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Trig word problem: stars

Sal solves a word problem about the distance between stars using the law of cosines. Created by Sal Khan.

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  • female robot ada style avatar for user Alyssa Johnson
    What's the difference between law of cosine and law of sine?
    (18 votes)
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    • mr pants teal style avatar for user Karen Adams
      One defines the ratios between angles and their opposite sides as a constant (the Law of Sines) while the other does not. Look at the basic formulae:
      Law of Cosines:
      a^2 = b^2 + c^2 - 2bc * cos(theta)
      Law of Sines:
      sin(a)/A = sin(b)/B = sin(c)/C
      The Law of Cosines incorporates the Pythagorean Theorem with an "escape hatch" for triangles without a right angle (the "-2bc*cos(theta)" half of the expression). It is best utilized when you are given 2 sides with an angle between them, or 3 sides of a triangle and asked to solve for the angles.
      The Law of Sines, on the other hand, is a set of straight-up, constant ratios. It states that the ratio between any angle and the side opposite it is going to be the same ratio as another angle in the same triangle and the side *it* opposes. So if the ratio for angle (a) is 3/4, it's going to be 3/4 for angles (b) and (c) as well. The Law of Sines is best used when you're given 2angles and a side -- and the side doesn't necessarily have to be between the angles.
      Make sense? Sorry I can't add pictures -- they always make it clearer for me :(
      (83 votes)
  • starky ultimate style avatar for user Justin Orji
    When I worked out this problem, I went through all the steps of simplifying the equation before I actually square-rooted it. Here's what I wrote:
    x^2 = (736)^2 + (915)^2 - 2(736)(915)cos3
    x^2 = 1378921 - 1346880cos3
    x^2 = 32041cos3
    (then square rooted both sides of the equation)
    x = 179cos3
    x ≈ 178.8

    Why did I get a different answer?
    (5 votes)
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  • aqualine ultimate style avatar for user deepesh k
    A bit unrelated but how does she calculate the distance between her home and a star 700+ light years away ?
    (5 votes)
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  • male robot hal style avatar for user andre.gosteli
    Hello, is this application of trigonometry allowed when calculating such long distances in the universe or is is just a simplification? I am pointing to a physics question and just wonder if the relativity theory would have to be applied in real astronomy of this question.
    (6 votes)
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    • leaf blue style avatar for user Matthew Daly
      You're right to assume that this is a simplification. I'm not sure if relativity is precisely the issue here, but it's assuming that the stars aren't moving. After all, it took 736 and 915 years for the light from the two stars to reach us, so there really isn't any legitimacy in saying that the two stars were 184 light years away from each at any specific point in history. I'm also not sure if it makes sense to say that this is the "width" of Orion's belt in any way that makes sense, although I'm not an astronomer.
      (8 votes)
  • scuttlebug yellow style avatar for user Adam St. Amand
    Is there a way to determine if an angle should be obtuse or acute when using the law of cosines? One of the exercise questions gives 3 side lengths and no angle measures. I had to find the measure of one of those angles. The question doesn't specify if the angle is obtuse or acute. I ended up getting 121 degrees as my answer, and the answer to the question was 59 degrees, so I would have needed to do 180-121 to get the acute angle. On some of the other questions, they say if the angle should be obtuse or acute. How do I find out if it's obtuse or acute if that information isn't given?
    (4 votes)
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    • mr pink green style avatar for user David Severin
      The biggest angle is always opposite the biggest side, and only 1 angle in a triangle can be obtuse. Therefore, if the angle is opposite one of the shorter sides, it cannot be obtuse. However, if the angle is opposite the largest side, then you can use Pythagorean theorem to determine. If a^2+b^2=c^2, then it is a right triangle. If c^2<a^2+b^2, it is acute triangle, and if c^2>a^2+b^2, it is obtuse triangle, thus c has to be obtuse. c is always the longest side in these equations and inequalities.
      (6 votes)
  • aqualine ultimate style avatar for user nicolas.onstott
    I don't really understand the concept of modifying the formula to solve for ( for example ) cos(A). Why don't we always modify it? Is it just for the finding of the angles?
    (3 votes)
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    • starky ultimate style avatar for user Eric Ramesh
      If I understand your question correctly, you are asking about the Law of Cosines (which is used in the video) and how it is able to be modified depending on the situation. The answer is that it completely depends on what information is given in the problem as to how you modify it. If the triangle is a right triangle, you would of course use Pythagorean Theorem. But with a triangle that does not have a right angle (90 deg.), you can use this law. You can use it for finding the interior angles. But you can also use it to find one of the sides that is missing (length of side b for example). Basically you have to modify the formula based on what you are solving for: sideneeded = sidegiven^2 + othersidegiven^2 - 2*sidegiven*othersidegiven*Cos(angleofsideneeded). In other words, you take the side you need (say side 'c' ) and take the angle across from it (in this case 'C' ) and then plug those in as 'sideneeded' and 'angleofsideneeded' into the formula. I hope this is not too confusing. You don't have to modify the formula if the information given in the problem is straight forward and fits nicely. However if it is different (say they don't give you any interior angles) then you must modify accordingly.
      (7 votes)
  • winston baby style avatar for user Ethan
    Is it possible to do this except with only one side measure and two angle measures? (i.e. H--->Mintaka = 915, 3 degrees, and angle Alnitako, Mintaka, Home is say 85 degrees). If so, what would the template equation be?
    (1 vote)
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    • mr pink orange style avatar for user Izek
      Yes, it would be possible.
      After drawing a diagram, we would see that we actually would use the Law of Sines.
      915 / sin(85°) = x / sin(3°)
      Solve for x. Then find the other angle measure, and then use the Law of Sines again to find the last side length. Hope this helps!
      (11 votes)
  • area 52 purple style avatar for user Cutie Math
    Is this how we found out the size between stars?? 😍
    (4 votes)
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  • blobby green style avatar for user Patrick Rusk
    Was anyone else a bit queasy that we are measuring the distance between where Alnitak was 736 years ago and where Mintaka was 179 years earlier?
    (5 votes)
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  • male robot johnny style avatar for user X-Drip
    I went to do the test again and the problem was:
    Devora explored a secret cave. 48m from the entrance, she found an empty chest with a map drawn on it. The map showed treasure buried 95m from the entrance to the cave. From the treasure, the map showed a 30 degree angle between the chest and the entrance, as shown below.
    Devora faced the entrance, then turned a certain number of degrees to her left and walked until she got to the treasure. How many degrees did she turn?
    The picture showed the information for a tangent problem and was 48m/tan30 degrees = 95m/tan. I then changed it to tan-1(tan 30*(95/48)= 48.8094846 or 49 degrees. But I got it wrong and apparently I was supposed to use sine. Did anyone else do this and get 82 degrees or 49 degrees and how cause I am confused? Also I'm a hundred percent positive that opposite over adjacent is tangent and not sine.
    (3 votes)
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Video transcript

