Right Triangle Trigonometry

This section covers:

You may have been introduced to Trigonometry in Geometry, when you had to find either a side length or angle measurement of a triangle.  Trigonometry is basically the study of triangles, and was first used to help in the computations of astronomy.  Today it is used in engineering, architecture, medicine, physics, among other disciplines.

The 6 basic trigonometric functions that you’ll be working with are sine (rhymes with “sign”), cosine, tangent, cosecant, secant, and cotangent.  (Don’t let the fancy names scare you; they really aren’t that bad).

With Right Triangle Trigonometry, we use the trig functions on angles, and get a number back that we can use to get a side measurement, as an example.    Sometimes we have to work backwards to get the angle measurement back so we have to use what a call an inverse trig function But basically remember that we need the trig functions so we can determine the sides and angles of a triangle that we don’t otherwise know.

Later, we’ll see how to use trig to find areas of triangles, too, among other things.

You may  have been taught SOH – CAH – TOA  (SOHCAHTOA)  (pronounced “so – kuh – toe – uh”) to remember these.  Back in the old days when I was in high school, we didn’t have SOHCAHTOA, nor did we have fancy calculators to get the values; we  had to look up trigonometric values in tables.

Remember that the definitions below assume that the triangles are right triangles, meaning that they all have one right angle (90°).   Also note that in the following examples, our angle measurements are in degrees; later we’ll learn about another angle measurement unit, radians, which we’ll discuss here in the Angles and Unit Circle section.

Basic Trigonometric Functions (SOH – CAH – TOA)

Here are the 6 trigonometric functions, shown with both the SOHCAHTOA and Coordinate System Methods.   Note that the second set of three trig functions are just the reciprocals of the first three; this makes it a little easier!

Note that the cosecant (csc), secant (sec), and cotangent (cot) functions are called reciprocal functions, or reciprocal trig functions, since they are the reciprocals of sin, cos, and tan, respectively.

For the coordinate system method, assume that the vertex of the angle in the triangle is at the origin (0, 0):

Trigonometric Functions

Here are some example problems.  Note that we commonly use capital letters to represent angle measurements, and the same letters in lower case to represent the side measurements opposite those angles.  We also use the theta symbol θ  to represent angle measurements, as we’ll see later.

Note also in these problems, we need to put our calculator in the DEGREE mode.

And don’t forget the Pythagorean Theorem (\({{a}^{2}}+{{b}^{2}}={{c}^{2}}\),  where a and b are the “legs” of the triangle, and c is the hypotenuse), and the fact that the sum of all angles in a triangle is 180°.

Right Triangle Trig

Trigonometry Word Problems

Here are some types of word problems that you might see when studying right angle trigonometry.

Note that the angle of elevation is the angle up from the ground; for example, if you look up at something, this angle is the angle between the ground and your line of site.

The angle of depression is the angle that comes down from a straight horizontal line in the sky.  (For example, if you look down on something, this angle is the angle between your looking straight and your looking down to the ground).   For the angle of depression, you can typically use the fact that alternate interior angles of parallel lines are congruent (sorry, too much Geometry!) to put that angle in the triangle on the ground (we’ll see examples).

shadow

Note that shadows in these types of problems are typically on the ground.   When the sun casts the shadow, the angle of depression is the same as the angle of elevation from the ground up to the top of the object whose shadow is on the ground.

Also, the grade of something (like a road) is the tangent (rise over run) of that angle coming from the ground.  Usually the grade is expressed as a percentage, and you’ll have to convert the percentage to a decimal to use in the problem.

And, as always, always draw pictures!

Angle of Elevation Problem:

Devon is standing 100 feet from the Eiffel Tower and sees a bird land on the top of the tower (she has really good eyes!).   If the angle of elevation from Devon to the top of the Eiffel Tower is close to 84.6°, how tall is the tower?

Solution:

This is a good example how we might use trig to get distances that are typically difficult to measure.   Note that the angle of elevation comes up off of the ground.

Angle of Elevation Word Problem

Angle of Depression Problem:

From the top of a building that is 200 feet tall, Meryl sees a car coming towards the building.  (Somehow she knows that) the angle of depression when she first saw the car was 20° and when she stopped looking at it was 40° degrees.  How far did the car travel?

Solution:

The first step is to draw a picture, and note that we can sort of “reflect” the angles of depression down to angles of elevation, since the horizon and ground are parallel.  Then we get to use trig!

Angle of Depression Word Problem

Right Triangle Systems Problem:

Here’s a problem where it’s easiest to solve it using a System of Equations:

Two girls are standing 100 feet apart.  They both see a beautiful seagull in the air between them.  The angles of elevation from the girls to the bird are 20° and 45°, respectively.   How high up is the seagull?

