1.3 Applications and Solving Right Triangles

Throughout its early development, trigonometry was often used as a means of indirect measurement, e.g. determining large distances or lengths by using measurements of angles and small, known distances. Today, trigonometry is widely used in physics, astronomy, engineering, navigation, surveying, and various fields of mathematics and other disciplines. In this section we will see some of the ways in which trigonometry can be applied. Your calculator should be in degree mode for these examples.

Example 1.11

A person stands $150$ ft away from a flagpole and measures an angle of elevation of $32\Degrees$ from his horizontal line of sight to the top of the flagpole. Assume that the person’s eyes are a vertical distance of 6 ft from the ground. What is the height of the flagpole? [r]

Solution: The picture on the right describes the situation. We see that the height of the flagpole is $h + 6$ ft, where

\begin{displaymath} \frac{h}{150} ~=~ \tan\;32\Degrees \quad\Rightarrow\quad h ~=~ 150\;\tan\;32\Degrees ~=~ 150\;(0.6249) ~=~ 94 ~. \end{displaymath}

How did we know that $\tan\;32\Degrees = 0.6249\,$? By using a calculator. And since none of the numbers we were given had decimal places, we rounded off the answer for $h$ to the nearest integer. Thus, the height of the flagpole is $\,h + 6 = 94 + 6 = \boxed{100 ~\text{ft}}$ .

Example 1.12

A person standing $400$ ft from the base of a mountain measures the angle of elevation from the ground to the top of the mountain to be $25\Degrees$. The person then walks $500$ ft straight back and measures the angle of elevation to now be $20\Degrees$. How tall is the mountain? [r]

Solution: We will assume that the ground is flat and not inclined relative to the base of the mountain. Let $h$ be the height of the mountain, and let $x$ be the distance from the base of the mountain to the point directly beneath the top of the mountain, as in the picture on the right. Then we see that

\begin{align*} \frac{h}{x + 400} ~=~ \tan\;25\Degrees \quad &\Rightarrow \quad h ~=~ (x + 400)\;\tan\;25\Degrees ~,~\text{and}\\[5pt] \frac{h}{x + 400 + 500} ~=~ \tan\;20\Degrees \quad &\Rightarrow \quad h ~=~ (x + 900)\;\tan\;20\Degrees ~,~\text{so} \end{align*}

$(x + 400)\;\tan\;25\Degrees ~=~ (x + 900)\;\tan\;20\Degrees$, since they both equal $h$. Use that equation to solve for $x$:

\begin{displaymath} x\;\tan\;25\Degrees ~-~ x\;\tan\;20\Degrees ~=~ 900\;\tan\;20\Degrees ~-~ 400\;\tan\;25\Degrees \quad\Rightarrow\quad x ~=~ \frac{900\;\tan\;20\Degrees ~-~ 400\;\tan\;25\Degrees}{\tan\;25\Degrees ~-~ \tan\;20\Degrees} ~=~ 1378~ \text{ft} \end{displaymath}

Finally, substitute $x$ into the first formula for $h$ to get the height of the mountain:

\begin{displaymath} h ~=~ (1378 + 400)\;\tan\;25\Degrees ~=~ 1778\;(0.4663) ~=~ \boxed{829~ \text{ft}} \end{displaymath}

Example 1.13

A blimp $4280$ ft above the ground measures an angle of depression of $24\Degrees$ from its horizontal line of sight to the base of a house on the ground. Assuming the ground is flat, how far away along the ground is the house from the blimp? [r]

Solution: Let $x$ be the distance along the ground from the blimp to the house, as in the picture to the right. Since the ground and the blimp’s horizontal line of sight are parallel, we know from elementary geometry that the angle of elevation $\theta$ from the base of the house to the blimp is equal to the angle of depression from the blimp to the base of the house, i.e. $\theta = 24\Degrees$. Hence,

\begin{displaymath} \frac{4280}{x} ~=~ \tan\;24\Degrees \quad\Rightarrow\quad x ~=~ \frac{4280}{\tan\;24\Degrees} ~=~ \boxed{9613 ~\text{ft}}~. \end{displaymath}

Example 1.14

An observer at the top of a mountain $3$ miles above sea level measures an angle of depression of $2.23\Degrees$ to the ocean horizon. Use this to estimate the radius of the earth.

