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13-6 the law of cosines warm up lesson presentation lesson quiz

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13-6 The Law of Cosines Warm Up Lesson Presentation Lesson Quiz - ppt video online download

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Objectives Use the Law of Cosines to find the side lengths and angle measures of a triangle. Use Heron's Formula to find the area of a triangle.

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This Calculation Equation & Triangle

\( A = \cos^{-1} \left[ \dfrac{b^2+c^2-a^2}{2bc} \right]\)

A = angle A
B = angle B
C = angle C
a = side a
b = side b
c = side c
P = perimeter
s = semi-perimeter
K = area
r = radius of inscribed circle
R = radius of circumscribed circle

*Length units are for your reference-only since the value of the resulting lengths will always be the same no matter what the units are.

Calculator Use

Uses the law of cosines to calculate unknown angles or sides of a triangle. In order to calculate the unknown values you must enter 3 known values.

To calculate any angle, A, B or C, enter 3 side lengths a, b and c. This is the same calculation as Side-Side-Side (SSS) Theorem. To calculate side a for example, enter the opposite angle A and the two other adjacent sides b and c. Using different forms of the law of cosines we can calculate all of the other unknown angles or sides. This is the same calculation as Side-Angle-Side (SAS) Theorem.

Law of Cosines

If a, b and c are the lengths of the legs of a triangle opposite to the angles A, B and C respectively; then the law of cosines states:

\( a^2 = b^2 + c^2 - 2bc \cos A \)

\( b^2 = a^2 + c^2 - 2ac \cos B \)

\( c^2 = a^2 + b^2 - 2ab \cos C \)

Law of Cosines solving for sides a, b, and c

\( a = \sqrt{b^2 + c^2 - 2bc \cos A } \)

\( b = \sqrt{a^2 + c^2 - 2ac \cos B } \)

\( c = \sqrt{a^2 + b^2 - 2ab \cos C } \)

Law of Cosines solving for angles A, B, and C

\( A = \cos^{-1} \left[ \dfrac{b^2+c^2-a^2}{2bc} \right] \)

\( B = \cos^{-1} \left[ \dfrac{a^2+c^2-b^2}{2ac} \right] \)

\( C = \cos^{-1} \left[ \dfrac{a^2+b^2-c^2}{2ab} \right] \)

Triangle Characteristics

Triangle perimeter, P = a + b + c

Triangle semi-perimeter, s = 0.5 * (a + b + c)

Triangle area, K = √[ s*(s-a)*(s-b)*(s-c)]

Radius of inscribed circle in the triangle, r = √[ (s-a)*(s-b)*(s-c) / s ]

Radius of circumscribed circle around triangle, R = (abc) / (4K)

References/ Further Reading

Weisstein, Eric W. "Law of Cosines" From -- A Wolfram Web Resource. Law of Cosines.

Zwillinger, Daniel (Editor-in-Chief). CRC Standard Mathematical Tables and Formulae, 31st Edition New York, NY: CRC Press, p. 512, 2003.
http://hyperphysics.phy-astr.gsu.edu/hbase/lcos.html http://hyperphysics.phy-astr.gsu.edu/hbase/lsin.html

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( A = \cos^{-1} \left[ \dfrac{b^2+c^2-a^2}{2bc} \right]\) A = angle A B = angle B C = angle C a = side a b = side b c = side c P = perimeter s = semi-perimeter K = area r = radius of inscribed circle R = radius of circumscribed circle *Length units are for your reference-only since the value of the resulting lengths will always be the same no matter what the units are.

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This Calculation Equation & Triangle

\( A = \sin^{-1} \left[ \dfrac{a \sin B}{b} \right]\)

A = angle A
B = angle B
C = angle C
a = side a
b = side b
c = side c
P = perimeter
s = semi-perimeter
K = area
r = radius of inscribed circle
R = radius of circumscribed circle

*Length units are for your reference-only since the value of the resulting lengths will always be the same no matter what the units are.

Calculator Use

Uses the law of sines to calculate unknown angles or sides of a triangle. In order to calculate the unknown values you must enter 3 known values.

Some calculation choices are redundant but are included anyway for exact letter designations.

Calculation Methods

To calculate any angle, A, B or C, say B, enter the opposite side b then another angle-side pair such as A and a or C and c. The performed calculations follow the side side angle (SSA) method and only use the law of sines to complete calculations for other unknowns.

