Right Triangle Trig Stuff I really, really hope you remember some things about trigonometry but if not we’re going to start at the beginning and work our way up anyway, so that’s okay. The triangles we’re going to deal with are exclusively right triangles. They have two acute angles and one right angle. They have two legs and one hypotenuse. Here’s a figure.

For this triangle, the legs are segments AB and BC. The length of side AB is c. We often just call this side c. The length of side BC is a. We often just call this side a. The hypotenuse of the triangle is AC and it has length b. Just like with the legs, we frequently just call this side b. You probably noticed that side a is opposite angle A, side c is opposite angle C, and side b is opposite angle B. For reasons that will be clear later (much later…like, maybe next year or the year after), it’s a good idea to draw your triangles so they are oriented like the one shown above. You might have memorized the Pythagorean theorem as a 2 + b2 = c 2 , which is totally fine…but doesn’t apply to the triangle above! For that triangle we’d have to use a 2 + c 2 = b2 . (You just have to be flexible in your thinking about these things.) We use the Pythagorean Theorem to find missing sides in right triangles. Problem: Find the missing side given the following information about each right triangle. Sketch each triangle! (Sketch it “correctly!”) a. Triangle ABC with right angle A. Given c = 5 b. Triangle ABC with right angle C. Given c = 6 and a = 10 . and a = 4 .

c. Triangle ABC with right angle B. Given c = 5 and a = 10 .

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d. Triangle ABC with right angle A. Given c = 24 and a = 25 .

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That last triangle we found the missing side for worked to have really nice, integer sides. Anytime we end up with three sides that are all integers, we say that we’ve found a Pythagorean Triple. There are many famous Pythagorean triples and you should memorize them. Here they are: •

3, 4, 5

•

5, 12, 13

•

7, 24, 25

•

8, 15, 17

It’s not even remotely as famous but 20, 21, 29 is also a Pythagorean triple and it’s my favorite. Memorize all of those! Since the numbers in those triples have a greatest common factor of 1, we say those triples are primitive. You can multiply any primitive by a non-negative integer and get another Pythagorean triple. (You can actually multiply by any non-zero number and get a set of three numbers that still satisfy the Theorem.) Okay, let’s get into some trig! First we’re going to build something in GeoGebra. 1. Start a new document. 2. Along the top find the segment tool, , and select it. 3. Click on the origin to create a point, A, then put a second point, B, somewhere on the x-axis. A segment should have been created between the points. 4. Find the perpendicular tool,

, and select it. Create a line perpendicular to AB through B.

5. Find the angle with given size tool, , and select it. Click on point A and then segment AB. Type in 30° and make sure counterclockwise is selected. Press enter and you should have created a point, B′ . Right click (however you do that) on the angle and hide its label. 6. Find the ray tool, , and select it. Select the point A and then the point B′ to create a ray with initial point A through B′ . 7. Find the intersection of tool, , and select it. Click on the ray and the perpendicular through B to create the intersection of the two objects. GeoGebra should name it C if things have gone well. 8. Find the show/hide tool, , and select it. Everything you click with this tool will switch between visible and hidden depending on its current state. Click on the perpendicular through B and the ray to hide them. 9. Find the segment tool, again,

, and create segments between B and C and between A and C.

10. Find the angle tool, , and select it. Click on the points C, B, and then A in that order. You should now see a right angle symbol at angle B. 11. On the right side of the screen, find the names of the segments, they should be something like f, i, and j. When you arrow over them, GeoGebra will highlight the segments on the figure. 12. Figure out what segment name corresponds to BC and what corresponds to AC. Use GeoGebra to calculate BC AC . Also use GeoGebra to calculate AB AC and BC AB . 13. Now grab point B and drag it back and forth. Watch the calculations as you move B. If you did things correctly, your triangle stays a right triangle and your ratios, maybe surprisingly, don’t vary. Problem: Follow along with me as we create a more general type of right triangle to explore these ratios.

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Relative to angle A in our triangle, we say that: • BC is opposite angle A. • AB is adjacent to angle A. • AC is the hypotenuse. There are special names for the ratios that we calculated. • The ratio of the side opposite the angle to the hypotenuse of the triangle is called the sine of the angle. • The ratio of the side adjacent to the angle to the hypotenuse of the triangle is called the cosine of the angle. • The ratio of the side opposite the angle to the side adjacent to the angle is called the tangent of the angle. Sine, cosine, and tangent are called trigonometric functions. For our purposes, if we have an acute angle in a right triangle and substitute it into one of the trig functions the function returns the correct ratio of the correct sides. Sine, cosine, and tangent are the three most famous trig functions we have. We need to learn how to use them—and how to use our calculator to evaluate them. The most common way these ratios are remembered is in the box below. For an acute angle A in a right triangle. opp adj opp sin ( A) = cos ( A) = tan ( A) = hyp hyp adj Most people remember these using the mnemonic SOH-CAH-TOA. If you somehow already know all of the sides of a right triangle, you can just start listing off the trigonometric ratios of each of the acute angles. This is actually a weird situation but we might as well do it. Problem: List the trigonometric ratios for each of the acute angles in the triangles shown below. a. b.

