# 2.1: Trigonometry and Right Angles

**At Grade**Created by: CK-12

## Basic Functions

**In-Text Examples**

1) Students may get mixed up and think that a relation is a function if it has a unique \begin{align*}x-\end{align*}value for every \begin{align*}y-\end{align*}value instead of the other way around. Stress that if the same \begin{align*}x-\end{align*}value shows up twice, paired with two different \begin{align*}y-\end{align*}values, then the relation is not a function (so relation A is not), but if every \begin{align*}x-\end{align*}value has only one corresponding \begin{align*}y-\end{align*}value, the relation is a function (so relation B is).

2) On part a, students may think the domain is restricted to positive numbers; explain that that restriction applies to the range, not the domain. (We can find the square of any real number \begin{align*}x\end{align*}, but the \begin{align*}y-\end{align*}value we get back will always be positive.)

On part c, students might think of trying to draw a line or curve connecting those points, or might just interpolate to assume that the domain includes all the numbers between \begin{align*}2\end{align*} and \begin{align*}5\end{align*}. Explain if necessary that those points aren’t representative samples of the function, they are all the points in the function, so the domain consists only of the three \begin{align*}x-\end{align*}values given.

4) Students are likely to try plugging the numbers \begin{align*}15\end{align*} and \begin{align*}30\end{align*} into their equations; this will lead to error because those are the times in minutes, not hours, while the speeds given are in miles per hour. Remind them to convert minutes to hours before solving the problem.

6) Of course, the most common mistake here is to forget that only three sides of the enclosure need to be fenced because the barn will make the fourth side.

**Review Questions**

1) Part a is subject to the same error as Example 1 above. For students stuck on part b, suggest drawing a graph, or just ask them to consider whether it seems possible to plug in the same \begin{align*}x-\end{align*}value twice and get two different \begin{align*}y-\end{align*}values.

3) Part d is a possible sticking point, as students may not know what to do with the equation from part c. Remind them that they need their profit to be greater than zero, so they must find the smallest value of \begin{align*}x\end{align*} that will make \begin{align*}P(x)\end{align*} greater than zero. Also, remind them that their answer should be a whole number.

4) On part b, students may jump to the conclusion that the range is just the positive real numbers, because the equation resembles \begin{align*}y = x^2\end{align*}. When they graph the function, point out to them that part of the graph is below the \begin{align*}x-\end{align*}axis, so the range actually includes those \begin{align*}y-\end{align*}values as well. Specifically, since the vertex has \begin{align*}y-\end{align*}coordinate \begin{align*}-3.25\end{align*}, the range consists of all real numbers greater than or equal to \begin{align*}-3.25\end{align*}.

5) Students may identify the vertical asymptote at \begin{align*}x = -3\end{align*} but miss the horizontal one at \begin{align*}y = 1\end{align*}, especially because it’s not as easy to derive from the equation.

6) Students may get confused and set up the equation as if the cost per person were \begin{align*}\$500\end{align*} and they were trying to figure out the total cost for \begin{align*}p\end{align*} people. Stress that \begin{align*}\$500\end{align*} is the total cost and that they are trying to figure out how much each person will pay if the bill is split among \begin{align*}p\end{align*} people.

## Angles in Triangles

**In-Text Examples**

1) Students shouldn’t get hung up on trying to figure out if the triangles are right, acute, or obtuse; it’s harder to figure those out from the side lengths alone, but easy to tell if the triangle is equilateral, isosceles, or scalene. They may notice that \begin{align*}3-4-5\end{align*} is a Pythagorean triple (making that triangle a right triangle), and it may be worth pointing out that all equilateral triangles are acute (because all three of their angles must measure \begin{align*}60^\circ\end{align*}).

(If they really want to know how to tell if a triangle is acute or obtuse just from the side lengths, you can tell them this: Label the longest side \begin{align*}c\end{align*} and the other two sides \begin{align*}a\end{align*} and \begin{align*}b\end{align*}, and then compare \begin{align*}c^2\end{align*} to the quantity \begin{align*}(a^2 + b^2)\end{align*}. If \begin{align*}c^2\end{align*} is greater than the sum of the other two squares, the triangle is obtuse; if it is smaller, the triangle is acute; and if it is equal, of course, the triangle is right.)

2) Students should avoid jumping to the conclusion on part c that the \begin{align*}50^\circ\end{align*} angle is one of the two angles that are equal—although it could be, it might not be, and so there are two possible solutions.

**Review Questions**

1) This problem has two right answers, so don’t let students agonize over which one is correct.