tikzpicture[every node/.style=font=] [ball color=earthcolor,shading angle=-70,line width=0.5pt] (0,0) circle (2); [dashed] (0,0) -- (45:2) node[sloped,midway,below] r; [dashed] (0,0) -- (0,2) node[midway,left] r; (0,2) -- (0,2.5) node[midway,left] 3; [dashed] (0,2.5) -- (2.5,2.5); [dashed] (0,2.5) -- (2.5,0.635); [above] at (0,2.5) A; [above] at (2.5,2.5) B; [above right] at (45:2) H; [below] at (0,0) O; at (0.9,2.3) 2.23 ; [dashed,-latex] (1.3,2.5) arc (360:323:1.3); at (68:0.4) ; [dashed] (0,0.6) arc (90:45:0.6); [rotate=45] (1.75,0) -- (1.75,0.25) -- (1.98,0.25); (0,0) circle (2pt); (0,2.5) circle (2pt); (2.5,2.5) circle (2pt); (45:2) circle (2pt); tikzpicture — Figure 1.3.1

Solution: We will assume that the earth is a sphere.^[1] Let $r$ be the radius of the earth. Let the point $A$ represent the top of the mountain, and let $H$ be the ocean horizon in the line of sight from $A$, as in Figure 1.3.1. Let $O$ be the center of the earth, and let $B$ be a point on the horizontal line of sight from $A$ (i.e. on the line perpendicular to $\overline{OA}$). Let $\theta$ be the angle $\angle\,AOH$. Since $A$ is $3$ miles above sea level, we have $OA = r + 3$. Also, $OH = r$. Now since $\overline{AB} \perp \overline{OA}$, we have $\angle\,OAB = 90\Degrees$, so we see that $\angle\,OAH = 90\Degrees - 2.23\Degrees = 87.77\Degrees$. We see that the line through $A$ and $H$ is a tangent line to the surface of the earth (considering the surface as the circle of radius $r$ through $H$ as in the picture). So by Exercise 14 in Section 1.1, $\overline{AH} \perp \overline{OH}$ and hence $\angle\,OHA = 90\Degrees$. Since the angles in the triangle $\triangle\,OAH$ add up to $180\Degrees$, we have $\theta = 180\Degrees - 90\Degrees - 87.77\Degrees = 2.23\Degrees$. Thus,

\begin{displaymath} \cos\;\theta ~=~ \frac{OH}{OA} ~=~ \frac{r}{r+3} \quad\Rightarrow\quad \frac{r}{r+3} ~=~ \cos\;2.23\Degrees ~, \end{displaymath}

so solving for $r$ we get

\begin{align*} r ~=~ (r ~+~ 3)\;\cos\;2.23\Degrees \quad &\Rightarrow \quad r ~-~ r\;\cos\;2.23\Degrees ~=~ 3\;\cos\;2.23\Degrees\\[4pt] &\Rightarrow \quad r ~=~ \frac{3\;\cos\;2.23\Degrees}{1 ~-~ \cos\;2.23\Degrees}\\[5pt] &\Rightarrow \quad \boxed{r ~=~ 3958.3 ~\text{miles}} ~. \end{align*}

Note: This answer is very close to the earth’s actual (mean) radius of $3956.6$ miles.

Example 1.15

[r] tikzpicture[scale=0.8,every node/.style=font=] [line width=1pt] (0,0) circle (1.5); (0,0) -- (1.5,0) -- (1.5,3.5) -- cycle; (0,0) circle (2.5pt); [below] at (0,0) O; (1.5,0) circle (2.5pt); [right] at (1.5,0) A; (1.5,3.5) circle (2.5pt); [above] at (1.5,3.5) B; at (1.35,2.7) ; (1.5,2.4) arc (270:246.8:1.1); tikzpicture As another application of trigonometry to astronomy, we will find the distance from the earth to the sun. Let $O$ be the center of the earth, let $A$ be a point on the equator, and let $B$ represent an object (e.g. a star) in space, as in the picture on the right. If the earth is positioned in such a way that the angle $\angle\,OAB = 90\Degrees$, then we say that the angle $\alpha = \angle\,OBA$ is the equatorial parallax of the object. The equatorial parallax of the sun has been observed to be approximately $\alpha = 0.00244\Degrees$. Use this to estimate the distance from the center of the earth to the sun.

Solution: Let $B$ be the position of the sun. We want to find the length of $\overline{OB}$. We will use the actual radius of the earth, mentioned at the end of Example 1.14, to get $OA = 3956.6$ miles. Since $\angle\,OAB = 90\Degrees$, we have

\begin{displaymath} \frac{OA}{OB} ~=~ \sin\;\alpha \quad\Rightarrow\quad OB ~=~ \frac{OA}{\sin\;\alpha} ~=~ \frac{3956.6}{\sin\;0.00244\Degrees} ~=~ 92908394 ~, \end{displaymath}

so the distance from the center of the earth to the sun is approximately $93$ million miles .
Note: The earth’s orbit around the sun is an ellipse, so the actual distance to the sun varies.

