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# 4.4: Applications & Models

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Apply inverse trigonometric functions to real life situations.

The following problems are real-world problems that can be solved using the trigonometric functions. In everyday life, indirect measurement is used to obtain answers to problems that are impossible to solve using measurement tools. However, mathematics will come to the rescue in the form of trigonometry to calculate these unknown measurements.

Example 1: On a cold winter day the sun streams through your living room window and causes a warm, toasty atmosphere. This is due to the angle of inclination of the sun which directly affects the heating and the cooling of buildings. Noon is when the sun is at its maximum height in the sky and at this time, the angle is greater in the summer than in the winter. Because of this, buildings are constructed such that the overhang of the roof can act as an awning to shade the windows for cooling in the summer and yet allow the sun’s rays to provide heat in the winter. In addition to the construction of the building, the angle of inclination of the sun varies according to the latitude of the building’s location.

If the latitude of the location is known, then the following formula can be used to calculate the angle of inclination of the sun on any given date of the year:

Angle of sun=90latitude+23.5cos[(N+10)360365]\begin{align*}\text{Angle of sun} = 90^\circ - \text{latitude} + -23.5^\circ \cdot \cos \left [ (N+10) \frac{360}{365} \right ]\end{align*} where N\begin{align*}N\end{align*} represents the number of the day of the year that corresponds to the date of the year. Note: This formula is accurate to ±12\begin{align*}\pm \frac{1^\circ}{2}\end{align*}

a. Determine the measurement of the sun’s angle of inclination for a building located at a latitude of 42\begin{align*}42^\circ\end{align*}, March 10th\begin{align*}10^{th}\end{align*}, the 69th\begin{align*}69^{th}\end{align*} day of the year.

Solution:

Angle of sunAngle of sunAngle of sunAngle of sun=9042+23.5cos[(69+10)360365]=48+23.5(0.2093)=484.92=43.08\begin{align*}\text{Angle of sun} & = 90^\circ - 42^\circ + -23.5^\circ \cdot \cos \left [(69+10) \frac{360}{365} \right ]\\ \text{Angle of sun} & = 48^\circ + -23.5^\circ (0.2093)\\ \text{Angle of sun} & = 48^\circ - 4.92^\circ\\ \text{Angle of sun} & = 43.08^\circ\end{align*}

b. Determine the measurement of the sun’s angle of inclination for a building located at a latitude of 20\begin{align*}20^\circ\end{align*}, September 21st\begin{align*}21^{st}\end{align*}.

Solution:

Angle of sunAngle of sunAngle of sun=9020+23.5cos[(264+10)360365]=70+23.5(0.0043)=70.10\begin{align*}\text{Angle of sun} & = 90^\circ - 20^\circ + -23.5^\circ \cdot \cos \left [ (264+10) \frac{360}{365} \right ]\\ \text{Angle of sun} & = 70^\circ + -23.5^\circ (0.0043)\\ \text{Angle of sun} & = 70.10^\circ\end{align*}

Example 2: A tower, 28.4 feet high, must be secured with a guy wire anchored 5 feet from the base of the tower. What angle will the guy wire make with the ground?

Solution: Draw a picture.

tanθtanθtanθtan1(tanθ)θ=opp.adj.=28.45=5.68=tan1(5.68)=80.02\begin{align*}\tan \theta & = \frac{opp.}{adj.}\\ \tan \theta & = \frac{28.4}{5}\\ \tan \theta & = 5.68\\ \tan^{-1} (\tan \theta) & = \tan^{-1}(5.68)\\ \theta & = 80.02^\circ\end{align*}

The following problem that involves functions and their inverses will be solved using the property f(f1(x))=f1(f(x))\begin{align*}f(f^{-1}(x)) = f^{-1}(f(x))\end{align*}. In addition, technology will also be used to complete the solution.

Example 3: In the main concourse of the local arena, there are several viewing screens that are available to watch so that you do not miss any of the action on the ice. The bottom of one screen is 3 feet above eye level and the screen itself is 7 feet high. The angle of vision (inclination) is formed by looking at both the bottom and top of the screen.

a. Sketch a picture to represent this problem.

b. Calculate the measure of the angle of vision that results from looking at the bottom and then the top of the screen. At what distance from the screen does the maximum value for the angle of vision occur?

