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# 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.(39 votes)- 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.(40 votes)

- 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(35 votes)
- 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(29 votes)

- At around4:30, why do you need to take the reciprocal of both sides to solve the law of sines?(21 votes)
- 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.(22 votes)

- what is the difference between degree and radian mode?(8 votes)
- 360 degrees is equal to 2pi radians. They are simply different units for the same measurement.(16 votes)

- Hey, I'm quite confused.

at around3:20Sal says we might remember that:`sin 45 = (root of 2) / 2`

how do I get to that?(11 votes)- 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.(7 votes)

- At3:36, why can't Sal cross multiply 1 over 4 = sine 105 degrees over a to solve for a?(7 votes)
- Sal does that but shows his work. Cross multiply is essentially multiplying and dividing on both sides(8 votes)

- Just out of curiosity, how was the law of sines formula devised? Is it something that is set in place or is there a reasoning behind it? I'm just wondering.(5 votes)
- There is reasoning behind it.

Imagine a triangle. At the top of the triangle, there is an angle c. There are two angles at the base: ∠a (opposite to side A) and ∠b (opposite to side B).

Drop an altitude from ∠c. Let this altitude have a length of x.

Now you have two right triangles that share a side inside this triangle.

sin(a)=x/B → x=Bsin(a)

sin(b)=x/A → x=Asin(b)

Since x equals them both, you can say:

Bsin(a)=Asin(b), divide both sides by A*B

sin(a)/A=sin(b)/B

And that’s the Law of Sines!(10 votes)

- There is no real explanation at3:38as 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?(5 votes)
- 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.(9 votes)

- Isn't 1/2 over 2 technically 1?(2 votes)
- No, 1/2 over 2 is one half of one half, which is one quarter.(11 votes)

- Can somebody help me?

Why can you take the reciprocal and solve it??

Doesn't that change the value??(5 votes)- If you do the same operation to both sides of an equation, equivalence is held. So you can take the reciprocal of both sides of an equation, for the same reason you can add/subtract something to both sides of an equation.

Hope this helps!(5 votes)

## Video transcript

Voiceover:We've got a triangle here where we know two of the angles
and one of the sides. And what I claim, is that I
can figure out everything else about this triangle just
with this information. You give me two angles and a side, and I can figure out what the other two sides are going to be. And I can, of course,
figure out the third angle. So, let's try to figure that out. And the way that we're going to do it, we're going to use something
called the Law of Sines. In a future video, I will
prove the Law of Sines. But here, I am just going to show you how we can actually apply it. And it's a fairly straightforward idea. The Law of Sines just tells us that the ratio between
the sine of an angle, and the side opposite to it, is going to be constant for any
of the angles in a triangle. So for example, for this
triangle right over here. This is a 30 degree angle, This is a 45 degree angle. They have to add up to 180. So this right over here
has to be a, let's see, it's going to be 180 minus 45 minus 30. That's 180 minus 75, so
this is going to equal 105 degree angle, right over here. And so applying the Law of Sines, actually let me label the different sides. Let's call this side right over here, side A or has length A. And let's call this side,
right over here, has length B. So the Law of Sines tells us that the ratio between
the sine of an angle, and that the opposite side is going to be constant through this triangle. So it tells us that sine of
this angle, sine of 30 degrees over the length of the side opposite, is going to be equal to
sine of a 105 degrees, over the length of the
side opposite to it. Which is going to be equal
to sine of 45 degrees. equal to the length of the side opposite. So sine of 45 degrees over B. And so if we wanted to figure out A, we could solve this
equation right over here. And if we wanted to solve for B, we could just set this equal
to that right over there. So let's solve each of these. So what is the sine of 30 degrees? Well, you might just remember
it from your unit circles or from even 30, 60, 90
triangles and that's 1/2. And if you don't remember it, you can use a calculator to verify that. I have already verified that this is in degree mode, so it's 0.5. So this is going to be
equal to 1/2 over two. So another way of thinking about it, that's going to be equal to 1/4, this piece is equal to 1/4 is equal to sine of a 105 degrees over A. Let me write this, this is equal to sine of 105 degrees over A. And actually, we could also say, since we could actually
do both at the same time, that this is equal to that. That 1/4 is equal to sine
of 45 degrees over B. Actually, sine of 45 degrees
is another one of those that is easy to jump out of unit circles. You might remember it's
square of two over two. Let's just write, that's
square root of two over two. And you can use a calculator, but you'll get some decimal
value right over there. But either case, in
either of these equations, let's solve for A then let's solve for B. So one thing we could do is we could take the reciprocal of both
sides of this equation. The reciprocal of 1/4 is four. And the reciprocal of this right-hand side is A over the sine of 105 degrees. And then to solve for A,
we could just multiply both sides times the
sine of a 105 degrees. So we get four times the sine
of 105 degrees is equal to A. Let's get our calculator out, so four times the sine of 105 gives us, it's approximately equal to, let's just round to the
nearest 100th, 3.86. So A is approximately equal to 3.86. Which looks about right if this is two, and I have made my angles appropriately, that looks like about 3.86. Let's figure out what B is. We could once again take the reciprocal of both sides of this and we get four is equal to B over
square root of two over two, we could multiply both sides times square root of two over two. And we would get B is equal to four times the square
root of two over two. Come to think of it, B is four times the sine of 45 degrees. Let's figure out what that is. If we wanted actual numerical value, we could just write this
as two square roots of two. But let's actually
figure out what that is. Two square roots of two is equal to 2.83. So B is approximately equal to 2.83. So [I'm] be clear, this
four divided by two is two square roots of two, which is 2.8. Which is approximately equal to 2.83 if we round to the nearest 100th, 2.83, which also seems pretty reasonable here. So the key of the Law of Cosines is if you have two angles and a side, you're able to figure out
everything else about it. Or if you actually had
two sides and an angle, you also would be able to figure out everything else about the triangle.