4) For part a, students may need to be reminded what a complement is. (Complementary angles are angles that add up to \begin{align*}90^\circ\end{align*}.) Then, they may try to solve part b by working out \begin{align*}180 - 90 - 23 = 67\end{align*}. This method does yield the correct answer, but misses the point of the problem: once we know the two non-right angles are complementary, all we have to do is subtract \begin{align*}23\end{align*} from \begin{align*}90\end{align*} to get the missing angle.

5) Drawing a picture may help students who get stuck on this problem, even though the problem is really more algebra than geometry. The important thing they may forget is that they know all three angles must add up to \begin{align*}180^\circ\end{align*}, which means they can set up the equation \begin{align*}D + O + G = 180\end{align*}, substitute 2D and 3D for \begin{align*}O\end{align*} and \begin{align*}G\end{align*} respectively, and solve for \begin{align*}D\end{align*}.

8) The triangles certainly look similar, and students may think they are because they have two sides in proportion and one angle the same. However, since the angle isn’t between the two sides, we can’t actually tell if the triangles are similar.

9) The fact that the numbers \begin{align*}100\end{align*} and \begin{align*}20\end{align*} appear next to each other may tempt students to set up the proportion \begin{align*}\frac{20}{100} = \frac{24}{x}\end{align*}, or \begin{align*}\frac{100}{20} = \frac{24}{x}\end{align*}. Have them draw a diagram to see which distances they are actually comparing, or simply remind them that they must pair the flagpole with *its* shadow and the building with its shadow to get the right proportion: \begin{align*}\frac{20}{24} = =\frac{x} {100}\end{align*}.

10) Make sure answers to this problem demonstrate an understanding that similar triangles are not necessarily congruent—i.e. they do not necessarily have the same side lengths.

## Measuring Rotation

**In-Text Examples**

1) For students who have trouble keeping the terms “acute” and “obtuse” straight, the mnemonic “a cute little angle” may help remind them that acute angles are the smaller ones.

2) Since protractors like the one shown here display two different numbers for the measure of an angle, students may read off the wrong one. Remind them to think about whether the angle appears to be acute or obtuse, and figure out which number makes sense based on that. (Another way to check is to note that the end of the protractor they placed against one side of the angle either reads \begin{align*}0^\circ\end{align*} on the inside “track” and \begin{align*}180^\circ\end{align*} on the outside track, or vice versa. The track on which it reads \begin{align*}0^\circ\end{align*} is the one from which they should read off the measure of the other side of the angle. For example, in the illustration shown in the text, the end of the protractor placed against the bottom side of the angle reads \begin{align*}0^\circ\end{align*} on the inside track, so the inside track is the one to use for determining the angle measure, and so we know it is \begin{align*}50^\circ\end{align*} and not \begin{align*}130^\circ\end{align*}.)

3) Students may get their calculations backwards here, possibly due to a vague notion that the larger wheel should rotate a greater number of times. Explain if necessary that since the wheels both travel the same distance along their circumference and the larger wheel has more circumference, it doesn't have to make as many rotations to travel that distance. (A possibly useful analogy is that of a shorter person having to take more steps to keep up with a taller person.)

**Review Questions**

3) On part c, students may think they are done when they have converted the decimal portion to minutes and forget about converting the remainder to seconds, or they may “convert” to seconds by just copying the number after the decimal point—e.g., expressing \begin{align*}57.6\text{'}\end{align*} as \begin{align*}57\text{'}6\text{''}\end{align*}. Remind them that \begin{align*}57.6\text{'}\end{align*} is equal to \begin{align*}57 \frac{6}{10}\;\mathrm{minutes}\end{align*}, and they need to figure out how many seconds are in \begin{align*}\frac{6}{10}\end{align*} of a minute if each second is \begin{align*}\frac{1}{60}\end{align*} of a minute.

4) Possible errors here include converting the seconds but not the minutes to decimal form, or the minutes but not the seconds. Referring back to page 33 should help students remember how to perform these conversions.

7) Students may get the diameter of the wheels mixed up with the distance between them, and plug in the wrong one at the wrong time.

9) \begin{align*}-120^\circ\end{align*} is a possible wrong answer for part a. Demonstrate that an angle of \begin{align*}-120^\circ\end{align*} falls in quadrant III, while an angle of \begin{align*}120^\circ\end{align*} falls in quadrant II, so they are not co-terminal.

10) The length of the axles is a red herring, and so is the distance between them; students may think they need to find which one of the four wheels makes the most rotations, but really they only need to find whether the front or back wheels rotate more. (Also, the answers given in the text are incorrect; the numbers of revolutions should be \begin{align*}200\end{align*} and \begin{align*}66.67\end{align*} respectively, and the difference in the number of degrees should be \begin{align*}48000\end{align*}.)