To calculate any side, a, b or c, say b, enter the opposite angle B and then another angle-side pair such as A and a or C and c. The performed calculations follow the angle angle side (AAS) method and only use the law of sines to complete calculations for other unknowns.

Law of Sines

If a, b and c are the lengths of the legs of a triangle opposite to the angles A, B and C respectively; then the law of sines states:

\( \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \)

Equations from Law of Sines solving for angles A, B, and C

\( A = \sin^{-1} \left[ \dfrac{a \sin B}{b} \right] \)

\( A = \sin^{-1} \left[ \dfrac{a \sin C}{c} \right] \)

\( B = \sin^{-1} \left[ \dfrac{b \sin A}{a} \right] \)

\( B = \sin^{-1} \left[ \dfrac{b \sin C}{c} \right] \)

\( C = \sin^{-1} \left[ \dfrac{c \sin A}{a} \right] \)

\( C = \sin^{-1} \left[ \dfrac{c \sin B}{b} \right] \)

Equations from Law of Sines solving for sides a, b, and c

\( a = \dfrac{b \sin A}{\sin B} \)

\( a = \dfrac{c \sin A}{\sin C} \)

\( b = \dfrac{a \sin B}{\sin A} \)

\( b = \dfrac{c \sin B}{\sin C} \)

\( c = \dfrac{a \sin C}{\sin A} \)

\( c = \dfrac{b \sin C}{\sin B} \)

Triangle Characteristics

Triangle perimeter, P = a + b + c

Triangle semi-perimeter, s = 0.5 * (a + b + c)

Triangle area, K = √[ s*(s-a)*(s-b)*(s-c)]

Radius of inscribed circle in the triangle, r = √[ (s-a)*(s-b)*(s-c) / s ]

Radius of circumscribed circle around triangle, R = (abc) / (4K)

References/ Further Reading

Weisstein, Eric W. "Law of Sines" From -- A Wolfram Web Resource. Law of Sines.

Zwillinger, Daniel (Editor-in-Chief). CRC Standard Mathematical Tables and Formulae, 31st Edition New York, NY: CRC Press, p. 512, 2003.
http://hyperphysics.phy-astr.gsu.edu/hbase/lcos.html http://hyperphysics.phy-astr.gsu.edu/hbase/lsin.html

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( A = \sin^{-1} \left[ \dfrac{a \sin B}{b} \right]\) A = angle A B = angle B C = angle C a = side a b = side b c = side c P = perimeter s = semi-perimeter K = area r = radius of inscribed circle R = radius of circumscribed circle *Length units are for your reference-only since the value of the resulting lengths will always be the same no matter what the units are.

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The Law of Sines (or Sine Rule) is very useful for solving triangles:

It works for any triangle:

a, b and c are sides.

A, B and C are angles.
(Side a faces angle A,
side b faces angle B and
side c faces angle C).

And it says that:

When we divide side a by the sine of angle A
it is equal to side b divided by the sine of angle B,
and also equal to side c divided by the sine of angle C

Sure ... ?

Well, let's do the calculations for a triangle I prepared earlier:

a 8 8 0.885... = sin(62.2°) = sin A = 9.04...

b 5 5 0.552... = sin(33.5°) = sin B = 9.06...

c 9 9 0.995... = sin(84.3°) = sin C = 9.04...

The answers are almost the same!
(They would be exactly the same if we used perfect accuracy).

So now you can see that:

Is This Magic?

Not really, look at this general triangle and imagine it is two right-angled triangles sharing the side h:

The sine of an angle is the opposite divided by the hypotenuse, so:

sin(A) = h/b

b sin(A) = h

sin(B) = h/a

a sin(B) = h

a sin(B) and b sin(A) both equal h, so we get:

a sin(B) = b sin(A)

Which can be rearranged to:

We can follow similar steps to include c/sin(C)

How Do We Use It?

Let us see an example:

Example: Calculate side "c"

Now we use our algebra skills to rearrange and solve:

Swap sides: c/sin(105°) = 7/sin(35°)

Calculate: c = ( 7 / 0.574... ) × 0.966...

Finding an Unknown Angle

In the previous example we found an unknown side ...