In the above example we used a very small angle ($0.00244\Degrees$). A degree can be divided into smaller units: a minute is one-sixtieth of a degree, and a second is one-sixtieth of a minute. The symbol for a minute is $'$ and the symbol for a second is $''$. For example, $4.5\Degrees = 4\Degrees\;30'$. And $4.505\Degrees = 4\Degrees\;30'\;18''$:

\begin{displaymath} 4\Degrees\;30'\;18'' ~=~ 4 ~+~ \frac{30}{60} ~+~ \frac{18}{3600} ~\text{degrees} ~=~ 4.505\Degrees \end{displaymath}

In Example 1.15 we used $\alpha = 0.00244\Degrees \approx 8.8''$, which we mention only because some angle measurement devices do use minutes and seconds.

Example 1.16

[r] $tikzpicture[scale=1.1,every node/.style=font=] [line width=1pt] (3,0) circle (0.7); (0,0) circle (1.4pt); [left] at (0,0) E; (3,0) circle (1.4pt); [right] at (3,0) S; [dashed] (0,0) -- ([shift=(3,0)] 256.507:0.7) -- (3,0) -- ([shift=(3,0)] 103.493:0.7) -- (0,0) -- (3,0); [below] at ([shift=(3,0)] 256.507:0.7) A; [above] at ([shift=(3,0)] 103.493:0.7) B; [latex-latex] (-13.493:1.7) arc (-13.493:13.493:1.7); [fill=white] at (1.7,0) 32'\;4"; tikzpicture$ An observer on earth measures an angle of $32'\;4''$ from one visible edge of the sun to the other (opposite) edge, as in the picture on the right. Use this to estimate the radius of the sun.

Solution: Let the point $E$ be the earth and let $S$ be the center of the sun. The observer’s lines of sight to the visible edges of the sun are tangent lines to the sun’s surface at the points $A$ and $B$. Thus, $\angle\,EAS = \angle\,EBS = 90\Degrees$. The radius of the sun equals $AS$. Clearly $AS = BS$. So since $EB = EA$ (why?), the triangles $\triangle\,EAS$ and $\triangle\,EBS$ are similar. Thus, $\angle\,AES = \angle\,BES = \frac{1}{2}\; \angle\,AEB = \frac{1}{2}\;(32'\;4'') = 16'\;2'' = (16/60) + (2/3600) = 0.26722\Degrees$.

Now, $ES$ is the distance from the surface of the earth (where the observer stands) to the center of the sun. In Example 1.15 we found the distance from the center of the earth to the sun to be $92,908,394$ miles. Since we treated the sun in that example as a point, then we are justified in treating that distance as the distance between the centers of the earth and sun. So $ES = 92908394 - ~\text{radius of earth} = 92908394 - 3956.6 = 92904437.4$ miles. Hence,

\begin{displaymath} \sin\;(\angle\,AES) ~=~ \frac{AS}{ES} \quad\Rightarrow\quad AS ~=~ ES \;\sin\;0.26722\Degrees ~=~ (92904437.4)\;\sin\;0.26722\Degrees ~=~ \boxed{433,293 ~\text{miles}} ~. \end{displaymath}

Note: This answer is close to the sun’s actual (mean) radius of $432,200$ miles.

You may have noticed that the solutions to the examples we have shown required at least one right triangle. In applied problems it is not always obvious which right triangle to use, which is why these sorts of problems can be difficult. Often no right triangle will be immediately evident, so you will have to create one. There is no general strategy for this, but remember that a right triangle requires a right angle, so look for places where you can form perpendicular line segments. When the problem contains a circle, you can create right angles by using the perpendicularity of the tangent line to the circle at a point^[2] with the line that joins that point to the center of the circle. We did exactly that in Examples 1.14, 1.15, and 1.16.

Example 1.17

[r] The machine tool diagram on the right shows a symmetric V-block, in which one circular roller sits on top of a smaller circular roller. Each roller touches both slanted sides of the V-block. Find the diameter $d$ of the large roller, given the information in the diagram.

Solution: The diameter $d$ of the large roller is twice the radius $OB$, so we need to find $OB$. To do this, we will show that $\triangle\,OBC$ is a right triangle, then find the angle $\angle\,BOC$, and then find $BC$. The length $OB$ will then be simple to determine.

Since the slanted sides are tangent to each roller, $\angle\,ODA = \angle\,PEC = 90\Degrees$. By symmetry, since the vertical line through the centers of the rollers makes a $37\Degrees$ angle with each slanted side, we have $\angle\,OAD = 37\Degrees$. Hence, since $\triangle\,ODA$ is a right triangle, $\angle\,DOA$ is the complement of $\angle\,OAD$. So $\angle\,DOA = 53\Degrees$.