**Additional Problems**

1) What is the angle between the hands of a clock at 6:30? (Remember, the hour hand is not directly on the 6.)

2) Name an angle that is coterminal with \begin{align*}-180^\circ.\end{align*}

**Answers to Additional Problems**

1) \begin{align*}15^\circ\end{align*}

2) Answers will vary. Possible answers include \begin{align*}180^\circ\end{align*} and \begin{align*}540^\circ\end{align*}.

## Defining Trigonometric Functions

**In-Text Examples**

3) Students may mix up the definitions of secant and cosecant. Emphasize that secant is the reciprocal of *co*sine and *co*secant is the reciprocal of sine, so there is exactly one “co-” function per pair of reciprocals (and since tangent and cotangent are reciprocals, they too fit this pattern).

5) Students who are still thinking in terms of angles in triangles may get stuck here. Remind them to think of the trig functions as ratios of \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}coordinates instead; using the definitions above, they can plug in any values for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, even values for which it isn’t possible to draw a triangle and measure the side lengths.

Also, because students first learned the definition of sine before the definition of cosine, they can easily get confused and think that the sine value is the \begin{align*}x-\end{align*}coordinate and the cosine value is the \begin{align*}y-\end{align*}coordinate. Even if they “know” that’s wrong, it’s still an easy trap to fall into any time they’re not thinking very hard about it (sometimes even after they’ve been studying trigonometry for quite a while!). Remind them to watch out for this error and to double-check their answers whenever they are finding sines and cosines with this method.

**Review Questions**

6) The answer to part b, of course, is not simply \begin{align*}2y\end{align*}; it’s true that the length of \begin{align*}BD\end{align*} is \begin{align*}2y\end{align*}, but the point is that it is also 1 because triangle \begin{align*}ABD\end{align*} is equilateral. Similarly, the answer to 6c is not just \begin{align*}y\end{align*}, but \begin{align*}\frac{1}{2}\end{align*}; this means that \begin{align*}y = \frac{1}{2}\end{align*}, and that’s important for solving the rest of the problem.

8) The ratios for \begin{align*}60^\circ\end{align*} angles are easy to mix up with the ratios for \begin{align*}30^\circ\end{align*} angles, especially since the values of a given trig function for a \begin{align*}60^\circ\end{align*} angle is the same as the value of the corresponding “co-” function for a \begin{align*}30^\circ\end{align*} angle.

9) Students may give the knee-jerk answer “quadrants I and II” because that’s where the \begin{align*}y-\end{align*}values are positive, or “quadrants I and IV” because that’s where the \begin{align*}x-\end{align*}values are positive. Remind them that the value of the tangent function depends on both the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}value: since the tangent is \begin{align*}\frac{y}{x}\end{align*}, is it positive or negative when \begin{align*}x\end{align*} and \begin{align*}y\end{align*} are both positive? Both negative? How about when one is positive and the other is negative? In which quadrant(s) does each of those conditions hold?

10) Possible wrong answers include “it’s five times \begin{align*}30^\circ\end{align*},” “it’s the supplement of \begin{align*}30^\circ\end{align*},” and “it’s \begin{align*}30^\circ\end{align*} plus \begin{align*}120^\circ\end{align*}.” Although these are all technically true, they aren’t what we’re looking for because they aren’t useful in this case. The correct answer is along the lines of “it’s like a \begin{align*}30^\circ\end{align*} angle, but reflected across the \begin{align*}y-\end{align*}axis,” because noticing this fact helps us figure out what the ordered pair for a \begin{align*}150^\circ\end{align*} angle is.

**Additional Problems**

1) Sketch the angle \begin{align*}210^\circ\end{align*} on the unit circle. What do you think its ordered pair is?

**Answers to Additional Problems**

1) \begin{align*}\left (\frac{-\sqrt{3}} {2}, \frac{-1}{2}\right )\end{align*}

## Trigonometric Functions of Any Angle

*In-Text Examples*

1) Make sure students remember that the reference angle is always the distance to the closest part of the \begin{align*}x-\end{align*}axis, never the \begin{align*}y-\end{align*}axis—even if the \begin{align*}y-\end{align*}axis is closer.

4) Students may be momentarily thrown by part b; remind them that an angle of *negative* \begin{align*}300^\circ\end{align*} is co-terminal with an angle of positive \begin{align*}60^\circ\end{align*}.

