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Solving for a side in right triangles with trigonometry

Sal is given a right triangle with an acute angle of 65° and a leg of 5 units, and he uses trigonometry to find the two missing sides. Created by Sal Khan and Monterey Institute for Technology and Education.

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  • leaf green style avatar for user Alex VanHulle
    At he says " you could of solved this using the Pythagorean theorem... But there is an issue if im not mistaken:

    10.7*10.7 + 5*5 does not equal to 11.8?
    (0 votes)
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    • leaf green style avatar for user chris ponce
      I don't think there is an issue he just wanted to solve using the trigonometric functions because that is what we had just been working on.

      Just to show, 10.7*10.7 = 114.49
      5*5 = 25
      114.49+25=139.49
      And the square root of 139.49 = 11.8

      a^2+b^2=c^2 so don't forget to square root everything.
      (216 votes)
  • duskpin ultimate style avatar for user RanyaM
    How would you find tan65 degrees without a calculator?
    (50 votes)
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    • piceratops ultimate style avatar for user Just Keith
      You will not be asked to do that computation: It is exceedingly difficult and requires knowing advanced mathematics. Thus, the actual computations are not taught at this level of study. If you are not allowed a calculator on a test or something, you should be given the needed values.

      If you are given the sin and cosine of an angle, then the tangent is just sin x/ cos x.

      Anyway, just for reference, here is the actual computation for tan 65°:
      -i [e^(-25iπ/180) + e^(25iπ/180)] ÷ [ e^(-25iπ/180)-e^(25iπ/180)]
      Obviously, doing the calculation by hand is going to be too difficult.

      You will, however, be expected to memorize the sin, cos, and tan of certain special angles such as 0, 30°, 45°, 60°, 90°, etc.
      (84 votes)
  • blobby green style avatar for user strengthofthepen
    Are the tangent, sine, and cosine "fixed" ratios depending on the angle? Is that why you can just put in tan 65 into a calculator and have it give you a number?
    (36 votes)
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  • leafers ultimate style avatar for user J.L.Savard
    I'm having trouble understanding the solution to a problem in the section on Trigonometry 2 questions. I took a screenshot of the problem with all its hints revealed: http://s14.postimg.org/gq06og00h/Screenshot_2014_01_11_at_11_24_57_AM.png.

    The issue I'm having is that I can't seem to wrap my head around the last part of the hints. How does one get from [10 / (10 times the square root of 109) / 109] to an answer of "square root of 109"? I must be missing something algebraically, or in my order of operations. I've been stuck on this for little while now, so I'd appreciate any clarification you can offer. Thanks!
    (7 votes)
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    • starky ultimate style avatar for user gregg salomon
      The tricky part of the answer above is remembering that √(109)*√(109) in the denominator is simply 109 because if you take the square root of a number and multiply it by the square root of that same number, you get the number you started with.
      Also remember to always rationalize the denominator (remove the √)
      (7 votes)
  • mr pants teal style avatar for user Joshua
    Is there a way to find the sides is if you have just angles given, or the measure of the angles if just the sides are given?
    (9 votes)
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    • male robot hal style avatar for user Sid
      If you only have the angles, there is no way to find the sides. There is an infinite number of triangles fitting any three angles that add up to 180 degrees.

      If you know all the sides, you can figure out everything about that triangle.
      (15 votes)
  • leaf blue style avatar for user Alekya  Bheemreddy
    What would you do if your were not allowed to use a calculator on the test?
    (4 votes)
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  • blobby green style avatar for user agastyasreenivasan
    how can you derive values for cos and sin and other trig ratios without using a calculator? after all when no calculators were there trig was invented and solved to utmost presicion?
    (8 votes)
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    • leaf blue style avatar for user Matthew Daly
      It wasn't really utmost precision through much of history, but you're right that it was remarkably good. We credit that to Ptolemy, who devised his trig tables in the second century by half-degrees that turned out to be correct to the fifth or sixth decimal place for the most part. His exact path is hard to reproduce and would involve geometry that is more detailed than a typical high-school education, but a modern person following in Ptolemy's footsteps would get the same results from using the sum-angle and half-angle identities that you will learn shortly plus the known values of several critical values like 60 degrees and 72 degrees.
      (13 votes)
  • piceratops ultimate style avatar for user ykingrocks
    how to calculate these values without using a calculator?
    (5 votes)
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    • piceratops ultimate style avatar for user Just Keith
      Which values? If you mean the values of the trigonometric functions, then that math is too difficult to do by hand. The reason the actual formulas for sine, cosine and tangent are not given as this level of study is that the computations are nightmarishly difficult.

      You will be expected to memorize the values for sine, cosine, and tangent at some commonly used angles such as 30°, 45°, 60°, etc. There is a method for finding the values of sine and cosine for angles that are multiples of 3°, but it is quite tedious and takes a long time.