Solution:

a.

b. θ2tanθθ2=tanθtanθ1=10x and tanθ1=3x=tan1(10x)tan1(3x)\begin{align*}\theta_2 &= \tan \theta - \tan \theta_1\\ \tan \theta &= \frac{10}{x} \ \text{and} \ \tan \theta_1 = \frac{3}{x}\\ \theta_2 &= \tan^{-1} \left ( \frac{10}{x} \right ) - \tan^{-1} \left ( \frac{3}{x} \right )\end{align*}

To determine these values, use a graphing calculator and the trace function to determine when the actual maximum occurs.

From the graph, it can be seen that the maximum occurs when x5.59 ft\begin{align*}x \approx 5.59 \ ft\end{align*}. and θ32.57\begin{align*}\theta \approx 32.57^\circ\end{align*}.

Example 4: A silo is 40 feet high and 12 feet across. Find the angle of depression from the top edge of the silo to the floor of the opposite edge.

Solution: tanθ=4012θ=tan14012=73.3\begin{align*}\tan \theta = \frac{40}{12} \rightarrow \theta = \tan^{-1} \frac{40}{12} = 73.3^\circ\end{align*}

Example 5: The pilot of an airplane flying at an elevation of 5000 feet sights two towers that are 300 feet apart. If the angle between the point directly below him and the base of the tower closer to him is 30\begin{align*}30^\circ\end{align*}, determine the angle y\begin{align*}y\end{align*} between the bases of the two towers.

Solution: Draw a picture. First we need to find x\begin{align*}x\end{align*} in order to find y\begin{align*}y\end{align*}.

tan30tan(30+y)tan(30+y)30+y30+yy=x5000x=5000tan30x=2886.75=2886.75+3005000=3186.755000=tan13186.755000=32.51=2.51\begin{align*}\tan 30^\circ & = \frac{x}{5000} \rightarrow x = 5000 \tan 30^\circ \rightarrow x = 2886.75\\ \tan (30^\circ + y) & = \frac{2886.75+300}{5000}\\ \tan (30^\circ + y) & = \frac{3186.75}{5000}\\ 30^\circ +y & = \tan^{-1} \frac{3186.75}{5000} \\ 30^\circ + y & = 32.51^\circ \\ y & = 2.51^\circ\end{align*}

Which means that the bases of the two towers are 2.51\begin{align*}2.51^\circ\end{align*} apart.

## Review Questions

1. The intensity of a certain type of polarized light is given by the equation I=I0sin2θcos2θ\begin{align*}I = I_0 \sin 2\theta \cos 2\theta\end{align*}. Solve for θ\begin{align*}\theta\end{align*}.
2. The following diagram represents the ends of a water trough. The ends are actually isosceles trapezoids, and the length of the trough from end-to-end is ten feet. Determine the maximum volume of the trough and the value of θ\begin{align*}\theta\end{align*} that maximizes that volume.
3. A boat is docked at the end of a 10 foot pier. The boat leaves the pier and drops anchor 230 feet away 3 feet straight out from shore (which is perpendicular to the pier). What was the bearing of the boat from a line drawn from the end of the pier through the foot of the pier?
4. The electric current in a certain circuit is given by i=Im[sin(wt+α)cosφ+cos(wt+α)sinφ]\begin{align*}i = I_m[\sin(wt + \alpha) \cos \varphi + \cos(wt + \alpha) \sin \varphi]\end{align*} Solve for t\begin{align*}t\end{align*}.
5. Using the formula from Example 1, determine the measurement of the sun’s angle of inclination for a building located at a latitude of:
1. 64\begin{align*}64^\circ\end{align*} on the 16th\begin{align*}16^{th}\end{align*} of November
2. 15\begin{align*}15^\circ\end{align*} on the 8th\begin{align*}8^{th}\end{align*} of August
6. A ship leaves port and travels due east 15 nautical miles, then changes course to N 20 W\begin{align*}N\ 20^\circ\ W\end{align*} and travels 40 more nautical miles. Find the bearing to the port of departure.
7. Find the maximum displacement for the simple harmonic motion described by d=4cosπt\begin{align*}d = 4 \cos \pi t\end{align*}.
8. The pilot of an airplane flying at an elevation of 10,000 feet sights two towers that are 500 feet apart. If the angle between the point directly below him and the base of the tower closer to him is 18\begin{align*}18^\circ\end{align*}, determine the angle y\begin{align*}y\end{align*} between the bases of the two towers.