Solution:

Systems of Equations with Trig

Trig Shadow Problem:

The length of a tree’s shadow is 20 feet when the angle of elevation to the sun is 40°.  How tall is the tree?

Solution:

Again, note that shadows in these types of problems are on the ground.   When the sun casts the shadow, the angle of depression is the same as the angle of elevation from the ground up to the top of the tree.

So let’s solve using trig:

Shadow Trig Problem

Trig Grade Problem:

Chelsea walked up a road that has a 20% grade (she could feel it!) to get to her favorite store.  At what angle does the road come up from the ground (at what angle is the road inclined from the ground)?

Solution:

Remember that the grade of a road can be thought of as  \(\frac{\text{rise}}{\text{run}}\)  and you usually see it as a percentage.   So a 20% grade is the same as a grade of  \(\frac{\text{20}}{\text{100}}\); for every 20 feet the road goes up vertically, it goes 100 feet horizontally.

Elevated Road Problem

Understand these problems, and practice, practice, practice!


Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software.  You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Angles and the Unit Circle – you’re ready! 

27 thoughts on “Right Triangle Trigonometry

  1. In the seagull problem, the 45 degree angle gives you that y=100-x. (So your answer is off by 0.01) Your method can be used for any angle, so it might be better to change the problem to, say, 20 and 40 degrees.

    • Thanks so much for the suggestion! That’s a good point. I changed it to make it a little more accurate by leaving more decimal places in the computations. Thanks again – let me know if you see more! Lisa

  2. Hi, I know I keep pestering you for help (though none of them problems seem to be difficult for you 😉 ), but I have another problem that I was wondering you could help me with.

    It’s very similar to your angle of depression problem.
    In this one, though, we draw a right angled triangle of any size, using any angles (we’ll call the angle we’re interested in “A” – which is at the top apex of the triangle), and we define the opposite side (the height “h”) to be a single unit – just to keep things simple.
    Obviously, we can tell that the length of the adjacent side (which we’ll call “X”) will be sin(A)h/sin(90-A), which is sin(A)/cos(A), or tan(A).

    Now, we draw a number of triangles within this first triangle, which all use the same opposite side (height “h”), but whose angles are all 1 degree less than the triangle before it.
    I hope I’m not losing you here – I can try and supply a diagram, if needs be.

    We shall call the lengths described by sections of the adjacent line that are intersected by each new triangle (beginning from the section next to the right angle itself) “X0”, “X1”, “X2”, “X3”, “X4” (these are basically analogous to your “y” and “x” in the angle of depression problem.

    Now, obviously the length of these sections will increase the further away from the original right angle they are – in other words the distance between to lines separated at the top apex by a degree will grow larger the higher the value of “A” (and the lower the value of the angle of depression, or declination).
    What I want to know is how to figure out the ratio between two adjacent sections, and to plot a graph to show how these distances will grow as we move further away from the original right angle.

    I do so hope I’m explaining this properly. Apologies if I’m losing you.

    The way I worked it out is that
    Xn/Xn-1 = ((sin(A)/cos(A))-(sin(A-1)/cos(A-1)))/((sin(A-1)/cos(A-1))-(sin(A-2)/cos(A-2)))

    Which, of course, is just (tan(A)-tan(A-1))/(tan(A-1)-tan(A-2))

    (I’m sure there’s a way of simplifying this equation, but head is swimming at this point).

    Does this formula look right to you? It seems to agree with the values that I get when I plug in real values for the lengths and angles, when I draw the triangles themselves, but I wanted to check if I’m on the right track at least.

    A second question, kind of unrelated, is about the graph for tan(x). I’ve noticed that it looks a lot like I would expect a graph of f(x)=x^3….
    Is there a reason for this, or is it just a strange little coincidence?

    • Thanks for writing!
      What you’ve done looks good, but (and you may want to send me a drawing), wouldn’t the TanA = h/x (opposite over adjacent)? Actually, I did it this way, and got the same thing you did – good work! Lisa

      As far as your other question is concerned, I think it’s just coincidence that the tanx graph looks like the x^3. Good observation though!
      Sorry I can’t help you more at this point…Lisa

      • Thanks for the reply. What’s the simplest way to send a picture of what I’m talking about?

        The reason I was using tan(A), was because the side rule stated that (if the opposite is “h” and equals 1, and the adjacent is “X”) then X=sin(A)/sin(90-A). Since sin(90-A) = cos(A), this would surely be sin(A)/cos(A), which in turn is the same as tan(A).
        Have I made a mistake here?

        Anyhow, whichever is the best way to send a picture of the problem I’m dealing with, tell me, and I’ll be happy to try and clarify what I’m trying to work on 😉

        Thank you so much again.

        And thank you for the answer regarding the tan(x) graph and the cubic graph.

