---
title: "Precalculus: The Unit Circle"
---
# Precalculus: The Unit Circle
## Quick Preliminary
I'll be referring to this unit circle diagram throughout:
<center><img src="https://i.imgur.com/6MejELD.png" width="66%" height="66%"/></center>
I like to think of the coordinates for the 3 points in each quadrant as follows:
<center>
$$
\frac{\sqrt{3}}{2} = \textbf{long}, \frac{\sqrt{2}}{2} = \textbf{middle}, \frac{1}{2} = \textbf{short}
$$
</center>
This will help us keep track of what's going on in these problems, and the reasoning is as follows:
* If a coordinate is $\frac{\sqrt{3}}{2}$, I call the distance from the origin to that coordinate the **"long"** distance:
* Looking at the point where $\theta = \frac{\pi}{6}$, for example, the $x$ coordinate is $\frac{\sqrt{3}}{2}$, which is the **longest** distance from the $y$ axis of any of the three points
* If a coordinate is $\frac{\sqrt{2}}{2}$, I call the distance from the origin to that coordinate the "middle" distance, since it's **in-between** $\frac{\sqrt{3}}{2}$ and $\frac{1}{2}$
* Looking at the point where $\theta = \frac{\pi}{4}$, for example, the $x$ coordinate is $\frac{\sqrt{2}}{2}$, which is longer than $\frac{1}{2}$ but shorter than $\frac{\sqrt{3}}{2}$ in terms of distance from the origin.
* If a coordinate is $\frac{1}{2}$, I call the distance from the origin to that coordinate the **"short"** distance:
* Looking at the point where $\theta = \frac{\pi}{3}$, for example, the $x$ coordinate is $\frac{1}{2}$, which is the **shortest** distance from the $y$ axis to any of the three labeled points.
## Part 1 (No Calculator)
### 1. Which of the following values are positive? Circle all that qualify.
$$
\cos\left(-\frac{\pi}{11}\right)\text{: positive} \; \checkmark
$$
$$
\csc(214^\circ)\text{: not positive} \; \checkmark
$$
$$
\sec\left(-\frac{8\pi}{7}\right)\text{: }\color{red}{\text{not positive}}
$$
* First we remember the definition of the secant, that $\sec\theta = \frac{1}{\cos\theta}$.
* So, we can start by thinking about the cosine, $\cos\left(-\frac{8\pi}{7}\right)$, then at the end take 1/(that value) to get the secant.
* Since the cosine $\cos\theta$ represents the **x-coordinate** that we’re at when we’ve traveled $\theta$ radians around the unit circle, let’s think about this value $\theta = -\frac{8\pi}{7}$.
* Now we remember that an angle of $\pi$ represents traveling **halfway** around the unit circle going counterclockwise, and $-\pi$ represents traveling **halfway** around the unit circle going **clockwise.**
* So, if we’re traveling **clockwise** (since $\theta$ is negative here) by a little more than $-\pi$ (since $-\frac{7\pi}{7}$ would be exactly $-\pi$), that means we travel halfway around the unit circle, clockwise, then a liitle bit further, so that we end up in the 2nd quadrant (the upper-left portion of the $xy$ plane).
* Since this 2nd quadrant is to the left of the $y$-axis, we know that the $x$ coordinates for any points in this quadrant will be **negative**, and since the cosine of $\theta$ is the $x$ coordinate we arrive at after traveling (clockwise in this case) by $\theta$, this means that the **cosine** will be negative in this case.
* Since the cosine is negative, we can finally conclude that the **secant** is also negative, since $\sec\theta = \frac{1}{\cos\theta}$, and 1/(negative number) = (negative number)
So, the key here was to ignore the scariness of the secant, for most of the problem, and just think about the easier-to-work-with **cosine** instead.
$$
\tan\left(\frac{8\pi}{5}\right)\text{: not positive} \; \checkmark
$$
### Find the exact value of each. Box final answer.
### 2.
$$
\cot\left(\frac{29\pi}{6}\right) = \color{red}{-\sqrt{3}}
$$
* Here the first thing we do is ``unwrap'' that $\frac{29\pi}{6}$.
* We know that if we start at $\theta=0$ and travel (whether clockwise or counterclockwise) by $2\pi$, we end up back in the same place where we started (the point $(1,0)$).
