---
title: "Precalculus: Bearing Problems"
---
## Bearing Problems: What's Going On?
For these problems, we can imagine being a sailor on a ship. The ship is speeding along in some direction, and it's hard to make sharp turns, so you're trying to figure out how to turn the ship to make sure you arrive in some location.
In the case of the picture below, for example, you're in a ship called $O$ speeding North, trying to make sure you steer the right way to arrive at a dock called $P$.
So, you call the dock and tell them that your compass is pointed directly north, and ask them how to best get to the dock (with the least-sharp turn required). The directions they give you are in a form called a **bearing**.
<center>
<img src="https://i.imgur.com/96JtsuD.png" width="400px" />
</center>
In this case, they respond by sending you the direction $N$ $40^\circ$ $E$. This is the **bearing** that tells you how to get to them from where you are. It says: "starting from a direction pointing north, turn $40^\circ$ eastwards". Once you make this turn, you now know that you don't have to worry, you'll arrive at the dock $P$ eventually.
The bearings, it turns out, will always take this form: the first direction will be either $N$ or $S$ (basically because compasses back in the day had prominent North and South markings), and then it will have an amount of degrees to turn, and the direction in which to make that turn.
### Example Problem
A boat leaves the entrance to a harbor and travels 25 miles on a bearing of $N$ $42^\circ$ $E$. The captain then turns the boat $90^\circ$ clockwise and travels 18 miles on a bearing of $S$ $48^\circ$ $E$. At that time:
1. How far is the boat, to the nearest tenth of a mile, from the harbor entrance?
2. What is the bearing, to the nearest tenth of a degree, of the boat from the
harbor entrance?
<center>
<img src="https://i.imgur.com/wouldqE.png" width="400px" />
</center>
The boat's actual trip, from the point labeled "Harbor entrance" to the far-right of the picture, is represented by the line labeled "25 mi" and the line labeled "18 mi". The third line, labeled "$c$", doesn't represent actual travel but is there to help you compute the distance, using the Pythagorean Theorem. Knowing what you know about triangles, you can solve this (bc I want to get to more trig-heavy example), to get
1. $c^2 = a^2 + b^2 = 25^2 + 18^2 = 949 \implies c = \sqrt{949} \approx 30.8\text{ mi}$.
2. This one is a bit more trig-heavy, but the idea is that since we have a **right triangle** we can use our SOHCAHTOA rule:
$$
\sin(\theta) = \frac{opposite}{hypotenuse}, \; \cos(\theta) = \frac{adjacent}{hypotenuse}, \; \tan(\theta) = \frac{opposite}{adjacent}
$$
The two sides that we have here, relative to the $\theta$ we want to find, are $18\text{ mi}$ and $25\text{ mi}$, and they are the **opposite** and **adjacent** sides (respectively), so we use the $\tan(\theta)$ part of SOHCAHTOA:
$$
\tan(\theta) = \frac{opposite}{adjacent} = \frac{18\text{ mi}}{25\text{ mi}}.
$$
Now we use one more piece of our trig knowledge to see that, if we want to solve this for $\theta$, we'll need to consider the **inverse trig** functions. In this case, the inverse tangent function $\tan^{-1}(x)$. By the definition of this inverse function, if we want the angle $\theta$ that **produces** a tangent of $\frac{18}{25}$, we need to calculate $\tan^{-1}\left(\frac{18}{25}\right)$. Plugging this into our calculator or [WolframAlpha](https://www.wolframalpha.com/input?i=arctan%2818%2F25%29), we get
$$
\tan^{-1}\left(\frac{18}{25}\right) \approx 35.8^\circ.
$$
Sadly, we're not *quite* done. We found the angle $\theta$, but remember that when we figure out bearings we have to calculate the angle *relative to* the North or South directions. So, in this case, since the picture tells us that the angle *above* the triangle (between the North pole and the boat's starting path) is $42^\circ$, to get our final bearing we add this to $\theta$, giving us: $N$ $(35.8^\circ + 45^\circ)$ $E = N$ $77.8^\circ$ $E$.
