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## Trigonometry

### Course: Trigonometry>Unit 3

Lesson 3: Solving general triangles

# Trig word problem: stars

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

## Want to join the conversation?

• What's the difference between law of cosine and law of sine? •   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 :(
• 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? •  Two errors:
1. 1378921 and - 1346880cos3 are not like terms and so cannot be combined. You need to evaluate - 1346880cos3 before moving on.
2. you did not square root the cosine (but this step was wrong anyway).
• A bit unrelated but how does she calculate the distance between her home and a star 700+ light years away ? • 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. • 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.
• 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? • 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.
• 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) • when do i use law of sines or cosines? • Use the law of sines when you see a pair of a known angle and it's known opposite side, then if you are given another value you can solve for it's opposite angle/side.
Use the law of cosines when you see a known angle in between 2 known sides, where you can solve for the 3rd side. OR if you are given 3 sides then you can work out any angle of the triangle
• 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? • 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.  