Voiceover:Artemis seeks knowledge of the width of Orion's belt, which is a pattern of stars in the Orion constellation. She has previously discovered the distances from her house to Alnitak, 736 lights years, and to Mintaka, 915 light years, which are the endpoints of Orion's belt. She knows the angle between these stars in the sky is three degrees. What is the width of Orion's belt? That is, what is the distance between Alnitak and Mintaka? And they want us to the answer in light years. So let's draw a little diagram to make sure we understand what's going on. Actually, even before we do that, I encourage you to pause this and try this on your own. Now let's make a diagram. Alright, so let's say that this is Artemis' house right over here. This is Artemis' house. I'll say that's A for Artemis' house. And then... Alright, let me say H... Let me say this is home. This is home right over here. And we have these 2 stars. So she's looking out into the night sky and she sees these stars, Alnitak, which is 736 light years away, and obviously I'm not going to draw this to scale. So this is Alnitak. And Mintaka. So let's say this is Mintaka right over here. Mintaka. And we know a few things. We know that this distance between her home and Alnitak is 736 light years. So this distance right over here. So that right over there. Everything we'll do is in light years. That's 736. And the distance between her house and Mintaka is 915 light years. So it would take light 915 years to get from her house to Mintaka, or from Mintaka to her house. So this is 915 light years. And what we wanna do is figure out the width of Orion's belt, which is the distance between Alnitak and Mintaka. So we need to figure out this distance right over here. And the one thing that they did give us is this angle. They did give us that angle right over there. They said that the angle between these stars in the sky is three degrees. So this is three degrees right over there. So how can we figure out the distance between Alnitak and Mintaka? Let's just say that this is equal to X. This is equal to X. How do we do that? Well if we have two sides and an angle between them, we could use the law of cosines to figure out the third side. So the law of cosines, so let's just apply it. So the law of cosines tells us that X squared is going to be equal to the sum of the squares of the other two sides. So it's going to be equal to 736 squared, plus 915 squared, minus two times 736, times 915, times the cosine of this angle. Times the cosine of three degrees. So once again, we're trying to find the length of the side opposite the three degrees. We know the other two sides, so the law of cosines, it essentially... Sorry, I just had to cough off camera because I had some peanuts and my throat was dry. Where was I? Oh, I was saying, if we know the angle and we know the two sides on either side of the angle, we can figure out the length of the side opposite by the law of cosines. Where it essentially starts off not too different than the Pythagorean theorem, but then we give an adjustment because this is not an actual right triangle. And the adjustment... So we have the 736 squared, plus 915 squared, minus two times the product of these sides, times the cosine of this angle. Or another way we could say, think about it is, X, let me write that, X is to equal to the square root of all of this stuff. So, I can just copy and paste that. Copy and paste. X is going to be equal to the square root of that. And so let's get our calculator to calculate it. And let me verify that I'm in degree mode. Yes, I am indeed in degree mode. And so let's exit that. And so I wanna calculate the square root of 736 squared, plus 915 squared, minus two times 736, times 915, times cosine of three degrees. And we deserve a drum roll now. X is 100, if we round... Let's see, what did they want us to do? Round your answer to the nearest light years. So to the nearest light year is going to be 184 light years. So X is approximately equal to 184 light years. So it would take light 184 years to get from Mintaka to Alnitak. And so hopefully this actually shows you if you are going to do any astronomy, the law of cosines, law of sines, in fact all of trigonometry, becomes quite, quite handy.