MATHEMATICS CONTENT KNOWLEDGE (0061) TRIG TEST 1 ... · Trigonometry, loosely translated, means “the study of triangles”. Simplified even more: it’s the relationship between

MATHEMATICS: CONTENT KNOWLEDGE (0061) TRIG TEST 1 (PRAXIS2MATH.COM) ________________________________________________________________________

READ THE FOLLOWING BEFORE YOU BEGIN!

You are about to embark on a trigonometric journey. (Yeah that’s a real word, Google it.) Those reading this probably haven’t had a real Trig course in: 6 years for some, 7+ years for many, or in the case of myself, never at all. High school didn’t offer it, colleges combined it with other courses, and when all was said and done, some of us only had a vague idea of what Trigonometry really was. The following exam is going to test your comprehension on Trigonometry. What makes me qualified to test you? The high school that hired me gave me Trig as my first course. I have successfully taught dozens of students (nearing “hundreds” now), and as a result, our school’s NECAP scores have improved in that area. Enrollment in Trig has also increased during the time I have been teaching at my high school. According to one of my students: “You, like, totally rock and stuff!”

So before you begin this journey, I will give you a crash-course in Trigonometry. Consider the following two pages your miniature lesson in Trig. I will gladly answer questions via email if you ask. Half of the time I will be able to respond to you within minutes (unless I’m in school or sleeping, all Eastern time), so don’t hesitate to ask. Let’s begin…

Trigonometry, loosely translated, means “the study of triangles”. Simplified even more: it’s the relationship between the sides of a triangle. That’s it. Very simple, but there’s so much more than that.

The basic triangle is the right triangle, which is obviously a triangle that has a right angle (90 degrees). The side opposite the right angle is called the hypotenuse, which is always the longest side of a right triangle. Right triangles can easily vary in sizes, so the easiest right triangle for us to study is a triangle which has a hypotenuse of 1. Now if you were to draw a few right triangles on an xy-plane using a hypotenuse of 1, it would look something like:

Each triangle in the picture has a hypotenuse of 1, and a right angle that sits on the x-axis. The angle formed by the hypotenuse and the x-axis is called the “central angle”. It’s a pretty important angle for beginners in Trigonometry. Now , if we were to connect the end-points of all the hypotenuses (hmm, the plural version sounds

funny to say), we would get something like this:

Do you notice something? We ended up drawing a circle around the origin of the xy-axis. This circle is called the “unit circle”. If I were to draw any line that starts at coordinate (0,0), and draw it out towards the edge of this unit circle (a radius), it’s length would be 1 unit. From that line, I could draw another line straight back to the x-axis, and using that x-axis, we would form a triangle. I could draw an

infinite number of triangles with a hypotenuse of one, just by using the unit circle as a guide.

So here’s where the real trig comes into play. Using the unit circle as a guide, if we were to draw a triangle with a central angle of 30o, it would look like something below:

= 30o) Now we can study the relationship of sides using this triangle and as our guide. Let’s begin:

The ratio of the side opposite divided by the hypotenuse can be

written as: (

). This is

called the sin of (sin ). If you were to put sin(30) in your calculator, you will get

an answer of

. So: sin =

, and since our hypotenuse is

1, our opposite side (y) must be equal to one-half.

The ratio of the side adjacent to divided by the hypotenuse can

be written as: (

). This is called the cos of (cos ). If you were

to put cos(30) in your calculator, you will get an answer of

. So:

cos =

, and since our hypotenuse is 1, our adjacent side

(x) must be equal to

.

The ratio of the side opposite to divided by side adjacent to

can be written as: (

). This is called the tan of (tan ). If you

were to put tan(30) in your calculator, you will get an answer of

.

So: tan =

, written properly is:

.


OK so let’s recap:

sin =

cos =

tan =

Ever hear of a little acronym pronounced “so-cuh-toe-uh”? Well it looks like this: SOHCAHTOA. Look at it carefully, and then look at the recap I wrote just above it. SOH (Sin = O / H), CAH (Cos = A / H), TOA (Tan = O / A). Just remembering SOHCAHTOA will help you with a few problems on the 0061 exam itself.