Since the horizontal line segment $\overline{BC}$ is tangent to each roller, $\angle\,OBC = \angle\,PBC = 90\Degrees$. Thus, $\triangle\,OBC$ is a right triangle. And since $\angle\,ODA = 90\Degrees$, we know that $\triangle\,ODC$ is a right triangle. Now, $OB = OD$ (since they each equal the radius of the large roller), so by the Pythagorean Theorem we have $BC = DC$:

\begin{displaymath} BC^2 ~=~ OC^2 ~-~ OB^2 ~=~ OC^2 ~-~ OD^2 ~=~ DC^2 \quad\Rightarrow\quad BC ~=~ DC \end{displaymath}

Thus, $\triangle\,OBC$ and $\triangle\,ODC$ are congruent triangles (which we denote by $\triangle\,OBC \cong \triangle\,ODC$), since their corresponding sides are equal. Thus, their corresponding angles are equal. So in particular, $\angle\,BOC = \angle\,DOC$. We know that $\angle\,DOB = \angle\,DOA = 53\Degrees$. Thus,

\begin{displaymath} 53\Degrees ~=~ \angle\,DOB ~=~ \angle\,BOC ~+~ \angle\,DOC = \angle\,BOC ~+~ \angle\,BOC ~=~ 2\;\angle\,BOC \quad\Rightarrow\quad \angle\,BOC ~=~ 26.5\Degrees ~. \end{displaymath}

Likewise, since $BP = EP$ and $\angle\,PBC = \angle\,PEC = 90\Degrees$, $\triangle\,BPC$ and $\triangle\,EPC$ are congruent right triangles. Thus, $BC = EC$. But we know that $BC = DC$, and we see from the diagram that $EC + DC = 1.38$. Thus, $BC + BC = 1.38$ and so $BC = 0.69$. We now have all we need to find $OB$:

\begin{displaymath} \frac{BC}{OB} ~=~ \tan\;\angle\,BOC \quad\Rightarrow\quad OB ~=~ \frac{BC}{\tan\;\angle\,BOC} ~=~ \frac{0.69}{\tan\;26.5\Degrees} ~=~ 1.384 \end{displaymath}

Hence, the diameter of the large roller is $\,d = 2 \times OB = 2\,(1.384) = \boxed{2.768}$ .

Example 1.18

A slider-crank mechanism is shown in Figure 1.3.2 below. As the piston moves downward the connecting rod rotates the crank in the clockwise direction, as indicated.

tikzpicture[every node/.style=font=] [dashed] (0,-7.5) circle (2.5); [fill=white,line width=1pt,rounded corners,rotate around=15.255:(0,0)] (240:0.9) arc (240:-60:0.9) -- ([shift=(0,-5.7)] 37.875:1.1) arc (37.875:-217.875:1.1) -- cycle; [fill=white,line width=1pt,rotate around=53.13:(0,-7.5)] (0,-6.6) -- (2.5,-6.6) arc (90:-90:0.9) -- (0,-8.4) arc (270:90:0.9); [dashed] (0,0) -- (1.5,-5.5) node[black,pos=0.55,sloped,below] connecting rod node[midway,above right] b; [line width=1pt] (0,0) circle (0.6); (0,0) circle (2pt); [above] at (0,0) A; [line width=1pt] (1.5,-5.5) circle (0.6); (1.5,-5.5) circle (2pt); [below] at (1.5,-5.5) B; [line width=1pt] (0,-7.5) circle (0.6); (0,-7.5) circle (2pt); [below] at (0,-7.5) O; [dashed] (0,-7.5) -- (1.5,-5.5) node[sloped,midway,below] crank; [dashed] (0,0) -- (5.625,0) node[above,pos=0.7] a; [dashed] (1.5,-5.5) -- (5.625,0) node[midway,right] c; (5.625,0) circle (2pt); [right] at (5.625,0) C; [dashed,latex-latex] (-2.5,-7.5) -- (-0.05,-7.5) node[pos=0.4,fill=white] r; [dashed] (0,-7.5) -- (0,0); [dashed,-latex] (0,-6.4) arc (90:53.13:1.1); at (0.4,-6.25) ; [line width=1pt] (-1.3,1.2) -- (1.3,1.2) -- (1.3,-1) -- (1.7,-1) -- (1.7,2) -- (-1.7,2) -- (-1.7,-1) -- (-1.3,-1) -- cycle; at (0,1.6) piston; [pattern color=black,pattern=north east lines] (-2,2.4) -- (-1.8,2.4) -- (-1.8,-1.4) -- (-2,-1.4) -- cycle; [line width=1pt] (-1.8,2.4) -- (-1.8,-1.4); [pattern color=black,pattern=north east lines] (2,2.4) -- (1.8,2.4) -- (1.8,-1.4) -- (2,-1.4) -- cycle; [line width=1pt] (1.8,2.4) -- (1.8,-1.4); [linecolor,line width=1.5pt,-latex] (0,2.6) -- (0,2.1); [linecolor,line width=1.5pt,-latex] (2.7,-7.5) arc (360:330:2.7); tikzpicture — Figure 1.3.2Slider-crank mechanism