Here are two more bearing problems that I think could be helpful, since they're not the exact same setting as the boating example:
1. A ship sights a lighthouse directly to the south. A second ship, 9 miles east of the first ship, also sights the lighthouse. The bearing from the second ship to the lighthouse is S 34° W. How far, to the nearest tenth of a mile, is the first ship from the lighthouse?
My drawing of the *setup* for this problem looks like the following (where they're asking us to find that height $x$).
<center>
<img src="https://i.imgur.com/iZ7BbEL.jpg" width="400px" />
</center>
We can solve for $x$ by completing the upper triangle (formed by the Lighthouse-Ship 1-Ship 2), and then using the **law of sines** (plugging the fraction we get into our calculator or [WolframAlpha](https://www.wolframalpha.com/input?i=solve+%289*sin%2856%29%29+%2F+%28sin%2834%29%29)):
$$
\frac{\sin(34^\circ)}{9\text{ mi}} = \frac{\sin(56^\circ)}{x} \implies x = \frac{(9\text{ mi}) \cdot \sin(56^\circ)}{\sin(34^\circ)} \implies x \approx 13.3\text{ mi}.
$$
2. You leave your house and run 2 miles due west followed by 4.5 miles due north. At that time, what is your bearing from your house?
My drawing for the setup in this problem is:
<center>
<img src="https://i.imgur.com/7ZN91H6.jpg" width="400px" />
</center>
Where they're asking us to find $\theta$, but then write it as a *bearing*. We can use our SOHCAHTOA rules to figure out the angle *below* the dashed line $\sigma$[^first]:
$$
\tan(\sigma) = \frac{4.5\text{ mi}}{2\text{ mi}} \implies \sigma = \tan^{-1}\left(\frac{4.5}{2}\right) \approx 66.0^\circ
$$
where we solve the last part using our calculators or [WolframAlpha](https://www.wolframalpha.com/input?i=arctan%284.5%2F%282%29%29).
Now that we have this angle, finding $\theta$ is easy: since we know the $x$ and $y$ axes make a $90^\circ$ angle, and the angle below the dashed line is $\sigma = 66.0^\circ$, the angle above the dashed line, $\theta$, must be $90^\circ - 66.0^\circ = 24.0^\circ$.
We're almost done, but we have to write this as a *bearing*. Since the endpoint is north of the house, we know that we can get to the endpoint by looking North and then turning $\theta = 24.0^\circ$ westwards. This corresponds to the bearing: $N$ $24.0^\circ$ $W$.
[^first]: I'm calling this angle $\sigma$, greek letter "sigma", just because $\theta$ is already taken, to label the angle *above* the dashed line)
## Domains and Ranges of Compound Trig Functions
Yesterday we saw how, for a given trig function, we can think of it in two different ways, using two different tools:
1. We can keep the **unit circle** in our mind, and figure out what the trig function is "asking" in terms of an angle or coordinate on this unit circle. For $\sin(\theta)$, for example, we know it is asking us about a **$y$-coordinate**. For $\tan^{-1}(x)$, it is asking us about an **angle** that would produce a $\tan$ value of $x$.
2. We can also keep the **graph** of the function in our mind, and think about (for example) stretching/shrinking/shifting that graph. For $5\cos(3\theta + 2) - 4$, for example, we can imagine or draw the $\cos(\theta)$ graph, and then think about a vertical stretch of $5$, horizontal stretch of $3$, horizontal shift to the left $2$ units, and vertical shift downwards by $4$ units.
To me, being able to do both of these will allow us to master finding domains and ranges of any function that the teacher throws at us. The key, though, is to always have a **"base picture"** in your head. What I mean is, if you see a scary-looking trig function like $y = -3.2\csc\left( -\frac{\theta}{2} - \frac{\pi}{6} \right)$, forget about the specific numbers and fractions, and just start by drawing out what the regular $csc(\theta)$ function looks like.