* So, since we have a $6$ in our denominator here, we can think about what $2\pi$ is in terms of this denominator: $2\pi = \frac{12\pi}{6}$. This means that, wherever we are on the unit circle, we can travel $\frac{12\pi}{6}$ (clockwise or counterclockwise) and end up back in the same place.
* Since we're currently at $\frac{29\pi}{6}$, this means we can travel **clockwise** by $\frac{12\pi}{6}$ once to get to $\frac{29\pi}{6} - \frac{12\pi}{6} = \frac{17\pi}{6}$. And, since this value is still greater than $2\pi$, we can do the same full-circle travel again to get to $\frac{17\pi}{6} - \frac{12\pi}{6} = \frac{5\pi}{6}$.
* Finally, with $\frac{5\pi}{6}$, we’ve arrived at a value between $0$ and $2\pi$, which means we can think in terms of our regular unit circle.
* Remembering that $\cot\theta = \frac{1}{\tan\theta}$, we can make our lives easier by just thinking about $\tan\theta$ for now, and taking 1/(the value we find) to get our final answer.
* Since $\tan\theta = \frac{\sin\theta}{\cos\theta}$, we just need to find the $y$ coordinate of the point we arrive at after traveling $\frac{5\pi}{6}$, the sine, and the $x$ coordinate of that same point, the cosine.
* Of our three labeled points in the 2nd quadrant, $\frac{5\pi}{6}$ is the one closest to the leftmost point $(-1,0)$, which means it has "long" $x$ coordinate $-\frac{\sqrt{3}}{2}$ and "short" $y$ coordinate $\frac{1}{2}$. Solving the math now, we get
$$
\frac{\sin\theta}{\cos\theta} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{2}\cdot\left(-\frac{2}{\sqrt{3}}\right) = -\frac{1}{\sqrt{3}}.
$$
* And, since $\cot\theta = \frac{1}{\tan\theta}$ like we noticed above,
$$
\cot\theta = \frac{1}{\tan\theta} = \frac{1}{-\frac{1}{\sqrt{3}}} = 1\cdot \left(-\frac{\sqrt{3}}{1}\right) = -\sqrt{3}.
$$
### 3.
$$
8\csc(135^\circ)\cdot4\sin(300^\circ) = \color{red}{-16\sqrt{6}}
$$
Using the same types of things we did in the previous problems, and not letting the degrees scare us (since they’re always convertable to radians),
* We start by remembering that $\csc\theta = \frac{1}{\sin\theta}$. So for now we’ll just worry about $\sin\theta$, the $y$ coordinate after we travel by $\theta$, in this case $135^\circ$.
* Since $135^\circ$ is $90^\circ + 45^\circ$, this angle really represents traveling $90^\circ$, bringing us to the very top of the unit circle, then half of the way to the $(-1,0)$ point, so that we’re in the 2nd quadrant at a $45^\circ$ angle from the [left side of the] $x$-axis.
* Among our three labeled points on our unit circle in this quadrant, this corresponds to our nice “middle” point, where the $x$ and $y$ coordinates are negatives of each other: since we’re in the 2nd quadrant, $x = -\frac{\sqrt{2}}{2}$, while $y=+\frac{\sqrt{2}}{2}$.
* Since we’re interested in the **sine**, the $y$ coordinate, we take $y = +\frac{\sqrt{2}}{2}$, and since we **really** wanted to know the **cosecant**, we’ll take 1/(this value), to get $\frac{2}{\sqrt{2}}$.
* We do a similar set of steps for $\sin(300^\circ)$: since they want the **sine** we know we’re going to be looking for a $y$ coordinate.
* Since $300^\circ = 360^\circ - 60^\circ$, and since traveling by $360^\circ$ takes us right back to our starting point, we know that this is just the point we arrive at by traveling $-60^\circ$, or in other words, traveling (60/90)ths = (2/3)rds of the way through the 4th quadrant, moving **clockwise** (since $\theta$ is negative).
* This means that, of our labeled points in the 4th quadrant, $-60^\circ$ is the point closest to the $(0,-1)$ point, meaning that it’s the point with the “long” $y$ coordinate $-\frac{\sqrt{3}}{2}$, and “short” $x$ coordinate $\frac{1}{2}$.
* Since we want the **sine**, we’ll just take this $y$ coordinate, $y = -\frac{\sqrt{3}}{2}$.