The point $A$ is the center of the connecting rod’s wrist pin and only moves vertically. The point $B$ is the center of the crank pin and moves around a circle of radius $r$ centered at the point $O$, which is directly below $A$ and does not move. As the crank rotates it makes an angle $\theta$ with the line $\overline{OA}$. The instantaneous center of rotation of the connecting rod at a given time is the point $C$ where the horizontal line through $A$ intersects the extended line through $O$ and $B$. From Figure 1.3.2 we see that $\angle\,OAC = 90\Degrees$, and we let $a = AC$, $b = AB$, and $c = BC$. In the exercises you will show that for $0\Degrees < \theta < 90\Degrees$,

\begin{displaymath} c ~=~ \frac{\sqrt{b^2 ~-~ r^2 \;(\sin\,\theta)^2}}{\cos\;\theta} ~\qquad\text{and}\qquad~ a ~=~ r\;\sin\;\theta ~+~ \sqrt{b^2 ~-~ r^2 \;(\sin\,\theta)^2}~\tan\;\theta ~. \end{displaymath}

[r] $tikzpicture[scale=0.5,every node/.style=font=] [fill=fillcolor] (0,0) -- (4,0) -- (4,3) -- (0,0); at (1.05,0.38) ; (1.5,0) arc (0:37:1.5); [line width=0.5pt] (3.6,0) -- (3.6,0.4) -- (4,0.4); [linecolor,line width=1.5pt] (0,0) -- (4,0) -- (4,3) -- cycle; [below] at (2,0) r\; \; ; [right] at (4,1.5) r\; \; ; [above left] at (2,1.5) r; tikzpicture$ For some problems it may help to remember that when a right triangle has a hypotenuse of length $r$ and an acute angle $\theta$, as in the picture on the right, the adjacent side will have length $r\,\cos\;\theta$ and the opposite side will have length $r\,\sin\;\theta$. You can think of those lengths as the horizontal and vertical “components” of the hypotenuse.

Notice in the above right triangle that we were given two pieces of information: one of the acute angles and the length of the hypotenuse. From this we determined the lengths of the other two sides, and the other acute angle is just the complement of the known acute angle. In general, a triangle has six parts: three sides and three angles. Solving a triangle means finding the unknown parts based on the known parts. In the case of a right triangle, one part is always known: one of the angles is $90\Degrees$.

Example 1.19

tikzpicture[scale=0.4, every node/.style=font=] [fill=fillcolor] (0,0) -- (4,0) -- (4,3) -- (0,0); [line width=0.5pt] (3.6,0) -- (3.6,0.4) -- (4,0.4); [linecolor,line width=1.5pt] (0,0) -- (4,0) -- (4,3) -- cycle; [below left] at (0,0) A; [below right] at (4,0) C; [above right] at (4,3) B; [below] at (2,0) b; [right] at (4,1.5) a; [above left] at (2,1.5) c; tikzpicture — Figure 1.3.3

Solve the right triangle in Figure 1.3.3 using the given
information:

$c = 10$, $A = 22\Degrees$

Solution: The unknown parts are $a$, $b$, and $B$. Solving yields:

\begin{alignat*}{7} a ~ &= ~ c\;\sin\;A ~ &= ~ 10\;\sin\;22\Degrees ~ &= ~ 3.75\\ b ~ &= ~ c\;\cos\;A ~ &= ~ 10\;\cos\;22\Degrees ~ &= ~ 9.27\\ B ~ &= ~ 90\Degrees ~-~ A ~ &= ~ 90\Degrees ~-~ 22\Degrees ~ &= ~ 68\Degrees \end{alignat*}
$b = 8$, $A = 40\Degrees$

Solution: The unknown parts are $a$, $c$, and $B$. Solving yields:

\begin{alignat*}{3} \frac{a}{b} ~ &= ~ \tan\;A \quad &\Rightarrow \quad a ~ &= ~ b\;\tan\;A ~ = ~ 8\;\tan\;40\Degrees ~ = ~ 6.71\\[2mm] \frac{b}{c} ~ &= ~ \cos\;A \quad &\Rightarrow \quad c ~ &= ~ \frac{b}{\cos\;A} ~ = ~ \frac{8}{\cos\;40\Degrees} ~ = ~ 10.44 \end{alignat*}

\begin{displaymath} B ~ = ~ 90\Degrees ~-~ A ~ = ~ 90\Degrees ~-~ 40\Degrees ~ = ~ 50\Degrees \end{displaymath}
$a = 3$, $b = 4$