And if you haven’t noticed yet, we’ve only dealt with 3 ratios of sides, there are still 3 more to go. I’ll take it one step at a time:

sin =

If we are to take the reciprocal of this, we get

.

The reciprocal is called csc (cosecant). So: csc =

cos =


.

The reciprocal is called sec (secant). So: sec =

tan =


.

The reciprocal is called cot (cotangent). So: cot =

There’s no handy-dandy little acronym or anything for the other three trig functions, they are just the reciprocals of the first three. It should also be noted that a triangle can exist in any of the 4 quadrants on an xy-plane, thus resulting in negative fractions. The hypotenuse is never negative, but the adjacent and opposite sides of a triangle can be. Here’s a little chart explaining what I’m talking about below. Take a big look before you continue:

You should have noticed a little helpful reminder in that diagram for you. “All Students Take Calculus”, which means ALL trig functions are positive in the first quadrant, only Sin is positive in the second, only Tan is positive in the third, and only Cos is positive in the fourth. The reciprocals work the same way: All are positive in the first, csc is positive in the second, cot is positive in the third, sec is positive in the fourth.

So that’s the basics of Trig. I’m not going to go any farther, because much of what you need to know is explained in the answers of the following questions. I’m also not going to get into explaining the functions part of this exam before you take it. If you’ve taken my other practice exams, the functions part of this exam should be easy for you. Other components of trigonometry are shown below:

=

=

a2 = b2 + c2 – (2 ∙ b ∙ c ∙ cos A)

Cos A =

In a right triangle: a2 + b2 = c2

It should be noted that in triangles, capital letters (A, B, C, etc) are denoted as angles, while lower case letters (a, b, c, etc) are the sides opposite those angles. Don’t screw that up. Labeling a triangle incorrectly is an automatic wrong answer on your exam!

In an almost entirely unrelated note, cos spelled correctly is “cosine”, and sin is spelled “sine”. We use cos and sin for short. I just want to clear this up now in case there is confusion later on for whatever reason. You most likely will see sin, cos, tan, etc on your 0061 exam, but who knows really.

You should treat this exam just like the other ones you purchased from me. Give yourself a calculator ready to go. Rid yourself of any and all distractions for a total of 33 minutes. Pencils and scrap paper should be ready to go. Make sure you do not cheat, do not pause, nothing. Treat it like the actual exam. Don’t forget to use your equations sheet. Good luck!

(PS – The following questions are 0061 quality)

33 MINUTES! THAT’S WHAT YOU GET FOR THIS PART!


1) There exists a right triangle PQR (not shown), where angle

PQR is a right angle, angle RPQ is 29o, and side PR is 12 units. Find the length of side PQ.

(A) 10.50 units

(B) 13.72 units

(C) 24.75 units

(D) Not enough information provided. 2) There exists a triangle on an xy-plane that has a central angle

. If sin = , and tan in which quadrant is the

triangle located?

(A) Quadrant I

(B) Quadrant II

(C) Quadrant III

(D) Quadrant IV

3) A kid accidentally threw his Frisbee on top of a building that

has a flat roof. He needs to get a ladder, but first he needs to find out how tall the building is. He is 15 feet away from the building, and the angle of elevation to the top of the building is 58o. Which answer below best represents the height of the building? (A) 8 feet

(B) 13 feet

(C) 24 feet

(D) 36 feet

4) A summer camp is constructing a zip-line that will hang from

a 100-foot pole. The end of the line will be fastened to a platform that is 5 feet off the ground. If the angle of elevation from the platform to the top of the pole is 25o, then what is the amount of wire they will need in order to make the zip-line. (A) 203.73 feet

(B) 224.79 feet

(C) 236.62 feet

(D) Not enough information provided.