* Now we can finish up by solving out the math:
$$
8\csc(135^\circ)\cdot 4\sin(300^\circ) = 8\frac{2}{\sqrt{2}}\cdot4\left(-\frac{\sqrt{3}}{2}\right) = \frac{16}{\sqrt{2}}\cdot (-2\sqrt{3}) = -\frac{32\sqrt{3}}{\sqrt{2}},
$$
* and if we use our “multiply by 1” trick, multiplying this fraction by $\frac{\sqrt{2}}{\sqrt{2}}$ to get rid of the square root in the denominator, we get
$$
-\frac{32\sqrt{3}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = -\frac{32\sqrt{6}}{2} = -16\sqrt{6}.
$$
### 4.
$$
\sec\frac{11\pi}{3}+\tan\frac{5\pi}{4} = 3 \; \checkmark
$$
### 5.
$$
\tan405^\circ+\cos45^\circ\csc45^\circ = \color{red}{2}
$$
* Here, as you noticed, traveling by $405^\circ$ is the same as traveling $360^\circ + 45^\circ$, and since traveling $360^\circ$ gets you back where you started, this is the same as traveling by just $45^\circ$.
* And, since we know that at our $45^\circ$ point the $x$ and $y$ coordinates are the same ($\frac{\sqrt{2}}{2}$ and $\frac{\sqrt{2}}{2}$ in this case), the tangent will be exactly $1$.
* For the $\cos 45^\circ\csc 45^\circ$ term, first we see that we just found the $x$ coordinate at $45^\circ$ above, namely, $\frac{\sqrt{2}}{2}$, so we know that $\cos 45^\circ = \frac{\sqrt{2}}{2}$.
* Since we also know that $\csc\theta = \frac{1}{\sin\theta}$, for this term we just need the **sine**, the $y$ coordinate at $45^\circ$, which we saw is the same as the $x$ coordinate $\frac{\sqrt{2}}{2}$.
* But, since $\csc$ is the **inverse** of $\sin$, we have to make sure to take 1/(the sin value):
$$
\csc\theta = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}}
$$
* So now we can multiply these two values to obtain the right-side term:
$$
\cos 45^\circ\csc 45^\circ = \frac{\sqrt{2}}{2}\cdot\frac{2}{\sqrt{2}} = 1
$$
* So that our final answer is
$$
\tan 405^\circ + \cos 45^\circ\csc 45^\circ = 1 + 1 = 2.
$$
### Fill in each space to make a true statement. For angle answers use **radians**.
6. If $\sec \theta = 3$ then $\cos \theta = \underline{3} \; \checkmark$
7. If $\sin \theta = -\cos \theta$ then $\theta = \underline{\frac{3\pi}{4}}\text{ or }\underline{\frac{7\pi}{4}} \; \checkmark$
8. Name two angles for which $\tan\theta$ is undefined: $\underline{\frac{\pi}{2}}$ and $\underline{\frac{3\pi}{2}} \; \checkmark$
9. $\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8} = \underline{1} \; \checkmark$
10. $\sin 100^\circ = \underline{\sin(-260^\circ)} \; \checkmark$
11. True or False? $\cos 82^\circ = \cos(-82^\circ)$: $\underline{\text{True}} \; \checkmark$
12. $\sin \underline{\frac{3\pi}{2}} = -1 \; \checkmark$
13. $\cot39^\circ = \tan \underline{\color{red}{51^\circ}}$ (degrees)
* First we remember that $\cot\theta = \frac{1}{\tan\theta} = \frac{1}{\frac{\sin\theta}{\cos\theta}} = \frac{\cos\theta}{\sin\theta}$.
* So in our case we have $\cot39^\circ = \frac{\cos 39^\circ}{\sin 39^\circ}$, and we're close to an answer except, since we need a **tangent**, we want to figure out some **new** angle $\sigma$ so that $\frac{\cos 39^\circ}{\sin 39^\circ} = \frac{\sin\sigma}{\cos\sigma}$
* And, we have a formula for "converting" sines into cosines and vice-versa! (We derived them, by looking at the unit circle diagram, last meeting):
$$
\cos\theta = \sin(90^\circ - \theta)\text{ and }\sin\theta = \cos(90^\circ - \theta)
$$
* So we can use this to immediately convert our $\frac{\cos 39^\circ}{\sin 39^\circ}$ into $\frac{\sin}{\cos}$ form:
$$
\frac{\cos 39^\circ}{\sin 39^\circ} = \frac{\sin(90^\circ - 39^\circ)}{\cos(90^\circ - 39^\circ)} = \frac{\sin(51^\circ)}{\cos(51^\circ)}
$$
* Looking at this fraction, we see that it's exactly in the form that the **tangent** takes: since $\tan\theta = \frac{\sin\theta}{\cos\theta}$, we use this definition "in reverse" to finish the last step:
$$
\frac{\sin(51^\circ)}{\cos(51^\circ)} = \tan(51^\circ).