Solution: The unknown parts are $c$, $A$, and $B$. By the Pythagorean Theorem,

\begin{displaymath} c ~=~ \sqrt{a^2 ~+~ b^2} ~=~ \sqrt{3^2 ~+~ 4^2} ~=~ \sqrt{25} ~=~ 5 ~. \end{displaymath}

Now, $\tan\;A = \frac{a}{b} = \frac{3}{4} = 0.75$. So how do we find $A$? There should be a key labeled 1pt $\tan^{-1}$ on your calculator, which works like this: give it a number $x$ and it will tell you the angle $\theta$ such that $\tan\;\theta = x$. In our case, we want the angle $A$ such that $\tan\;A = 0.75$:

\begin{displaymath} \text{Enter: } 0.75 \quad \text{Press: {\setlength\fboxsep{1pt}\ovalbox{\footnotesize $\tan^{-1}$}}} \quad \text{Answer: } 36.86989765 \end{displaymath}

This tells us that $A = 36.87\Degrees$, approximately. Thus $B = 90\Degrees - A = 90\Degrees - 36.87\Degrees = 53.13\Degrees$.
Note: The 1pt $\sin^{-1}$ and 1pt $\cos^{-1}$ keys work similarly for sine and cosine, respectively. These keys use the inverse trigonometric functions, which we will discuss in Chapter 5.

Exercises

[r]

From a position $150$ ft above the ground, an observer in a building measures angles of depression of $12\Degrees$ and $34\Degrees$ to the top and bottom, respectively, of a smaller building, as in the picture on the right. Use this to find the height $h$ of the smaller building.

[r]

Generalize Example 1.12: A person standing $a$ ft from the base of a mountain measures an angle of elevation $\alpha$ from the ground to the top of the mountain. The person then walks $b$ ft straight back and measures an angle of elevation $\beta$ to the top of the mountain, as in the picture on the right. Assuming the ground is level, find a formula for the height $h$ of the mountain in terms of $a$, $b$, $\alpha$, and $\beta$.

As the angle of elevation from the top of a tower to the sun decreases from $64\Degrees$ to $49\Degrees$ during the day, the length of the shadow of the tower increases by $92$ ft along the ground. Assuming the ground is level, find the height of the tower. [r]
Two banks of a river are parallel, and the distance between two points $A$ and $B$ along one bank is $500$ ft. For a point $C$ on the opposite bank, $\angle\,BAC = 56\Degrees$ and $\angle\,ABC = 41\Degrees$, as in the picture on the right. What is the width $w$ of the river?
(Hint: Divide $\overline{AB}$ into two pieces.)

[r]

A tower on one side of a river is directly east and north of points $A$ and $B$, respectively, on the other side of the river. The top of the tower has angles of elevation $\alpha$ and $\beta$ from $A$ and $B$, respectively, as in the picture on the right. Let $d$ be the distance between $A$ and $B$. Assuming that both sides of the river are at the same elevation, show that the height $h$ of the tower is

\begin{displaymath} h ~=~ \frac{d}{\sqrt{(\cot\,\alpha)^2 ~+~ (\cot\,\beta)^2}} ~. \end{displaymath}

The equatorial parallax of the moon has been observed to be approximately $57'$. Taking the radius of the earth to be $3956.6$ miles, estimate the distance from the center of the earth to the moon. (Hint: See Example 1.15.)
An observer on earth measures an angle of $31'\;7''$ from one visible edge of the moon to the other (opposite) edge. Use this to estimate the radius of the moon. (Hint: Use Exercise 6 and see Example 1.16.) [r]
A ball bearing sits between two metal grooves, with the top groove having an angle of $120\Degrees$ and the bottom groove having an angle of $90\Degrees$, as in the picture on the right. What must the diameter of the ball bearing be for the distance between the vertexes of the grooves to be half an inch? You may assume that the top vertex is directly above the bottom vertex.

[r]

The machine tool diagram on the right shows a symmetric worm thread, in which a circular roller of diameter $1.5$ inches sits. Find the amount $d$ that the top of the roller rises above the top of the thread, given the information in the diagram. (Hint: Extend the slanted sides of the thread until they meet at a point.)
Repeat Exercise 9 using $1.8$ inches as the distance across the top of the worm thread.
In Exercise 9, what would the distance across the top of the worm thread have to be to make $d$ equal to $0$ inches?