OK so here’s the deal. If you found the first four questions to be troublesome, then you should skip to the answer sheet NOW and see if you got the first four right. There is no point in continuing with this exam if you have no clue what you’re doing, correct? Go to the answer sheet, see what the answers are, read the explanations, and make sure you know what you are doing right and what you are doing wrong. Only when you fully understand your mistakes and feel comfortable with yourself should you continue. But if you feel confident so far, then by all means go on to number 5. 5) A car is travelling up a straight road that has an

inclination of 12o. If the car initially started at sea level, then how high above sea level is the car after it has travelled for 6 miles?

(A) 1.25 miles

(B) 1.28 miles

(C) 5.87 miles

(D) 11.93 miles

6) A woman is skiing down a mountain with a vertical height of 1750 feet. The distance from the top of the mountain to the bottom is 3200 feet. What is the angle of elevation? (A) 28.67o

(B) 33.15 o

(C) 56.85 o

(D) 61.32 o

7) A man was standing 9 feet away from a sign that was resting on top of a post. He realized that the angle of elevation from where he stood to the bottom of the sign was 29o, and the angle of elevation to the top of the sign was 40o. How tall is the sign? (A) 1.75 ft

(B) 2.03 ft

(C) 2.27 ft

(D) 2.56 ft


8) There exists triangle JKL (not shown) that has side lengths of

21, 31, and 35. Which answer below best represents the biggest angle in triangle JKL? (A) 72o

(B) 76o

(C) 82o

(D) 89o 9) There exists a utility pole that normally sits at a 90 degree

angle with the ground. However, a wind storm has pushed the pole over an extra 8o directly towards the sun. The shadow (which is directly away from the sun) yields a measurement of 17 feet, and the angle from the end of the shadow to the top of the pole is 48o. Which answer best represents the length of the pole? (A) 13 feet

(B) 17 feet

(C) 18 feet

(D) 23 feet

10) Using information given in the following diagram, which answer best represents the measurement of side “a”?

(A) 9 cm

(B) 11 cm

(C) 13 cm


11) Me and a friend of mine met for coffee at a small town in

Vermont (where I live, true story). When we departed, he drove at a 37o angle above route 7 in a straight line headed north east. I drove 23o below route 7 in a straight line headed south east. If he drove a total distance of 16 miles, and I drove for 21 miles, which answer below best represents how far away we were from each other when we reached our destinations?

(A) 18 miles

(B) 19 miles

(C) 24 miles

(D) 26.5 miles 12) The sun shined down on a man who was 73 inches tall, which

produced a shadow that was 102 inches long on the ground. What is the angle of elevation of the sun? (A) 35.59o

(B) 45.70o

(C) 54.41o



13) An airplane takes off from an airport flying in a straight line at

a steady rate of 400 miles per hour, with an angle of elevation of 8o. After 2 minutes of flight, how high off the ground is the airplane? (A) 0.93 miles

(B) 0.98 miles

(C) 1.86 miles

(D) 13.20 miles

14) A pilot is sitting in an airplane that is 6,000 feet in the air. He

looked out his window and saw “Town A” with an angle of depression of 29o. Out of the same window, he saw “Town B” with an angle of depression of 67o. Which answer below best represents how far apart the two towns were from each other? (A) 4,688 feet

(B) 7,680 feet

(C) 8,277 feet

(D) 10,809 feet

The next problem is the MOAP (Mother Of All Problems). It won’t be easy and I don’t expect anybody but the best to get it right. So if you think you can conquer this beauty, then go ahead and continue. Otherwise, check your first 14 answers, see if they are right and if you know what you are doing. If you are all set after that, then continue:

15) The diagram below is of scalene triangle ABC. This triangle

has a line BD that bisects AC. Angle ABC = 58o, while AC = 17 cm, and BC = 12 cm. Using the values given, what is the approximate value of BD?

(A) 8.5 cm

(B) 10.5 cm

(C) 14 cm

(D) 15 cm

This is the end of the test. I do not have any questions that say “what is the minimum or maximum of such-and-such trig equation”. You could easily graph any such question on your calculator using your y= function, and simply zoom in at the answer. For example: What is the MAX of: y = 3 sin (x) – 4? If you go to y= on your calculator, put in the equation above, then hit “zoom” and “7” (Trig), it will graph the curve for you, and as you can see the MAX would be -1. Anytime you are asked to find the max and/or min of a sin/cos curve, just graph it and look. It’s too easy.