$$
14. $\sin(\theta) > 0$ and $\tan(\theta) < 0$ when $\theta$ is in quadrant $\underline{\color{red}{2}}$
* Let's start with the first part. Since $\sin(\theta)$ represents the **$y$ coordinate** of the point we arrive at after traveling by an angle of $\theta$, $\sin(\theta) > 0$ means that this $y$ coordinate will be positive.
* For the $y$ coordinate of a point to be positive, it needs to be in either the **1st** or **2nd** quadrant (the two quadrants above the $x$ axis, or in other words, the top half of the $xy$ plane).
* Since $\tan(\theta)$ doesn't have a nice rule in terms of directly inferring from coordinates, let's break it up into sine and cosine using the definition: $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.
* We already know that $\sin(\theta)$ is going to be **positive**, meaning that the numerator of this fraction will be positive.
* That means that, for the **whole fraction** to be **negative**, $\cos(\theta)$ has to be negative.
* Since $\cos(\theta)$ represents the **$x$ coordinate** of the point we arrive at after traveling $\theta$ radians, and since we know that this value must be negative, the point must be in either the **2nd** or **3rd** quadrant (in other words, the **left half** of the $xy$ plane).
* Gathering these two pieces of info we derived: from our first condition, we got that the point we travel to must be in either the **1st** or **2nd** quadrant. Then, from the second condition we got that the point must be in either the **2nd** or **3rd** quadrant.
* Putting these two together, we finally know the answer: the point must be in the **2nd** quadrant.
15. $\cos\frac{5\pi}{12} = \sin \underline{\frac{\pi}{12}} \; \checkmark$
16. $\cos24^\circ\cdot \sec24^\circ = \underline{1} \; \checkmark$
17. $\frac{\cos(-50^\circ)}{\sin40^\circ} = \underline{\color{red}{1}}$
* This one is tough! Let's start with the numerator. Since $\cos\theta$ represents the $x$ coordinate of the point we arrive at when we travel $\theta$ degrees, we know that in fact $\cos(-\theta) = \cos(\theta)$.
* You can verify this by looking closely at your unit circle diagram: the $x$ coordinate of the point we arrive at after traveling $50^\circ$ is the same as the $x$ coordinate of the point we arrive at if we travel $50^\circ$ **clockwise**, or in other words, if we travel by $-50^\circ$ (it is only the $y$ coordinate, or the **sine**, that will be different in these two cases).
* So, using this fact, we can rewrite the fraction as
$$
\frac{\cos(-50^\circ)}{\sin(40^\circ)} = \frac{\cos(50^\circ)}{\sin(40^\circ)}.
$$
* Now we can get somewhere by using our **cosine-to-sine** conversion formula from above, which states that $\cos(\theta) = \sin(90^\circ - \theta)$.
* Specifically, we can use this formula to convert the numerator into a sine, giving us
$$
\frac{\cos(50^\circ)}{\sin(40^\circ)} = \frac{\sin(90^\circ - 50^\circ)}{\sin(40^\circ)} = \frac{\sin(40^\circ)}{\sin(40^\circ)},
$$
* and now we can see the final answer, that
$$
\frac{\cos(-50^\circ)}{\sin(40^\circ)} = 1.
$$
18. If $\sin \theta = -\frac{2}{3}$, find:
a. $\theta$ is in quadrant $\underline{3}$ or $\underline{4} \; \checkmark$
b. $\sin(-\theta) = \underline{\color{red}{\frac{2}{3}}}$
* We can look at our unit circle to figure out that $\sin(-\theta) = -\sin(\theta)$ for a given $\theta$ value.
* For example, you could check $\theta = 90^\circ$: $\sin(90^\circ)$ is $+1$, since we're at the very **top** of the unit sircle, while $\sin(-90^\circ) = -1$, since in this case we're at the very **bottom** of the unit circle.