For $0\Degrees < \theta < 90\Degrees$ in the slider-crank mechanism in Example 1.18, show that

\begin{displaymath} c ~=~ \frac{\sqrt{b^2 ~-~ r^2 \;(\sin\,\theta)^2}}{\cos\;\theta} ~\qquad\text{and}\qquad~ a ~=~ r\;\sin\;\theta ~+~ \sqrt{b^2 ~-~ r^2 \;(\sin\,\theta)^2}~\tan\;\theta ~. \end{displaymath}

(Hint: In Figure 1.3.2 draw line segments from $B$ perpendicular to $\overline{OA}$ and $\overline{AC}$.) [r] $tikzpicture[every node/.style=font=] [line width=1pt] (0.4866,0) arc (0:63:0.4866) -- ++(153:0.8) -- ++(-1.3,0); [line width=1pt] (0.4866,0) arc (0:-63:0.4866) -- ++(-153:0.8) -- ++(-1.3,0); (0,0) circle (1pt); [-latex] (0,0) -- (60:0.4866) node[above,sloped,midway,above] r; scope[>=latex] [|<->|] (-2.2,0.7964) -- (-2.2,-0.7964) node [midway,fill=white] 2\, 18"; scope [line width=1pt] (-1.775,0.7964) -- (-1.775,-0.7964); [dashed] (-1.775,0) -- (0.4866,0); [dashed] (-0.492,1.2) -- (-0.492,0.9); [dashed] (0.4866,1.2) -- (0.4866,0.2); at (-0.0027,1.1) 1\, 12"; [-latex] (-0.892,1.1) -- (-0.492,1.1); [-latex] (0.8866,1.1) -- (0.4866,1.1); [dashed,latex-latex] ([shift=(1.071,0)] 153:1.5) arc (153:207:1.5); [above] at (-0.7,0.1) 54 ; tikzpicture$
The machine tool diagram on the right shows a symmetric die punch. In this view, the rounded tip is part of a circle of radius $r$, and the slanted sides are tangent to that circle and form an angle of $54\Degrees$. The top and bottom sides of the die punch are horizontal. Use the information in the diagram to find the radius $r$.

[r]

In the figure on the right, $\angle\,BAC = \theta$ and $BC = a$. Use this to find $AB$, $AC$, $AD$, $DC$, $CE$, and $DE$ in terms of $\theta$ and $a$.
(Hint: What is the angle $\angle\,ACD\,$?)

tikzpicture[scale=0.4, every node/.style=font=] [fill=fillcolor] (0,0) -- (4,0) -- (4,3) -- (0,0); [line width=0.5pt] (3.5,0) -- (3.5,0.5) -- (4,0.5); [linecolor,line width=1.5pt] (0,0) -- (4,0) -- (4,3) -- cycle; [below left] at (0.2,0) A; [below right] at (3.8,0) C; [above right] at (3.8,2.9) B; [below] at (2,0) b; [right] at (4,1.5) a; [above left] at (2,1.5) c; tikzpicture — Figure 1.3.4

For Exercises 15-23, solve the right triangle in Figure 1.3.4 using the
given information.

$a = 5$, $b = 12$
$c = 6$, $B = 35\Degrees$
$b = 2$, $A = 8\Degrees$

3
$a = 2$, $c = 7$
$a = 3$, $A = 26\Degrees$
$b = 1$, $c = 2$

3
$b = 3$, $B = 26\Degrees$
$a = 2$, $B = 8\Degrees$
$c = 2$, $A = 45\Degrees$

In Example 1.10 in Section 1.2, we found the exact values of all six trigonometric functions of $75\Degrees$. For example, we showed that $\cot\;75\Degrees = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}$. So since $\tan\;15\Degrees = \cot\;75\Degrees$ by the Cofunction Theorem, this means that $\tan\;15\Degrees = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}$. We will now describe another method for finding the exact values of the trigonometric functions of $15\Degrees$. In fact, it can be used to find the exact values for the trigonometric functions of $\frac{\theta}{2}$ when those for $\theta$ are known, for any $0\Degrees < \theta < 90\Degrees$. The method is illustrated in Figure 1.3.5 and is described below.