5) Some students may try dividing the given angles by some number, instead of subtracting \begin{align*}360^\circ\end{align*} from them. Also, on later problems they may forget that they aren’t done when they reduce the angle down to one that’s less than \begin{align*}360^\circ\end{align*}, and that they still have to find the reference angle for that angle.

6) Note that the cosine column comes before the sine column; this may confuse students momentarily.

7) Make sure calculators are in degree mode; in radian mode the answer will appear to be \begin{align*}0.8589.\end{align*}

**Review Questions**

7) “Between \begin{align*}10\end{align*} and \begin{align*}15\;\mathrm{degrees}\end{align*}” or “between \begin{align*}165\end{align*} and \begin{align*}170\;\mathrm{degrees}\end{align*}” is as precise as the answer needs to be. (Either of those approximations is correct, and students should be made aware of this, as the fact that there is more than one angle for a given sine value will be important later.)

8) Students may think of choosing \begin{align*}60^\circ\end{align*} as the “special angle.” The value of \begin{align*}\tan(50^\circ)\end{align*} is fairly close to the value of \begin{align*}\tan(60^\circ)\end{align*}, but you’d need a calculator to figure out the value of \begin{align*}\tan(60^\circ)\end{align*} in decimal form in order to compare the two tangent values. The value of \begin{align*}\tan(45^\circ)\end{align*}, though, is simply 1, and in any case \begin{align*}50^\circ\end{align*} is closer to \begin{align*}45^\circ\end{align*} than to \begin{align*}60^\circ\end{align*}, so it makes more sense to use \begin{align*}45^\circ\end{align*} as the special angle.

9) Leaving calculators in radian mode will yield the wrong answers \begin{align*}-0.9820\end{align*} and \begin{align*}45.1831\end{align*}.

10) Students may end up thinking a little too hard about this problem. All they’re supposed to conjecture is that the two expressions are not equal, so if any of them struggle with this problem, find out if they’ve got that much figured out and reassure them they can stop there.

**Additional Problems**

1) Use a calculator to find the tangent of \begin{align*}86^\circ, 87^\circ, 88^\circ,\end{align*} and \begin{align*}89^\circ\end{align*}. Then, find the tangent of \begin{align*}94^\circ, 93^\circ, 92^\circ,\end{align*} and \begin{align*}91^\circ\end{align*}. Now make a conjecture about the behavior of the tangent function as \begin{align*}x\end{align*} approaches \begin{align*}90^\circ.\end{align*}

**Answers to Additional Problems**

1) The values of the tangent function are as follows:

\begin{align*}& X && \tan x\\ & 86^\circ && 14.3007\\ & 87^\circ && 19.0811\\ & 88^\circ && 28.6363\\ & 89^\circ && 57.2900\\ & 90^\circ && \text{undefined}\\ & 91^\circ && -57.2900\\ & 92^\circ && -28.6363\\ & 93^\circ && -19.0811\\ & 94^\circ && -14.3007\end{align*}

The tangent function approaches infinity as \begin{align*}x\end{align*} approaches \begin{align*}90^\circ\end{align*} from below, and approaches negative infinity as \begin{align*}x\end{align*} approaches \begin{align*}90^\circ\end{align*} from above.

## Relating Trigonometric Functions

**In-Text Examples**

4) Demonstrating that \begin{align*}\cot \theta = \frac{\cos \theta}{\sin \theta}\end{align*} because cot is the reciprocal of tan and \begin{align*}\tan \theta = \frac{\sin \theta}{\cos \theta}\end{align*} is an equally valid answer.

5) Some students may try to simply subtract \begin{align*}\cos \theta\end{align*} from \begin{align*}1\end{align*} to get \begin{align*}\sin \theta\end{align*}; others may subtract \begin{align*}\cos^2 \theta\end{align*} from \begin{align*}1\end{align*} but then forget to take the square root of the answer.

6) The biggest problem here is that students may simply not have any idea where to begin. Telling them the first step may help, but it may also be easier for them to work the problem out “backwards” instead. When you start with \begin{align*}\cot^2 \theta + 1 = \csc^2 \theta\end{align*}, the logical first step is to rewrite \begin{align*}\cot^2 \theta\end{align*} as \begin{align*}\frac{\cos^2 \theta}{\sin^2 \theta}\end{align*} and \begin{align*}\csc^2 \theta\end{align*} as \begin{align*}\frac{1}{\sin^2 \theta}\end{align*}; this suggests the idea of also writing \begin{align*}1\end{align*} as \begin{align*}\frac{\sin^2 \theta}{\sin^2 \theta}\end{align*} and then dividing through by \begin{align*}sin^2 \theta.\end{align*}

**Review Questions**

2) You may need to remind students here not to use the \begin{align*}``\sin^{-1}\text{''}\end{align*} function on their calculators to find the cosecant, but instead to find the sine and then use the \begin{align*}``x^{-1}\text{''}\end{align*} key to find the reciprocal of the sine.