DETAILED ANSWERS

1) ANSWER: (A) 10.50 units This is your basic trigonometric question:

I. The first thing we should do is draw this thing up so we know

what it looks like. Make sure you label everything correctly. Just to make it a little simpler, I also labeled the side we are looking for with an “x”:

II. Alright here we go. What we first need to do is figure out which trig function we are going to use to find x. If we are to use the angle RPQ, then we are dealing with the sides which are adjacent and hypotenuse to our angle. SOHCAHTOA, tells us to use Cosine.

III. cos =

Now fill it in:

cos 29 =

Multiply sides by 12, you will see that: 12 (cos 29) = x Put this in your calculator and you will get: x = 10.50 units, choice (B)

2) ANSWER: (B) Quadrant II This kind of question can be difficult unless you know the process. Let’s look at what we know:

I. sin = , and tan . Since sin is equal to the

opposite side divided by the hypotenuse, and that the hypotenuse can NEVER be negative, we know that our opposite side must be positive (because our fraction is positive). The opposite side is the side that goes up and down (see page 1), which is the y value. So if the y value must be positive, then the triangle must be in either quadrants I or II.

II. If tan is negative (tan , so it must be negative), then either the opposite or adjacent sides must be negative. Since we know the opposite side is positive, then the adjacent side must be negative. The only quadrant where the opposite side is positive and adjacent is negative is in Quadrant II, choice (B). For more clarification, check out the work I did for you on page 2. But if you are lazy, here’s the picture I gave you:


3) ANSWER: (C) 24 feet

First step, draw a picture. Noticing a pattern are we?

I. Here is the picture that I drew:

II. Next, we have to know what trig function to use. Well, we know the value of the side adjacent to the angle, and we need to figure out the value of the side opposite to the angle. Going over our acronym (SOHCAHTOA) helps us figure out that we need to use the tangent function.

III. Tan 58 =

Which can be re-written as: 15 Tan 58 = x And if you put that in your calculator, you will get 24.005, which is closest to the answer (C), 24 feet.

4) ANSWER: (B) 224.79 feet

Guess what my first step is:

I. Picture:

II. See what I did there? This question is a little different than the first three. Yes, the pole is 100 feet high, but the triangle we are using as a height of 95 feet because of the platform down below.

III. Next, we need to figure out which trig function to use. We are dealing with the side opposite our angle, and the hypotenuse. Using SOHCAHTOA, we see that we need to use sin.

IV. Sin 25 =

Cross multiply

95 = (Sin 25)(x) Divide by Sin 25

= x

x = 224.79 feet, choice (B)


5) ANSWER: (A) 1.25 miles

Gee, I wonder what I’ll do to start this one… diagram? Yes please.

I. Here we go:

II. Just like the other questions before this one, we need to figure out which trig function to use. Well, we’re looking at the side opposite the angle, and the hypotenuse. This means we’re using sin.

III. sin 12 =

Re-write it

6 sin 12 = x, so x = 1.25 miles, choice (A).

6) ANSWER: (B) 33.15o

Finally we have a problem in which we have to determine an angle, not a side measurement:

I. Take a stab at what I’m doing first….. yup:

II. Once again, we need to first determine which trig function we are to use. Well, I see a side opposite angle “x”, and we also know the hypotenuse. This tells me to use the sin function.

III. Set it up: sin x =

Find the inverse of sin to solve

The inverse of sin is called “arcsin”, and is found on most TI-Calculators by doing: 2nd (blue button), sin. This will show you sin-1. So do: sin-1 (1750/3200) Don’t forget to end your parenthesis. It won’t screw up this problem, but it’s a good habit to get into. Your calculator will show you 33.15o, choice (B)


7) ANSWER: (D) 2.56 ft Ahh, the first real tough trig problem. However, this is a relatively simple problem disguised as a tough one. Did you make it harder than it has to be? Let’s take a look:

I. First, you have to make sure you know what’s being asked. The question asks how tall the sign is, not how tall off the ground it is or how tall the sign & post are, just the sign. The second thing you should know is “angle of elevation”. Angle of elevation (also known as “angle of depression”, we’ll get to that later) is the measurement from the bottom to the top of something. In this case, it’s the measurement from the ground to the bottom and top of the sign. I would draw a picture next.