      At this level of study, if you are not allowed a calculator, it is the custom either for you to be given a chart containing the values of the trigonometric functions you will need OR for you just to express your answer in terms of the trig function -- you might have an answer like x = 17.4 sin 17.35° or perhaps something like θ=arcin (0.804)
      (13 votes)
  • blobby green style avatar for user Rohit Matta
    In my calculator, when I do 5/cos(65). I am getting approx. -8.8896, and Sal got approximately 11.8. Can somebody tell me what I am doing wrong? Do you need to do something with calculator for this to work?
    (7 votes)
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  • piceratops ultimate style avatar for user Tanveer
    I want to know how to calculate tan,sin,cos without calculator
    (3 votes)
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    • duskpin ultimate style avatar for user Chayansudha Biswas
      Here's a method to know some of the values:
      sin0= under root(0/4)=0= cos 90
      sin30= under root(1/4)= 1/2= cos 60
      sin 45= under root(2/4)= root 2/2= 1/root2= cos 45
      sin 60= under root(3/4)= root3/4= cos 30
      sin 90= under root(4/4)= 2/2=1= cos 0
      Did you get the pattern?

      tan 0= sin0/cos0
      tan 30= sin30/cos30... etc

      Hope that helped!! :)
      (7 votes)

Video transcript

We're asked to solve the right triangle shown below. Give the lengths to the nearest tenth. So when they say solve the right triangle, we can assume that they're saying, hey figure out the lengths of all the sides. So whatever a is equal to, whatever b is equal to. And also what are all the angles of the right triangle? They've given two of them. We might have to figure out this third right over here. So there's multiple ways to tackle this, but we'll just try to tackle side XW first, try to figure out what a is. And I'll give you a hint. You can use a calculator, and using a calculator, you can use your trigonometric functions that we've looked at a good bit now. So I'll give you a few seconds to think about how to figure out what a is. Well, what do we know? We know this angle y right over here. We know the side adjacent to angle y. And length a, this is the side that's the length of the side that is opposite to angle Y. So what trigonometric ratio deals with the opposite and the adjacent? So if we're looking at angle Y, relative to angle Y, this is the opposite. And this right over here is the adjacent. Well if we don't remember, we can go back to SohCahToa. Sine deals with opposite and hypotenuse. Cosine deals with adjacent and hypotenuse. Tangent deals with opposite over adjacent. So we can say that the tangent of 65 degrees, of that angle of 65 degrees, is equal to the opposite, the length of the opposite side, which we know has length a over the length of the adjacent side, which they gave us in the diagram, which has length five. And you might say, how do I figure out a? Well we can use our calculator to evaluate what the tangent of 65 degrees are. And then we can solve for a. And actually if we just want to get the expression explicitly solving for a, we could just multiply both sides of this equation times 5. So let's do that. 5 times, times 5. These cancel out, and we are left with, if we flip the equal around, we're left with a is equal to 5 times the tangent of 65 degrees. So now we can get our calculator out and figure out what this is to the nearest tenth. That's my handy TI-85 out and I have 5 times the tangent-- I didn't need to press that second right over there, just a regular tangent-- of 65 degrees. And I will get, if I round to the nearest tenth like they ask me to, I get 10.7. So a is approximately equal to 10.7. I say approximately because I rounded it down. This is not the exact number. But a is equal to 10.7. So we now know that this has length 10.7, approximately. There are several ways that we can try to tackle b. And I'll let you pick the way you want to. But then I'll just do it the way I would like to. So my next question to you is, what is the length of the side YW? Or what is the value of b? Well there are several ways to do it. This is the hypotenuse. So we could use trigonometric functions that deal with adjacent over hypotenuse or opposite over hypotenuse. Or we could just use the Pythagorean theorem. We know two sides of a right triangle. We can come up with the third side. I will go with using trigonometric ratios since that's what we've been working on a good bit. So this length of b, that's the length of the hypotenuse. So this side WY is the hypotenuse. And so what trigonometric ratios-- or we can decide what we want to use. We could use opposite and hypotenuse. We could use adjacent and hypotenuse. Since we know that XY is exactly 5 and we don't have to deal with this approximation, let's use that side. So what trigonometric ratios deal with adjacent and hypotenuse? Well we see from SohCahToa cosine deals with adjacent over hypotenuse. So we could say that the cosine of 65 degrees is equal to the length of the adjacent side, which is 5 over the length of the hypotenuse, which has a length of b. And then we can try to solve for b. You multiply both sides times b, you're left with b times cosine of 65 degrees is equal to 5. And then to solve for b, you could divide both sides by cosine of 65 degrees. This is just a number here. So we're just dividing-- we have to figure it out what our calculator, but this is just going to evaluate to some number. So we can divide both sides by that, by cosine of 65 degrees. And we're left with b is equal to 5 over the cosine of 65 degrees. So let us now use our calculator to figure out the length of b. Length of b is 5 divided by cosine of 65 degrees. And I get, if I round to the nearest tenth, 11.8. So b is approximately equal to, rounded to the nearest tenth, 11.8. So b is equal to 11.8. And then we're almost done solving this right triangle. And you could have figured this out using the Pythagorean theorem as well, saying that 5 squared plus 10.7 squared should be equal to b squared. And hopefully you would get the exact same answer. And the last thing we have to figure out is the measure of angle W right over here. So I'll give you a few seconds to think about what the measure of angle W is. Well here we just have to remember that the sum of the angles of a triangle add up to 180 degrees. So angle w plus 65 degrees, that's this angle right up here, plus the right angle, this is a right triangle, they're going to add up to 180 degrees. So all we need to do is-- well we can simplify the left-hand side right over here. 65 plus 90 is 155. So angle W plus 155 degrees is equal to 180 degrees. And then we get angle W-- if we subtract 155 from both sides-- angle W is equal to 25 degrees. And we are done solving the right triangle shown below.