... but we can also use the Law of Sines to find an unknown angle.

In this case it is best to turn the fractions upside down ( sin A/a instead of a/sin A, etc):

Example: Calculate angle B

Sometimes There Are Two Answers !

There is one very tricky thing we have to look out for:

Two possible answers.

Imagine we know angle A, and sides a and b.

We can swing side a to left or right and come up with two possible results (a small triangle and a much wider triangle)

Both answers are right!

This only happens in the " Two Sides and an Angle not between" case, and even then not always, but we have to watch out for it.

Just think "could I swing that side the other way to also make a correct answer?"

Example: Calculate angle R

The first thing to notice is that this triangle has different labels: PQR instead of ABC. But that's OK. We just use P,Q and R instead of A, B and C in The Law of Sines.

But wait! There's another angle that also has a sine equal to 0.9215...

The calculator won't tell you this but sin(112.9°) is also equal to 0.9215...

So, how do we discover the value 112.9°?

Easy ... take 67.1° away from 180°, like this:

180° − 67.1° = 112.9°

So there are two possible answers for R: 67.1° and 112.9°:

Both are possible! Each one has the 39° angle, and sides of 41 and 28.

So, always check to see whether the alternative answer makes sense.

... sometimes it will (like above) and there are two solutions

... sometimes it won't (see below) and there is one solution

We looked at this triangle before.

As you can see, you can try swinging the "5.5" line around, but no other solution makes sense.

So this has only one solution.

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Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents.

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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.

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Law of sines: solving for a side | Trigonometry (video) | Khan Academy

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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.

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Voiceover:Say you're out flying kites with a friend and right at this moment you're 40 meters away from your friend and you know that the length of the kite's string is 30 meters, and you measure the angle between the kite and the ground where you're standing and you see that it's a 40 degree angle. What you're curious about is whether you can use your powers of trigonometry to figure out the angle between the string and the ground. I encourage you to pause the video now and figure out if you can do that using just the information that you have. Whenever I see, I guess, a non right triangle where I'm trying to figure out some lengths of sides or some lengths of angles, I immediately think maybe the Law of Cosine might be useful or the Law of Sines might be useful. So, let's think about which one could be useful in this case. Law of Cosines, and I'll just rewrite them here. The Law of Cosine is c squared is equal to a squared plus b squared minus 2ab cosine of theta. So what it's doing is it's relating 3 sides of a triangle. So a, b, c to an angle. So, for example, if I do 2 sides and the angle in between them, I can figure out the third side. Or if I know all 3 sides, then I can figure out this angle. But that's not the situation that we have over here. We're trying to figure out this question mark and we don't know 3 of the sides. We're trying to figure out an angle but we don't know 3 of the sides. The Law of Cosine just doesn't seem, at least in an obvious way, that it's going to help me. I could also try to find this angle. Once again, we don't know all 3 sides to be able to solve for the angle. So maybe Law of Sines could be useful. So the Law of Sines, the Law of Sines. Let's say that this is, the measure of this angle is a, the measure of this angle is lower case b, the measure of this angle is lower case c, length of this side is capital C, length of this side is capital A, length of this side is capital B. The Law of Sine tells us the ratio between the sine of each of these angles and the length of the opposite side is constant. So sine of lower case a over capital A is the same as lower case b over capital B, which is going to be the same as lower case c over capital C. Let's see if we can leverage that somehow right over here. We know this angle and the opposite side so we can write that ratio. Sine of 40 degrees over 30. Let's see. Can we say that that's going to be equal to the sine of this angle over that? Well it would be, but we don't know either of these so that doesn't seem like it's going to help us. But, we do know this side. Maybe we could use the Law of Cosines to figure out this angle, because if we know 2 angles of a triangle, then we can figure out the third angle. So let's do that. Let's say that this angle right over here is theta. We know this distance right over here is 40 meters, so we can say that the sine of theta over 40, this ratio is going to be the same as the sine of 40 over 30. Now we can just solve for theta. Multiplying both sides times 40, you're going to get, let's see. 40 divided by 30 is 4/3. 4/3 sine of 40 degrees is equal to sine of theta, is equal to sine of theta. Now to solve for theta, we just need to take the inverse sine of both sides. So inverse sine of 4 over 3 sine of 40 degrees. Put some parentheses here, is equal to theta. That will give us that angle here and we can use that information and this information to figure out the angle that we really care about. So, let's get a calculator out and see if we can calculate it. Let me just verify, I am in degree mode. Very important. All right, now I'm going to take the inverse sine of 4/3 times sine of 40 degrees, and that gets me, and I deserve a little bit of a drum roll, 58, well if we round to the nearest, let's just maintain our precision here. So 58.99 degrees roughly. This is approximately equal to 58.99 degrees. So, if that is 58.99 degrees, what is this one? It's going to 180 minus this angle's measure minus that angle's measure. Let's calculate that. It's going to 180 degrees minus this angle, so minus 40, minus the angle that we just figured out. Actually I can get all of our precision by just typing in second answer. So that just says our previous answer so I get all that precision there and so I get 81.01 degrees. So, if I want to round to the nearest, let's say I want to round to nearest hundredth of a degree, then I'd say 81.01 degrees. So this, this right over here, is approximately 81.01 degrees and we're done.