II. The picture(s) should look something like this:

III. As you can see, the angles technically make two triangles. If we can somehow find the height of the shorter triangle (x) and subtract it from the height of the taller triangle (y), we should be able to find out how tall the sign is.

IV. We have two parts of the triangle, an angle and a side. That’s a good start, so now we need to decide which trig function to use in order to find x and y. Well, we have the side opposite the angle, and the side adjacent to the angle. Opposite/Adjacent = Tangent (TOA). Now we start solving.

V. Since we’re using tangent, we get the following

equations:

tan 40 =

which can be re-written: y = 9 tan (40)

tan 29 =

which can be re-written: x = 9 tan (29)

Simplified, we get: y = 7.55 ft x = 4.99 ft Subtract them from each other and we get 2.56 ft, choice (D).


8) ANSWER: (C) 82o

Looks like a lot of work to figure out the answer, but it should actually take one step:

I. We are given a triangle, and the measurement of all three sides of that triangle. Your first instinct might be to find all three angles, and then pick the biggest angle of the three as your answer. However, your biggest angle is always the angle opposite the biggest side. Our biggest side in this triangle is 35 units.

II. We could draw a picture if we wanted, or in this case, we could jump straight to the equation since we already know what we want to find. We want to use an equation that uses all three sides, and finds an angle. This is the best one to use:

Cos A =

III. If we’re finding angle “A”, then side “a” must be 35 units. The

other two (b and c) are 21 and 31 units. It doesn’t matter which is which, so we can label it accordingly:

Cos A =

Cos A =

Cos A = .136 Take the arcos of both sides and you get: A = 82.19o, which is closest to choice (C).

9) ANSWER: (D) 23 feet

This question isn’t as obvious as some of the others.

I. The first thing I will do this time is something completely and totally different… draw a picture:

II. Do you notice something about the diagram that isn’t mentioned in the problem? I labeled the top angle 34o. I know it’s 34 because 180 – 48 – 98 = 34. With that, I can now use the law of sines to figure out the answer to x.

III.

=

Cross multiply.

17 (sin 48) = x (sin 34) Divide both sides by (sin 34).

= x Do it out.

22.59 = x Which is closest to 23 feet, choice (D).


10) ANSWER: (C) 13 cm

I’m hoping to maybe catch a few people off-guard with this question.

I. You should have looked at the diagram, tried to figure out what to do, and then realized something

This triangle has 2 angle measurements, so in order to get the third, we could just simply subtract the other two measurements from 180. In doing so, we know that the top angle is 76

o. We can’t figure out the answer without it.

II. Knowing all angle measurements, we can now use the law of sines:

III.

=

Cross multiply.

13.5 (sin 65) = x (sin 76) Divide both sides by (sin 34).

= x Do it out.

12.61 = x Which is closest to 13 feet, choice (C).

11) ANSWER: (B) 19 miles

Did you find this to be tricky? Some have, but if you haven’t, good for you!

I. Let’s take a look at the diagram:

II. What we really have here is one big triangle disguised as perhaps two triangles. What we want to find out is the distance between two points. We know two sides of a triangle, and an angle that is opposite of the third side (37o + 23o = 60o). Time to set this up:

III. c2 = 162 + 212 – (2 ∙ 16 ∙ 21 ∙ cos 60) One step at a time. c

2 = 697 – (336)

c2 = 361 Take the square root of both sides. c = 19 miles, choice (B).


12) ANSWER: (A) 35.59o

All the information you needed is right there.

I. Let’s take a look at the diagram:

II. Looks like we have ourselves a right triangle. The angle we want to find (x) is shown on the diagram. We know the measurements of the side opposite (73 inches) and adjacent (102 inches) to the angle. Opposite / Adjacent = Tangent.