3) On this and the previous problem, some students may still be confusing the domain with the range. They may also not realize that an input value that makes a function undefined is an input value that must be excluded from the function’s domain.

8) Students may get the functions here mixed up with their reciprocals, or may get the Pythagorean identity backwards.

**Additional Problems**

1) If \begin{align*}\cos \theta = \frac{24}{25}\end{align*}, what is the value of tan \begin{align*}\theta\end{align*}?

2) If \begin{align*}\sin \theta = \frac{5}{13}\end{align*}, what is the value of cot \begin{align*}\theta\end{align*}?

**Answers to Additional Problems**

1) \begin{align*}\frac{7}{24}\end{align*}. Solving this problem takes two steps: first finding \begin{align*}\sin \theta = \frac{7}{25}\end{align*} using the Pythagorean identity, and then finding \begin{align*}\frac{\sin \theta}{\cos \theta}\end{align*}.

2) \begin{align*}\frac{12}{5}\end{align*}. The problem is similar to the previous one, but students must remember that \begin{align*}\cot \theta\end{align*} equals \begin{align*}\frac{\cos \theta}{\sin \theta}\end{align*} and not \begin{align*}\frac{\sin \theta}{\cos \theta}\end{align*}.

## Applications of Right Triangle Trigonometry

**In-Text Examples**

1) Some students may have a hard time understanding which ratios to use to solve which triangles. Try to make clear to them that they should pick a ratio for which they know the angle measure and one of the two sides involved, and then use the ratio to find the other side—or pick a ratio for which they know the two sides, and use the ratio to find the angle. A trig ratio for which they only know one of those three pieces of information won’t do them any good, and a ratio for which they already have all three will only tell them what they already know.

2) Using the sine to find the second leg might seem like a good idea, but it would involve plugging in the value we just found for the hypotenuse, which is an approximation. Using the tangent is better because it allows us to plug in the exact value we were given for the length of the first leg, and plugging in an exact value instead of an approximation will yield a more accurate result.

**Review Questions**

1) As in example 2, students should try to use the given numbers whenever possible rather than plugging the approximate values they’ve found earlier into later parts of the problem. In this case, that means they should use the sine to find b and the cosine to find a, rather than finding one of them and then using the tangent or the Pythagorean Theorem to find the other.

3) Note that this problem refers to Example 2 in the earlier part of the lesson, not to either of the problems directly above.

7) Students may try to plug in \begin{align*}100^\circ\end{align*} as one of the angles in the triangle, when the relevant angles are actually \begin{align*}80^\circ\end{align*} and \begin{align*}10^\circ.\end{align*}

8) Some students may not see what this problem has to do with solving triangles; others may see that they have to divide the quadrilateral into triangles, but may pick the wrong way to do it—drawing a diagonal from upper right to lower left instead of vice versa, which makes the problem unsolvable using the tools they currently have.

9) Don’t let students get hung up on trying to figure out how to find the width of the pond at its widest point; “how wide” here just means “how far is it from \begin{align*}A\end{align*} to \begin{align*}B\end{align*}?”

10) Students may jump to the conclusion that \begin{align*}\triangle PAN\end{align*} is a right triangle and try to find \begin{align*}x\end{align*} based on the sine or tangent of \begin{align*}50^\circ.\end{align*}

**Additional Problems**

1) a) What value would you get for the height of the tree in Example 3 if you did not take the height of the person into account?

b) In example 4, why couldn’t we just add \begin{align*}5\;\mathrm{feet}\end{align*} to the answer we found in the first part (where we didn’t take the person’s height into account) to get the answer to the second part (where we did)?

**Answers to Additional Problems**

1) a) \begin{align*}15.63\;\mathrm{feet.}\end{align*}

b) As you can see from the diagrams, taking the height of the viewer into account in example 3 just required us to shift the triangle up \begin{align*}5\;\mathrm{feet}\end{align*}. In example 4, however, accounting for the person’s height required us to actually lengthen one of the sides of the triangle by \begin{align*}5\;\mathrm{feet}\end{align*}, which resulted in the proportions of the whole triangle being different. This is why drawing accurate diagrams is important in trigonometry; sometimes we need them to make it absolutely clear what information we can just assume and what information we can’t.

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