* This identity gives us
$$
\sin(-\theta) = -\sin(\theta) = -\left(-\frac{2}{3}\right) = \frac{2}{3}.
$$
c. $\sin(\theta + 2\pi) = \underline{\color{red}{-\frac{2}{3}}}$
* Here the trick is to remember what $2\pi$ represents when we're talking about the unit circle.
* If we start at the zero point (the coordinate $(1,0)$) and travel $\theta$ radians, we'll end up at some point on the unit circle. But then, if we travel $2\pi$ radians from that point in either direction, we'll end up right back at that point.
* So, whatever the $y$ coordinate was after we traveled $\theta$ radians, the point we get to after traveling $2\pi = 360^\circ$ counterclockwise will be that same point, with the same $y$ coordinate.
* Mathematically, this means that $\sin(\theta + 2\pi) = \sin(\theta)$, so that in our specific case
$$
\sin(\theta + 2\pi) = \sin(\theta) = -\frac{2}{3}.
$$
d. $\sin(\theta + \pi) = \underline{\color{red}{\frac{2}{3}}}$
* This one is less easy than the previous one, since we can no longer depend on the "shortcut" that traveling $2\pi$ takes us right back where we started.
* But, we do have the following **addition formulas** for sine and cosine:
$$
\sin(a+b) = \cos(a)\sin(b) + \cos(b)\sin(a), \\
\cos(a+b) = \sin(a)\cos(b) + \sin(b)\cos(a).
$$
* So n this case, we can apply the first formula with $a = \theta$ and $b = \pi$:
$$
\sin(\theta + \pi) = \cos(\theta)\sin(\pi) + \sin(\theta)\cos(\pi) = \cos(\theta)\cdot 0 + \sin(\theta)\cdot -1 = -\sin(\theta)
$$
* Therefore, for our particular case where $\sin \theta = -\frac{2}{3}$, this means
$$
\sin(\theta + \pi) = -\sin(\theta) = -\left(-\frac{2}{3}\right) = \frac{2}{3}.
$$
19. If $\csc(\theta) = \frac{7}{3}$ and $\tan(\theta) < 0$, then $\cos \theta = \underline{\color{red}{-\frac{2\sqrt{10}}{7}}}$ and $\cot\theta = \underline{\color{red}{-\frac{2\sqrt{10}}{3}}}$
* Here we use the definition of the cosecant, $\csc(\theta) = \frac{1}{\sin(\theta)}$.
* Since we know that $\csc(\theta) = \frac{7}{3}$, we can equate the two right-hand sides of these two equations:
$$
\frac{1}{\sin(\theta)} = \frac{7}{3},
$$
and using cross-multiplication here we have
$$
7\sin(\theta) = 3 \implies \sin(\theta) = \frac{3}{7}.
$$
* We see that this is a positive number, which helps us incorporate the second fact, that $\tan(\theta) < 0$.
* Since we know $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ is negative, but $\sin(\theta)$ is positive, $\cos(\theta)$ must be negative.
* Now we can combine what we know: $\sin(\theta)$ is a positive number, which means that the angle $\theta$ must be between $0$ and $\pi$. But, since we also know that $\cos(\theta)$ must be **negative**, the angle must be between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$. Putting these together we know that the angle must be between $\frac{\pi}{2}$ and $\pi$, or in other words, in the **2nd** quadrant.
* Here I couldn't find any way to proceed without drawing a picture, but once the picture is drawn we have a triangle in the 2nd quadrant with base of length $x$, height of $\frac{3}{7}$, and hypotenuse of $1$. Using our Pythagorean Theorem,
$$
1^2 = \left(\frac{3}{7}\right)^2 + x^2 \implies x = \pm \frac{\sqrt{40}}{7} = \pm \frac{2\sqrt{10}}{7},
$$
and since we know that $\cos(\theta)$ is negative, we choose the **negative** solution, giving us that the $x$ coordinate, aka the **cosine**, is $-\frac{2\sqrt{10}}{7}$.