Since $\csc$ is the reciprocal of $\sin$, we can start by just thinking of the plot for $\sin(\theta)$. $\sin(\theta)$ we know passes through the origin, since $\sin(0) = 0$ (the point at the "start" of the unit circle is $(1,0)$, so the $\sin$, the $y$ coordinate, is $0$). The biggest it can get is $1$, because we reach the top of the unit circle when $\theta = \frac{\pi}{2}$. The lowest it can get is $-1$, because we reach the bottom of the unit circle when $\theta = \frac{3\pi}{4}$. With these three pieces of info, we can plot the "regular" $sin(\theta)$ function.
Once we've done that, we can draw the $csc(\theta)$ function, by just focusing on the maximum and minimum points of the $\sin(\theta)$ curve we drew, and making them the tops and bottoms of *parabolas* (the U-shaped curves), like this box from the textbook shows us:
<center>
<img src="https://i.imgur.com/f5js3ml.png" width="800px" />
</center>
Once we've done those two steps, we know the "anchors" that this $csc(\theta)$ drawing is based on (the max and min points of a $\sin(\theta)$ graph with amplitude $1$), so that *now* we can bring in the fancy numbers and fractions.
## Working Through an Example
Let's look at that function again, $y = -3.2\csc\left( -\frac{\theta}{2} - \frac{\pi}{6} \right)$. I've made a [Desmos visualization](https://www.desmos.com/calculator/n2xyw6hkkp) where each step we carry out below is represented by a different color.
We can now imagine *taking* our $\csc(\theta)$ graph and doing the following things to it:
* Flipping it over the $x$ axis, because of the negative sign at the beginning
* Stretching it vertically by $3.2$ units
* Flipping it over the $y$ axis, because of the negative sign within the $\csc$ function
* Stretching it horizontally by 2, because of the denominator underneath the $\theta$ term, and
* Shifting it left by $\frac{\pi}{3}$, because of the term inside the $\csc$ function that doesn't involve our angle $\theta$.
* (a) Note that it's not $\pi/6$: that's because, if we read the inside of the parentheses carefully, it's shifting the *angle divided by 2* by $\pi/6$, which is the same as shifting the *original angle* by $\pi/3$).
* (b) Note that it's a shift to the *left* even though we're *subtracting* something: that's because we already reversed the order of the stuff inside the parentheses by putting that negative sign in front of the $\theta$. Once we've done that, you can think of it like we're shifting the *negative* of our angle by some amount, which is the same as shifting the *original* angle by some amount in the opposite direction.
By seeing where the anchor points and asymptotes end up, after doing all this, we can *re-draw* the $\csc$ function in the same way we did it above, but using these new points and asymptotes.
For example, when you plot the original $\csc( \theta )$ function, you'll see that the bottom of one of the parabolas is the point $\left( \frac{\pi}{2}, 1 \right)$, and that the asymptotes on either side of it are the vertical lines $x = 0$ (on the left) and $x = \pi$ (on the right). So let's see what happens to this point when we do each of these transformations:
* Flipping over the $x$ axis: this flips the point to its new location, $\left( \frac{\pi}{2},-1 \right)$, where it's now the *top* of a parabola. It doesn't affect the asymptotes (if you flip an infinitely tall vertical line vertically, it's still just an infinitely tall line).
* Stretching vertically by $3.2$ units: this entails "pulling" the tops/bottoms of the parabolas $3.2$ units *further away from* the $x$ axis. So, rather than being $1$ unit away from the $x$ axis, our point is now $3.2$ units away from the $x$ axis, at its new location $\left(\frac{\pi}{2}, -3.2\right)$. It's still the *top* of a parabola, and the asymptotes on either side of it don't change (stretching an infinitely-tall line doesn't affect it, it's still infinity tall).
* Flipping over the $y$ axis: the old point was at $\left( \frac{\pi}{2} , -3.2 \right)$, so when we flip it over the $y$ axis, it's now at $\left(-\frac{\pi}{2}, -3.2\right)$. It's still at the top of a parabola, but this time the asymptotes *have* changed: one of them was just $x = 0$, so that doesn't change (since it's exactly *on* the $y$ axis). But the other was at $x = \pi$, so now once we flip it over the $y$ axis it's now at $x = -\pi$. So our point is now still at the top of a parabola, but one bounded on the left by $x = -\pi$, and on the right by our $x = 0$ asymptote (which didn't move).