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You can (and should) use your calculator to evaluate the trigonometric functions of an angle when you know the angle. Let’s see how to do that. The first thing to worry about is what mode your calculator is in. You’ve maybe noticed that your calculator has three letters in the top right corner of the screen. It either says DEG or RAD. We need it to say DEG for what we’re doing. DEG stands for degrees. RAD stands for radians. RAD is actually the default setting because that’s how almost all math is done…but not when you’re working in degrees! Below is a screen shot of the calculator currently in radian mode and what it looks like as you access the screen to change the settings. The other screen shot shows DEG in the top right, which means you’re ready to use degrees.

This is important…memorize the key sequence! When you want to change the angle settings of your calculator you press DOC, 7, 2. Problem: Change your settings to Degrees. Change to radians. Change back to degrees. Problem: With your calculator in degree mode, evaluate each of the following. a. sin ( 35° ) b. cos ( 24° ) c. tan (87° ) d. sin (1° )

e. cos (88.5° )

f.

tan (10° )

There are too many interesting things about trig to get into in an introductory set of notes, but the next few problems are just designed to make you think a little bit about it. We’re not going to go very in depth with any of it—in fact, we probably won’t really discuss anything about these problems. We’ll just do them and if you’re interested, you’ll explore a little more. Problem: Evaluate each of the following pairs of trig functions. a. sin ( 20° ) and cos ( 70° ) b. sin ( 41° ) and cos ( 49° ) c. sin (1° ) and cos (89° )

d. sin (89.1° ) and cos ( 0.9° )

Problem: For what angle x, 0° < x < 90° , does cos ( 37° ) = sin ( x ) ? MR Notes 17

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When you press the TRIG button, you’ll notice there are a lot of options—because there are several other trig functions as well as the inverses of each of them. We’ve only pressed three of them so far. We’re not really going to go into it, but let’s see what happens when we press some of the other buttons. Problem: Evaluate each of the following pairs of trig functions. a. sin ( 30° ) and csc ( 30° ) b. cos ( 60° ) and sec ( 60° )

c. tan ( 30° ) and cot ( 30° )

The three new functions we just punched in are called cosecant, secant, and cotangent, respectively. Problem: What (do you think) is the relationship between sine and cosecant, cosine and secant, and tangent and cotangent, respectively?

Problem: Find the missing side of the triangle and then list the sine, cosine, and tangent of the acute angles of the triangle. a. b.

Problem: Evaluate each of the following. a. sin (87° ) b. sin (88° )

c. sin (89° )

Problem: What do you think sin ( 90° ) is? (Based just on right triangles it’s weird that this value would even exist…there’s a lot more to trig functions than right triangles!)

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Now let’s actually do some trig! At the outset almost all trig is about right triangles. Let’s see what we can do. Problem: Find the measure of the missing angle and the length of each side of the triangle (correct to three decimal places). Show work. Summarize your results by making a list of angles and sides. (Get into this habit and your future teachers will love your work!) a. b.

c.

d.

e.

f.

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We’ve looked at a bunch of examples where we knew an angle and one side of a right triangle. In that scenario you have enough information to find every other part of the triangle. What we don’t currently know how to do is find an angle given a two sides of a triangle. We need to be able to do that as well. Question: What do the trig functions take as inputs and what do they give as outputs?

Question: What concept have we worked with this year that essentially switches the inputs and outputs of a function? What was the correct notation for this concept?

Now we want to talk about this idea for trig functions. I’ll summarize it in a box. I’m going to use a made up function called trig for this. It helps to generalize the concept. The basic relationship between a trig function and its inverse function is:

trig ( angle ) = ratio ⇒ trig −1 ( ratio ) = angle where trig can be the sine, cosine, or tangent functions (so far…). Problem: Find the acute angle satisfying each of the following. Write out a step or two. 3 b. cos ( x ) = 0.485 c. tan ( x ) = 105.254 a. sin ( x ) = 5

Problem: Find the measure of both acute angles in the triangles below. Show work! a. b.

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Problem: Solve the triangles below. (To solve a triangle means to find all angles and sides.) Summarize the results. a. b.

c.

d.

e.

f.

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Trig is probably most famous at the pre-calculus level for being associated with a ton of “word problems.” Word problems are actually just problems, though, so let’s look at some and solve them. There are two very related concepts that you need to understand to do a lot of trig problems: angle of elevation and angle of depression. Angles of elevation involving tilting your eyes up. Angles of depression involve tilting your eyes down. Both angles are measured from the horizontal through the observer’s point of view. Because of properties of parallel lines being cut by a transversal the angles of depression and elevation for observers who are observing each other are equal in measure. Problem: Two buildings are 40 meters apart. The height of the taller building is 220 meters. The angle of depression from the top of the taller building to the top of the shorter building is 14° . Find, to the nearest meter, the height of the shorter building. Include a sketch.

Problem: A drone is filming a herd of wildebeest from an altitude of 200 meters. The drone sights the herd at a 12° angle of depression. What is the line of sight distance from the drone to the herd?

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Problem: One method of measuring the height of a cloud layer involves shining a spotlight directly up toward the clouds and then measuring the angle of elevation from a location a known distance from the spotlight. Suppose a team shines a spotlight on a layer of clouds and then a second team, 630 meters from the spotlight (at the same elevation) measure the angle of elevation to the light on the clouds to be 29° . How high are the clouds?

Problem: (One of my personal favorites.) To measure the height of an inaccessible mountain peak a surveyor does the following. Standing at a point certain point she measures the angle of elevation to the peak as 42° . She then backs up 200 meters from the peak, in a straight line, and measures the angle of elevation again, this time getting 17° . How high is the peak? (Include a sketch or you’ll probably never solve this correctly.)

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