  3. I noticed in your problem under the Basic Trig Functions Heading, you wrote “To get side b we need to use, cos(b) = Adjacent/Hypotenuse, where b is 55 degrees”.
    Shouldn’t this read “cos(A) = Adjacent/Hypotenuse, where A is 55 degrees”?
    I thought b is the side length adjacent to angle A. I apologize if I’m incorrect, but I thought you’d want to fix that. Thanks so much for putting together such a detailed and organized website. It has been very helpful for me as a math teacher.

    • Another small typo I found is that in the Trig Shadow problem, you called the height of the tree y, but wrote “cross multiply to get x =”. I doubt anyone would be confused by this, but I thought you’d like to have consistency 🙂

      • Thanks so much for finding these things; I think I’ve corrected them! Please keep reading, so you might find more. Thanks again, Lisa

    • can you please help me with that problem?
      A carpenter is preparing to put a roof on a garage that is 20ft by 40ft by 20ft. A steel support beam h=32ft in length is positioned in the center of the garage. To support the roof, another beam will be attached to the top of the center beam. At what angle of elevation is the new beam? In other words, what is the pitch of the roof?
      thanks

      • Thanks for writing! Here’s how I’d do this problem: We will use trig to find the hypotenuse of a triangle with base 10 and height 12 (32 – 20), so we have tan(x) = 12/10. So the pitch of the roof (x) is .88 (very “flat” roof). Here’s a video of how to do this, with using 46 ft as the support beam:
        https://www.youtube.com/watch?v=tUXdzvZXmzw Does that help? Lisa

  4. Hello 🙂
    So I’m a Sophomore and I’m currently taking Honors Math II. ( Algebra II) Let’s just say it hasn’t been easy…The teacher I have doesn’t provide us with notes, which is pretty much how I learn because I love to read. Your website has been a blessing!!! It’s everything I’ve been looking for! It’s very detailed and organized (just the way I like it). So for that, I want to say thank you…

    I have a math problem I need some help on…
    ” A person flying a kite has released 176 m of string. The string makes an angle of 27 degrees with the ground. How high is the kite? How far away is the kite horizontally? Answer to the nearest meter.”

    I’ve always been pretty good at math but this year, it has kicked my butt. My teacher didn’t really teach me much about Trig. So I have been completely lost. I was just wondering what steps I’d take to solve this problem…

    Thanks 🙂

    • Thanks for writing! Here’s how I’d do this problem: to get the height of the kite, use sine: sin(27) = y/176; y = 80 meters. To get the distance horizontally, use cosine: cos(27) = x/176; x = 157 meters. Does that made sense? Lisa

  5. i have trouble in understanding this. my teacher told gave us this problem to find a . triangle ABC, <B has a <67 degree and segmet CA measures 10.6. i first solve for c , c=10.6/sin67degree. he let's us find A using all the 6trigonometric . if is use sin. the soulution would be ..sinB=opp/hyp..sin67=b/c. but there is no a in it, it is all b and c.
    . what is wrong with it? i really need to understand this lesson. im so bad at my math subject

    • Thanks for writing! I have a question: were you only given angle B and segment CA? I don’t think you can solve the triangle with just those two things. Can you give me more information about the problem? Lisa

    • just wanted to share about this kind of problem. if angle B= 67 degree and segment CA measures 10.6 and you’re assigned to look for angle A. solve first for side c using sine 67=10.6 /c, so c=10.6/sin67, c then is equal to 11.515. Then solve for A using cos A. cos A =adj/hyp, cos A= b/c, substituting the value obtain get the inverse cosine, A= inverse cos 10.6/11.515 so the answer is 23 degree.. angleA=23degrees

  6. From the seagull problem can you elaborate where the x = 73.315 from? My brain cells are bleeding but i cannot figure out how it become 73.315. Thank you.

    • Thanks for writing! Here’s how I got the 73.315. We have 1=(.36397x)/(100-x) If we cross multiply we get 100 – x = .36397x. So we have 100 = .36397x + x, or 100 = 1.36397x. Then we divide 100 by 1.36397 to get 73.315. Does that make sense? Lisa

  7. Umm, can i ask a question about angle of depression. I was wondering if you can help me solve this problem, “A submarine traveling 8 miles per hour is descending at an angle of depression of 6 degrees. How many minutes does it take the submarine to reach a depth of 80 feet ? ” .. Thank you. And please reply ASAP..

    • Thanks for writing! Here’s how I’d do this: Use sin = opposite/hypotenuse to get sin(6) = 80/x. x = 765.34 feet, or .14495 miles (divide by 5280). Now the submarine travels at 8 miles per hour, or .1333 miles per minute.
      To get time, we have to divide distance by rate, which is .14495/.1333 = 1.087 minutes. Does that make sense? Lisa

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