inside

Figure 1.3.5

Draw a semicircle of radius $1$ centered at a point $O$ on a horizontal line. Let $P$ be the point on the semicircle such that $\overline{OP}$ makes an angle of $60\Degrees$ with the horizontal line, as in Figure 1.3.5. Draw a line straight down from $P$ to the horizontal line at the point $Q$. Now create a second semicircle as follows: Let $A$ be the left endpoint of the first semicircle, then draw a new semicircle centered at $A$ with radius equal to $AP$. Then create a third semicircle in the same way: Let $B$ be the left endpoint of the second semicircle, then draw a new semicircle centered at $B$ with radius equal to $BP$.
This procedure can be continued indefinitely to create more semicircles. In general, it can be shown that the line segment from the center of the new semicircle to $P$ makes an angle with the horizontal line equal to half the angle from the previous semicircle’s center to $P$.
1. Explain why $\angle\,PAQ=30\Degrees$. (Hint: What is the supplement of $60\Degrees$?)
2. Explain why $\angle\,PBQ=15\Degrees$ and $\angle\,PCQ=7.5\Degrees$.
3. Use Figure 1.3.5 to find the exact values of $\sin\;15\Degrees$, $\cos\;15\Degrees$, and $\tan\;15\Degrees$. (Hint: To start, you will need to use $\angle\,POQ = 60\Degrees$ and $OP = 1$ to find the exact lengths of $\overline{PQ}$ and $\overline{OQ}$.)
4. Use Figure 1.3.5 to calculate the exact value of $\tan\;7.5\Degrees$.
5. Use the same method but with an initial angle of $\angle\,POQ = 45\Degrees$ to find the exact values of $\sin\;22.5\Degrees$, $\cos\;22.5\Degrees$, and $\tan\;22.5\Degrees$.
[r]
A manufacturer needs to place ten identical ball bearings against the inner side of a circular container such that each ball bearing touches two other ball bearings, as in the picture on the right. The (inner) radius of the container is $4$ cm.
1. Find the common radius $r$ of the ball bearings.
2. The manufacturer needs to place a circular ring
  inside the container. What is the largest possible
  (outer) radius of the ring such that it is not on top
  of the ball bearings and its base is level with the
  base of the container?

[r] tikzpicture[scale=2,every node/.style=font=] (0,0) circle (1); in 1,...,8 (45* :1.0824) -- (45+45* :1.0824); [-latex] (0,0) -- (90:1) node[pos=0.5,fill=white] 1; (0,0) circle (0.8pt); tikzpicture

A circle of radius $1$ is inscribed inside a polygon with eight sides of equal length, called a regular octagon. That is, each of the eight sides is tangent to the circle, as in the picture on the right.
1. Calculate the area of the octagon.
2. If you were to increase the number of sides of the
  polygon, would the area inside it increase or decrease?
  What number would the area approach, if any? Explain.
3. Inscribe a regular octagon inside the same circle. That is,
  draw a regular octagon such that each of its eight vertexes
  touches the circle. Calculate the area of this octagon.

[r]

The picture on the right shows a cube whose sides are of length $a > 0$.
1. Find the length of the diagonal line segment $\overline{AB}$.
2. Find the angle $\theta$ that $\overline{AB}$ makes with the base of the cube.

tikzpicture[every node/.style=font=] [line width=1pt] (0,0) -- (2.2,0) -- (2.2,3) -- cycle; [line width=1pt] (1.2,0) -- (2.2,3); [below] at (0,0) A; [right] at (2.2,3) B; [below] at (2.2,0) C; [below] at (1.2,0) D; (2,0) -- (2,0.2) -- (2.2,0.2); at (0.35,0.2) ; (0:0.6) arc (0:53.746:0.6); at (1.4,0.2) ; ([shift=(1.2,0)] 0:0.52) arc (0:71.565:0.52); tikzpicture — Figure 1.3.6

In Figure 1.3.6, suppose that $\alpha$, $\beta$, and $AD$ are known. Show that:
1. $BC ~=~ \dfrac{AD}{\cot\;\alpha - \cot\;\beta}$
2. $AC ~=~ \dfrac{AD\;\cdot\;\tan\;\beta}{\tan\;\beta - \tan\;\alpha}$
3. $BD ~=~ \dfrac{AD\;\cdot\;\sin\;\alpha}{\sin\;(\beta - \alpha)}$
  (Hint: What is the measure of the angle $\angle\,ABD\;$?)

Persons A and B are at the beach, their eyes are $5$ ft and $6$ ft, respectively, above sea level. How many miles farther out is Person B’s horizon than Person A’s? (Note: $1$ mile = $5280$ ft)

Of course it is not perfectly spherical. The earth is an ellipsoid, i.e. egg-shaped, with an observed ellipticity of $1/297$ (a sphere has ellipticity $0$). See pp. 26-27 in W.H. Munk and G.J.F MacDonald, The Rotation of the Earth: A Geophysical Discussion, London: Cambridge University Press, 1960. ↩
This will often be worded as the line that is tangent to the circle. ↩