1. III0II02II02II0sin12II014sin12II0=I0sin2θcos2θ=I0I0sin2θcos2θ=sin2θcos2θ=2sin2θcos2θ=sin4θ=4θ=θ\begin{align*}I & = I_0 \sin2\theta \cos 2\theta\\ \frac{I}{I_0} & = \frac{I_0}{I_0} \sin 2\theta \cos 2\theta\\ \frac{I}{I_0} & = \sin 2\theta \cos 2\theta\\ \frac{2I}{I_0} & = 2 \sin 2\theta \cos 2\theta\\ \frac{2I}{I_0} & = \sin 4\theta\\ \sin^{-1}\frac{2I}{I_0} & = 4\theta\\ \frac{1}{4} \sin^{-1}\frac{2I}{I_0} & = \theta\end{align*}
2. The volume is 10 feet times the area of the end. The end consists of two congruent right triangles and one rectangle. The area of each right triangle is \begin{align*}\frac{1}{2} (\sin \theta) (\cos \theta)\end{align*} and that of the rectangle is \begin{align*}(1)(\cos \theta)\end{align*}. This means that the volume can be determined by the function \begin{align*}V(\theta)=10(\cos \theta + \sin \theta \cos \theta)\end{align*}, and this function can be graphed as follows to find the maximum volume and the angle \begin{align*}\theta\end{align*} where it occurs. Therefore, the maximum volume is approximately 13 cubic feet and occurs when \begin{align*}\theta\end{align*} is about \begin{align*}30^\circ\end{align*}.
3. \begin{align*}\cos x & = \frac{7}{230} \rightarrow x = \cos^{-1} \frac{7}{230}\\ x & = 88.26^\circ\end{align*}
4. \begin{align*}i & = I_m[\sin(wt + \alpha) \cos \varphi + \cos(wt + \alpha) \sin \varphi]\\ \frac{i}{I_m} & = \underbrace{\sin(wt+\alpha)\cos \varphi + \cos(wt+\alpha)\sin \varphi}_{\sin(wt+\alpha+\varphi)}\\ \frac{i}{I_m} & = \sin(wt+\alpha+\varphi)\\ \sin^{-1} \frac{i}{I_m} & = wt + \alpha + \varphi\\ \sin^{-1} \frac{i}{I_m}-\alpha-\varphi & = wt\\ \frac{1}{w} \left ( \sin^{-1} \frac{i}{I_m}-\alpha-\varphi \right ) & = t\end{align*}
1. \begin{align*}64^\circ\end{align*} on the \begin{align*}16^{th}\end{align*} of November \begin{align*}= 90^\circ-64^\circ-23.5^\circ \cos\left [ (320+10) \frac{360}{365} \right ] = 6.64^\circ\end{align*}
2. \begin{align*}15^\circ\end{align*} on the \begin{align*}8^{th}\end{align*} of August \begin{align*}= 90^\circ-15^\circ-23.5^\circ \cos\left [ (220+10) \frac{360}{365} \right ] = 91.07^\circ\end{align*}
5. We need to find \begin{align*}y\end{align*} and \begin{align*}z\end{align*} before we can find \begin{align*}x^\circ\end{align*}. \begin{align*}\sin 70^\circ & = \frac{y}{40} \rightarrow y = 40\sin 70^\circ = 37.59\\ \cos 70^\circ & = \frac{z}{40} \rightarrow z = 40\cos 70^\circ = 13.68\end{align*} Using 15-13.68 as the adjacent side for \begin{align*}x\end{align*}, we can now find the missing angle. \begin{align*}\tan x^\circ = \frac{37.59}{1.32} = 28.48 \rightarrow x^\circ = \tan^{-1} (28.48) = 87.99^\circ\end{align*}. Therefore, the bearing from the ship back to the point of departure is \begin{align*}W 87.99^\circ S\end{align*}.
6. The maximum displacement for this equation is simply the amplitude, 4.
7. You can use the same picture from Example 5 for this problem. \begin{align*}\tan 18^\circ =\frac{x}{10,000} \rightarrow x = 10,000 \tan 18^\circ \rightarrow x = 3249.2\\ \tan (18^\circ +y) = \frac{3249.2+500}{10,000} \rightarrow \tan (18^\circ +y) = \frac{3749.2}{10,000} \rightarrow 18^\circ +y = \tan^{-1} \frac{3749.2}{10,000} \rightarrow 18^\circ +y = 20.55^\circ \rightarrow y=2.55^\circ\end{align*} So, the towers are \begin{align*}2.6^\circ\end{align*} apart.

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