every 100 meters you go forward, the hill climbs 12 meters up. This concept helps us visualize and quantify the slope's steepness in terms you can easily measure using degrees.
Trigonometry
Trigonometry is a branch of mathematics that studies the relationships between the sides and angles of triangles. It's incredibly useful for problems involving slopes, like the hill grade scenario. Imagine a right triangle drawn with the hill's slope as one side. Trigonometry gives us tools to find unknown angles when some side lengths are known. By using trigonometric functions like sine, cosine, and tangent, we can solve problems involving triangular shapes, whether they come from a physical hill or are drawn on paper.
Tan Function
The tangent, or tan, function is one of the primary functions in trigonometry. It relates an angle in a right triangle to the ratio of the opposite side to the adjacent side. Mathematically, it's expressed as \[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]In our hill example, the opposite side is 12 meters (vertical rise), and the adjacent side is 100 meters (horizontal run). By inserting these values into the tangent function, you get \[ \tan(\theta) = \frac{12}{100} \]This value helps us find how steep our hill is by looking at the angle \( \theta \). The tangent function is crucial for translating linear measurements into angular terms.
Right Triangle Calculation
Right triangles are the cornerstone of trigonometry, and they're handy for visualizing slopes like the hill in our problem. A right triangle is a triangle with one angle of 90 degrees. When you break down a slope or any inclined surface into a right triangle, you can use the properties of right triangles to find missing angles or sides.
  • The longest side, opposite the right angle, is the hypotenuse.
  • The other two sides are known as the opposite (vertical rise) and the adjacent (horizontal run).
For the inclined hill, we use these sides to create a right triangle where the angle of inclination \( \theta \) is what we're solving for. Using the inverse tangent, which is another trigonometric tool, you can find \( \theta \), helping you piece together the real-world inclination from simple vertical and horizontal measurements.

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Most popular questions from this chapter

The route followed by a hiker consists of three displacement vectors \(\overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{B}},\) and \(\overrightarrow{\mathbf{C}}\) . Vector \(\overrightarrow{\mathbf{A}}\) is along a measured trail and is 1550 \(\mathrm{m}\) in a direction \(25.0^{\circ}\) north of east. Vector \(\overrightarrow{\mathbf{B}}\) is not along a measured trail, but the hiker uses a compaass and knows that the direction is \(41.0^{\circ}\) east of south. Similarly, the direction of vector \(\overrightarrow{\mathbf{C}}\) is \(35.0^{\circ}\) north of west. The hiker ends up back where she started. Therefore, it follows that the resultant displace- ment is zero, or \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}=0 .\) Find the magnitudes of \((\mathbf{a})\) vector \(\overrightarrow{\mathbf{B}}\) and \(\quad(\mathbf{b})\) vector \(\overrightarrow{\mathbf{C}}\)

Vector \(\vec{A}\) has a magnitude of 12.3 units and points due west. Vector \(\overrightarrow{\mathbf{B}}\) points due north. \(\quad\) (a) What is the magnitude of \(\overrightarrow{\mathbf{B}}\) if \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) has a magnitude of 15.0 units? (b) What is the direction of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) relative to due west? \(\quad\) (c) What is the magnitude of \(\overrightarrow{\mathbf{B}}\) if \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) has a magnitude of 15.0 units? (d) What is the direction of \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) relative to due west?

A student sees a newspaper ad for an apartment that has 1330 square feet \(\left(f t^{2}\right)\) of floor space. How many square meters of area are there?

Vector \(\overrightarrow{\mathrm{A}}\) has a magnitude of 63 units and points due west, while vector \(\overrightarrow{\mathbf{B}}\) has the same magnitude and points due south. Find the magnitude and direction of \((\mathbf{a}) \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) and \((\mathbf{b}) \overline{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) . Specify the directions relative to due west.

Vector \(\overrightarrow{\mathbf{A}}\) has a magnitude of 6.00 units and points due east. Vector \(\overrightarrow{\mathbf{B}}\) points due north. (a) What is the magnitude of \(\overrightarrow{\mathbf{B}},\) if the vector \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) points \(60.0^{\circ}\) north of east? (b) Find the magnitude of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) .
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