III. Tan x =

Take the arctan of both sides (Press: 2nd, tan)

x = 35.59o, choice (A).

13) ANSWER: (C) 1.86 miles

Did you do your conversions first?

I. Draw it:

Don’t make fun of my airplane...

II. We aren’t given the exact bits of information we need, but we can use what we have to get our “y” value as shown in the picture. Since the plane is travelling at 400 miles per hour, it’s we can take 400 and divide it by 60 minutes to see how fast it’s going per minute. 400 / 60 = 6.6666 miles per minute Multiply it by 2, since the airplane flew for 2 minutes. The plane has travelled 13.33 miles!

III. Since our y value = 13.33333, we can now setup our equation. We know our angle (8o), and our hypotenuse. Since we want to find out our opposite side, we must use sin to figure it out!

sin 8 =

Multiply both sides by 13.333.

13.333 sin 8 = x Simplify x = 1.86 miles, choice (C).


14) ANSWER: (C) 8,277 feet

Fun to do!

I. Draw it again:

II. The above drawing will make you go cross-eyed if you look at it long enough. “Angle of Depression” means the angle going downward from the horizontal line from the height of the plane. Actually, angle of depression = angle of elevation most of the time, but this Praxis II 0061 exam uses confusing wording, and I don’t trust it one bit.

III. Look closely at our drawing, and you should notice that there are two triangles. Both triangles share the same side (6,000 feet), so I’ll just redraw those two triangles separately:

IV. Now we need to determine which trig function to use.

We know the side opposite the angle, and want to find the side adjacent to the angle. Opposite / Adjacent = tangent, so:

V. Tan 67 =

and: Tan 29 =

Now here’s a little hint for you; when you have a fraction equal to a fraction, you can legally switch the diagonal values, and it won’t change the fact that they are still equal to each other. Watch:

=

This is true. Now switch the “3” and the “8”.

=

Still true! Now switch the “4” and the “6”.

=

Still true! This will save you time on the exam.

So anyway:

Tan 67 =

and: Tan 29 =

Tan 67 is like saying it’s over one, so like I did above, let’s switch the diagonals by switching the Tan 67 with x, and Tan 29 with x + y:

x =

and: x + y =

simplify.

x = 2,546.85 and x + y = 10,824.29 So if we subtract the two from each other, we get: y = 8,277.44, closest to choice (C).


15) ANSWER: (C) 14 cm

Did you get it?

I. Draw the original with more labels:

II. As you can see, I took side AC, and divided it into two equal parts since that’s what the problem said to do. I did not, however, divide the ABC angle into two equal parts. Just because a line divides a side of a triangle in half, doesn’t mean it divides the angle in half! It also doesn’t make a right angle with side AC. So if you made one of those two mistakes, learn now!

III. Next, we need to figure out which part of the triangle to solve for first. Well, we have the angle and its opposite side, so we can use that to find angle BAC while using its opposite side, using law of sines:

=

Cross multiply

17 (sin A) = 12 (sin 58) Divide both sides by 17

Sin A =

Simplify

Sin A = 0.598622 Take the arcsin of both sides Angle A = 36.77o

OK, that’s the first part. And since we now know two angles, we can subtract those angles from 180 and get angle C. Angle C = 85.23o

IV. I say we re-draw the triangle:

Since we are trying to find BD, we must look at the bottom triangle. We have two sides (8.5, 12) and an angle (85.23). We can use the law of cosines to find that last side!

V. BD2 = 122 + 8.52 – (2 ∙ 12 ∙ 8.5 ∙ cos 85.23) Do the parenthesis first. BD2 = 122 + 8.52 – (16.96384) Do the right side. BD2 = 199.286 Square root of both sides

BD = 14.12, which is closest to choice (C).

If you would like to contact me for questions, please do at: [email protected]. I’m pretty quick to respond, so don’t hesitate!

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MATHEMATICS CONTENT KNOWLEDGE (0061) TRIG TEST 1 ... · Trigonometry, loosely translated, means “the study of triangles”. Simplified even more: it’s the relationship between

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