* Since we know that $\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{\cos(\theta)}{\sin(\theta)}$, and we now have both $\cos(\theta)$ and $\sin(\theta)$, we can compute the cotangent as
$$
\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{-\frac{2\sqrt{10}}{7}}{\frac{3}{7}} = -\frac{2\sqrt{10}}{7} \cdot \frac{7}{3} = -\frac{2\sqrt{10}}{3}.
$$
20. If $(6,-8)$ is a point on the angle $\theta$, then $\sin \theta = \underline{\color{red}{-\frac{4}{5}}}$ and $\tan \theta = \underline{\color{red}{-\frac{4}{3}}}$
* For this one it also seems like we need to draw a picture, and once we do we see that it forms a Pythagorean 6-8-10 triple (with hypotenuse of length 10).
* The raises the additional complication that we now need to divide all the lengths of the triangle by $10$ to ensure that the hypotenuse is exactly $1$ (corresponding to the radius of the unit circle).
* Once we've done that, we get a triangle with sides of $\frac{6}{10}$, $\frac{8}{10}$, and $\frac{10}{10}$. Or in other words, a triangle with corners at $(0,0)$, $\left(\frac{6}{10},0\right)$, and $\left(\frac{6}{10},-\frac{8}{10}\right)$
* Given these coordinates, we can immediately get $\sin(\theta)$ by looking at the $y$ coordinate of the third corner (the point that is actually **on** the unit circle), namely, $-\frac{8}{10} = -\frac{4}{5}$.
* We can also get $\cos(\theta)$ from this point's $x$ coordinate, namely, $\frac{6}{10} = \frac{3}{5}$.
* And now that we have both $\sin(\theta)$ and $\cos(\theta)$, we can compute the tangent:
$$
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-\frac{4}{5}}{\frac{3}{5}} = -\frac{4}{5} \cdot \frac{5}{3} = -\frac{4}{3}.
$$
## Part 2 (Calculator OK)
Find the value of each.
21.
$$
\sin\left(-\frac{9\pi}{5}\right) \approx 0.588 \; \checkmark
$$
* This section allows calculators, but I'm going to try solving without one so we can get more practice with identities and simplification.
* Here let's start by applying our identity $\sin(-\theta) = -\sin(\theta)$, to get
$$
\sin\left(-\frac{9\pi}{5}\right) = -\sin\left(\frac{9\pi}{5}\right).
$$
* But, we also see that $\frac{9\pi}{5}$ is aaalmost $\frac{10\pi}{5} = 2\pi$, meaning, it's almost all the way around the unit circle, just $\frac{10\pi}{5} - \frac{9\pi}{5} = \frac{\pi}{5}$ short.
* But if it's $\frac{\pi}{5}$ short, that (by definition) means that going this distance brings us to the same point we get to by traveling $-\frac{\pi}{5}$:
$$
-\sin\left(\frac{9\pi}{5}\right) = -\sin\left(-\frac{\pi}{5}\right)
$$
and we can apply our $\sin(-\theta)$ identity once more to get
$$
-\sin(-\frac{\pi}{5}) = -\left(-\sin\left(\frac{\pi}{5}\right)\right) = \sin\left(\frac{\pi}{5}\right),
$$
since the double-negatives cancel each other out.
* At this point there's not much else we can do (without knowing fancier math tricks), so we just plug this into our calculator to get (rounding to three decimal places) $0.588$.
22. $\cot( 58^\circ ) \approx 0.625 \; \checkmark$
23. $\sec \frac{3\pi}{11} \approx 1.527 \; \checkmark$
24. Write an equation and then find the value of $x$.
* $\cos(x) = \frac{68}{84} \implies x = \cos^{-1}(68/84) \approx 35.951 \; \checkmark$
* ($\cos^{-1}(x)$ is sometimes also called $\arccos(x)$)
25. Draw and label a diagram. Write an equation and then solve.
To measure the height of Lincoln’s face on Mt. Rushmore, two sightings are taken **800 feet** from the base of the mountain. The angle of elevation to the bottom of Lincoln’s face is **32°**. The angle of elevation to the top is **35°**.
What is the height of Lincoln’s face?
* Let $a$ represent the height of the cliffside **up to the bottom of** Lincoln's head, then let $b$ represent the height of Lincoln's head itself. By drawing the picture, we get
$$
\tan(32) = \frac{a}{800} \implies a = 800\tan(32)\\
\tan(35) = \frac{a+b}{800} \implies a+b = 800\tan(35) \\
\implies b = 800\tan(35) - a = 800\tan(35)-800\tan(32) \\
\implies b = 800(\tan(35) - \tan(32)) \\
= 60.271 \; \checkmark
$$

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