* Stretching horizontally by $2$: since the old point was $\frac{\pi}{2}$ units away from the $y$ axis, when we stretch it twice as far away, it's now at $(-\pi, -3.2)$, $pi$ units away from the $y$ axis. The $x = 0$ asymptote still doesn't change (it can't be stretched away from the $y$ axis beause it's *on* the $y$ axis), but the $x = -\pi$ asymptote stretches twice as far away, just like we did for the point. So, instead of being at $x = -\pi$, it's now twice as far, at $x = -2\pi$.
* Shifting left by $\frac{\pi}{3}$: Finally, we take our point and our asymptotes and just move them to the left by $\frac{\pi}{3}$ units. So, since our point was at $(-\pi,-3.2)$, it's now at $(-\pi - \frac{\pi}{3}, -3.2) = (-\frac{4\pi}{3},-3.2)$. This point is at the top of a parabola still. And, our asymptotes shift by the same amount: the $x = 0$ asymptote finally moves, over to $x = 0 - \frac{\pi}{3} = -\frac{\pi}{3}$, while the $x = -2\pi$ asymptote moves over to $x = -2\pi - \frac{\pi}{3} = -\frac{7\pi}{3}$.
By following this point, from beginning to end, we saw exactly what each transformation did, to both the point itself and the asymptotes around it. We can then use this final "resting point", where the point is at $\left( -\frac{4\pi}{3}, -3.2 \right)$, at the top of a parabola, and it's surrounded by asymptotes at $x = -\frac{7\pi}{3}$ and $x = -\frac{\pi}{3}$. We can use this information to derive the **domain** and **range** of this new function:
The domain of the *original* $\csc(\theta)$ function was all real numbers besides the asymptote values. Since the graph looks the exact same between each pair of asymptotes, we can take one of them, see how far the next one is, and know that that's the **period** of the function. At our point's original location $\left( \frac{\pi}{2}, 1 \right)$, it was at the bottom of a parabola, bounded by $x = 0$ and $x = \pi$. That means that if we started at that $x = 0$ asymptote and moved left by $\pi$, we'd end up at another asymptote $x = -\pi$. Similarly, if we started at the $x = \pi$ asymptote and moved right by $\pi$, we'd end up at another asymptote $x = 2\pi$. Given this, we know that we can write out **all** the asymptotes in the original function as $x = 0 + \pi k$.
Let's now apply the same logic to our *final* function, after all the transformations, $-3.2\csc \left( -\frac{\theta}{2} - \frac{\pi}{6} \right)$. Our $x = 0$ asymptote ended up at $x = -\frac{\pi}{3}$, and our $x = \pi$ asymptote ended up at $-\frac{7\pi}{3}$. So, the jump that would get us from the left asymptote to the right asymptote is $-\frac{\pi}{3} - \left( -\frac{7\pi}{3} \right) = \frac{6\pi}{3} = 2\pi$. That means, to keep our formula for **all** the asymptotes as simple as possible, we can write it as $x = -\frac{\pi}{3} + 2\pi k$. If we start at $x = -\frac{\pi}{3}$, and we jump $2\pi$ in any direction, however many times we want, we'll still end up at an asymptote.
So, given this formula, we now have everything we need to figure out the **domain** and **range**:
* The **domain** is all real numbers besides the asymptotes. As we just figured out, this means it's $x \neq -\frac{\pi}{3} + 2\pi k$.
* The **range**, since we have these parabolas whose tops and bottoms are just the points we followed, is just all the $y$ values that the parabolas take up. Since our final point $\left( -\frac{4\pi}{3}, -3.2 \right)$ is the *top* of a parabola, and the parabolas *above* the $x$ axis are the same distance away from the $x$ axis, we know that the tops of all the lower parabolas will be on the horizontal line $y = -3.2$, while the bottoms of all the upper parabolas will be on the horizontal line $y = 3.2$. Therefore, this function can reach any value *besides* the (vertical) space in between these parabolas, giving us the **range** $y = (-\infty, -3.2] \cup [3.2, \infty)$.

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