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Chapter 5: Problem 42

A plane flies 810 miles from Franklin to Centerville with a bearing of \(75^{\circ}\). Then it flies 648 miles from Centerville to Rosemount with a bearing of \(32^{\circ}\). Draw a figure that visually represents the situation, and find the straight-line distance and bearing from Franklin to Rosemount.

Short Answer

Expert verified
After all calculations, we find the straight-line distance from Franklin to Rosemount and the required bearing. The distance could slightly vary due to the rounding of intermediate results, and the bearing must be a number between \(0^{\circ}\) to \(360^{\circ}\).

Step by step solution

01

Visual Representation

First of all, draw a diagram to visualize the situation. To represent the situation in the plane, denote Franklin as \(F\), Centerville as \(C\), and Rosemount as \(R\). Draw a direct line \(FR\) from Franklin to Rosemount. Now, we have two vectors \(FC\) and \(CR\), and we need to find the third vector \(FR\). The angle between \(FC\) and \(CR\) can be found by subtracting \(32^{\circ}\) from \(75^{\circ}\) which equals \(43^{\circ}\).
02

Calculate the Distance

Use the law of cosines to calculate the distance \(FR\). The law of cosines states that in any triangle, the square of any side is equal to the sum of squares of the other two sides minus twice the product of these two sides and the cosine of the included angle. So, apply this formula, \(FR^{2} = FC^{2} + CR^{2} - 2 \cdot FC \cdot CR \cdot cos(43^{\circ})\). Substituting the known values and solving will provide the length of \(FR\).
03

Calculate the Bearing

Now, we will calculate the bearing of \(FR\) from \(F\). For this, use the law of sines which states that the ratio of the length of a side of a triangle to the sine of the angle opposite is constant. So, applying this to triangle \(FCR\), we get \(sin(\angle{FRC}) = \frac{CR \cdot sin(\angle{FCR})}{FR}\). Calculate \(\angle{FRC}\) from the equation. The bearing angle from \(F\) to \(R\) is \(75^{\circ} - \angle{FRC}\).

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Key Concepts

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

Law of Sines
The Law of Sines is an essential principle in trigonometry, relating the lengths of sides of a triangle to the sines of its angles. This law can be stated as: \[\begin{equation}\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}\end{equation}\]where \( a \), \( b \), and \( c \) are the sides of the triangle, and \( A \), \( B \), and \( C \) are the angles opposite to these sides, respectively. When using the Law of Sines to find unknown angles or sides, it's essential to have either two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA) known.

In the context of our exercise, the Law of Sines helps to determine the bearing from Franklin to Rosemount after the plane has made its two trips. Once the side lengths are known through the Law of Cosines, the Law of Sines can be used to calculate the missing angles and thus, the final bearing required for navigation.

Application in Navigation

While navigating, identifying the correct bearing or direction is crucial. Using the Law of Sines, navigators can calculate precise angles required for plotting their course. In scenarios where GPS is not available, such knowledge becomes invaluable for pilots and sailors alike.
Bearing in Navigation
In navigation, bearing refers to the direction or path along which one moves or aims to reach a specific destination and is usually measured in degrees from a reference direction. Bearings are used to determine direction in navigation, commonly with respect to the compass north. There are different notations for bearings, such as true bearing, which is measured relative to true north, and magnetic bearing, measured concerning magnetic north.

The bearing is critical for accurately plotting the course of travel. In the exercise, two separate bearings are given from Franklin to Centerville and from Centerville to Rosemount. The bearings are based on compass directions with north as the reference point, and the angles are indicated in degrees.

Calculating Bearing Between Two Points

To find the bearing from Franklin to Rosemount, it's vital to first understand the triangle formed by the two legs of the flight (Franklin to Centerville and Centerville to Rosemount) and the direct path (Franklin to Rosemount). By calculating this direct path and the angle it makes with the north, the required bearing can be determined, guiding pilots along their journey.
Trigonometry in Precalculus
Trigonometry plays a foundational role in precalculus, as it introduces students to the relationship between angles and side lengths in triangles. This includes understanding various functions such as sine, cosine, and tangent, along with the rules and laws that relate these values, like the Law of Sines and the Law of Cosines.

Having a strong grasp on trigonometry is crucial for problem-solving in precalculus, as it provides the tools for analyzing and understanding geometric and real-world problems. In our exercise, trigonometry is used to calculate distances and angles based on the given bearings, leading to the application of both the Law of Cosines and the Law of Sines to find a direct path and its bearing.

Real-World Applications

Trigonometry is not just a mathematical concept but is widely used in various professions, including astronomy, engineering, physics, and as shown in our exercise, aviation navigation. Students will often see these trigonometric principles come to life in numerous applications, encouraging a deeper understanding and appreciation of its versatility and utility in everyday situations.

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Most popular questions from this chapter

A pilot has just started on the glide path for landing at an airport with a runway of length 9000 feet. The angles of depression from the plane to the ends of the runway are \(17.5^{\circ}\) and \(18.8^{\circ}\). (a) Draw a diagram that visually represents the situation. (b) Find the air distance the plane must travel until touching down on the near end of the runway. (c) Find the ground distance the plane must travel until touching down. (d) Find the altitude of the plane when the pilot begins the descent.

Verify the identity. $$\sin \left(\frac{\pi}{6}+x\right)+\sin \left(\frac{\pi}{6}-x\right)=\cos x$$

A flagpole at a right angle to the horizontal is located on a slope that makes an angle of \(12^{\circ}\) with the horizontal. The flagpole's shadow is 16 meters long and points directly up the slope. The angle of elevation from the tip of the shadow to the sun is \(20^{\circ}\). (a) Draw a triangle to represent the situation. Show the known quantities on the triangle and use a variable to indicate the height of the flagpole. (b) Write an equation that can be used to find the height of the flagpole. (c) Find the height of the flagpole.

The length \(s\) of a shadow cast by a vertical gnomon (a device used to tell time) of height \(h\) when the angle of the sun above the horizon is \(\theta\) (see figure) can be modeled by the equation \(s=\frac{h \sin \left(90^{\circ}-\theta\right)}{\sin \theta}\) (a) Verify that the equation for \(s\) is equal to \(h \cot \theta\). (b) Use a graphing utility to complete the table. Let \(h=5\) feet. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline \theta & 15^{\circ} & 30^{\circ} & 45^{\circ} & 60^{\circ} & 75^{\circ} & 90^{\circ} \\ \hline s & & & & & & \\ \hline \end{array} $$ (c) Use your table from part (b) to determine the angles of the sun that result in the maximum and minimum lengths of the shadow. (d) Based on your results from part (c), what time of day do you think it is when the angle of the sun above the horizon is \(90^{\circ} ?\)

Let \(R\) and \(r\) be the radii of the circumscribed and inscribed circles of a triangle \(A B C\), respectively (see figure), and let \(s=\frac{a+b+c}{2}\). (a) Prove that \(2 R=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). (b) Prove that \(r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\).
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