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## Integrated math 3

### Course: Integrated math 3>Unit 8

Lesson 1: Law of sines

# Solving for a side with the law of sines

Sal is given a triangle with two angle measures and one side length, and he finds all the missing side lengths and angle measures using the law of sines. Created by Sal Khan.

## Want to join the conversation?

• Another question here, too: is there some funky reason that the Law of Sines seems to be falling down when questions involving obtuse angles come up? I've encountered 2 problems this evening that come up the same way.

What you're given is an acute angle measurement and two sides that *don't* include that acute angle between them. (That, of course, precludes using the Law of Cosines to figure out the problem.) You're asked to find the measure of the obtuse angle.

But when you apply the Law of Sines, it yields an acute, not an obtuse, angle measurement; and secondly, simply subtracting the (wrong? inaccurate?) acute measurement without taking into account the one given angle measurement seems to violate the rules as well. What's the deal here? To what does this acute angle measurement yielded by the Law of Sines refer? And is all this hoo-hah the "ambiguous case" I've seen referred to here and there in the comments? I'm thoroughly confuzzled.
• The range of inverse sine is restricted to the first and fourth quadrants. So what this means is using the Law of Sines is only ever going to give you acute angles. If you want to find the obtuse angle, you have to subtract the acute angle from 180 or just use the Law of Sines on the smallest angle to ensure it works.
• Hey, everybody, this might sound like a dumb question, but since there is a Law of Sines and a Law of Cosines, is there also a Law of Tangents? Also, how would you use cosine and sine on a non-right triangle? There is no "hypotenuse" to base it off of. Thanks so much! :D
• In order the use sines and cosines in non-right triangles, we need to generalize our notion of sine and cosine. We do this by introducing the unit circle definitions of the trig functions, the details of which are covered in the precalc playlist on Khan Academy. With this, we turn sine and cosine into functions which accept an input and give an output. So we can use this to find the sine or cosine of any angle.

As for the Law of Tangents, apparently there is one! It's omitted from the US high school math curriculum, but you can read about it here:

https://en.wikipedia.org/wiki/Law_of_tangents
• At around , why do you need to take the reciprocal of both sides to solve the law of sines?
• The goal was to isolate the variable. There are several ways of accomplishing this, but since the variable was in the denominator, taking the reciprocal of both sides seemed a useful choice.
• what is the difference between degree and radian mode?
• 360 degrees is equal to 2pi radians. They are simply different units for the same measurement.
• Hey, I'm quite confused.
at around Sal says we might remember that:
``sin 45 = (root of 2) / 2 ``

how do I get to that?
• Sal is using special triangles. In this case, it is the 45° 45° 90° triangle. In this triangle, if the hypotenuse is one, then the other 2 sides would be √2/2. Image: http://zonalandeducation.com/mmts/miscellaneousMath/tri454590306090/t454590.gif

Since sine is opposite / hypotenuse, the sine of 45° would be

Opposite: √2/2
Over: -----------------------
Hypotenuse: 1

Which is √2/2/1 or just √2/2 since anything divided by one is just itself.
• At , why can't Sal cross multiply 1 over 4 = sine 105 degrees over a to solve for a?
• Sal does that but shows his work. Cross multiply is essentially multiplying and dividing on both sides
• There is no real explanation at as to why the reciprocal can be used in this case. Is there a standard situation for doing so? If so, what is the situation when using the reciprocal can be used. Also if the reciprocal is not used, will the answer be different and/or wrong?
• just so I don't have to write everything out I am going to use a generic set of fractions.

a/b = c/d if you multiply both sides by b and d it becomes

ad = cb then divide both sides by a and c

d/c = b/a

This shows why you can use the reciprocals in the law of sines. just use the sine terms and the sides as appropriate. Let me know if this doesn't make sense.
• Isn't 1/2 over 2 technically 1?
(1 vote)
• No, 1/2 over 2 is one half of one half, which is one quarter.
• What if you know three of the angles and one of the sides? How do you find one specific angle?