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Solving for an angle with the law of sines

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Voiceover:Say you're out flying kites with a friend and right at this moment you're 40 meters away from your friend and you know that the length of the kite's string is 30 meters, and you measure the angle between the kite and the ground where you're standing and you see that it's a 40 degree angle.

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Voiceover:Let's say you're studying some type of a little hill or rock formation right over here. And you're able to figure out the dimensions. You know that from this point to this point along the base, straight along level ground, is 60 meters. You know the steeper side, steeper I guess surface or edge of this cliff or whatever you wanna call it, is 20 meters. And then the longer side here, I guess the less steep side, is 50 meters long. So you're able to measure that. But now what you wanna do is use your knowledge of trigonometry, given this information, to figure out how steep is this side. What is the actual inclination relative to level ground? Or another way of thinking about it, what is this angle theta right over there? And I encourage you to pause the video and think about it on your own. Well it might be ringing a bell. Well you know three sides of a triangle and then we want to figure out an angle. And so the thing that jumps out in my head, well maybe the law of cosines could be useful. Let me just write out the law of cosines, before we try to apply it to this triangle right over here. So the law of cosines tells us that C-squared is equal to A-squared, plus B-squared, minus two A B, times the cosine of theta. And just to remind ourselves what the A, B's, and C's are, C is the side that's opposite the angle theta. So if I were to draw an arbitrary triangle right over here. And if this is our angle theta, then this determines that C is that side, and then A and B could be either of these two sides. So A could be that one and B could be that one. Or the other way around. As you can see, A and B essentially have the same role in this formula right over here. This could be B or this could be A. So what we wanna do is somehow relate this angle... We wanna figure out what theta is in our little hill example right over here. So if this is going to be theta, what is C going to be? Well C is going to be this 20 meter side. And then we could set either one of these to be A or B. We could say that this A is 50 meters and B is 60 meters. And now we could just apply the law of cosines. So the law of cosines tells us that 20-squared is equal to A-squared, so that's 50 squared, plus B-squared, plus 60 squared, minus two times A B. So minus two times 50, times 60, times 60, times the cosine of theta. This works out well for us because they've given us everything. There's really only one unknown. There's theta here. So let's see if we can solve for theta. So 20 squared, that is 400. 50 squared is 2,500. 60 squared is 3,600. And then 50 times... Let's see, two times 50 is 100, times 60, this is all equal to 6,000. So let's see, if we simplify this a little bit we're going to get 400 is equal to 2,500 plus 3,600. Let's see, that'd be 6,100. That's equal to 6,000... Let me do this in a new color. So when I add these two, I get 6,100. Did I do that right? Yeah. So it's 2,000 plus 3,000, plus 5,000. 500 plus 600 is 1,100. So I get 6,100 minus 6,000, times the cosine of theta. And let's see, now we can subtract 6,100 from both sides. So I'm just gonna subtract 6,100 from both sides so that I get closer to isolating the theta. So let's do that. So this is going to be negative 5,700. Is that right? 5,700 plus... Yes, that is right. Right, because if this was the other way around, if this was 6,100 minus 400 it would be positive 5,700. Alright. And then these two of course cancel out. And this is going to be equal to negative 6,000 times the cosine of theta. Now we can divide both sides by negative 6,000. And we get... I'm just gonna swap the sides. We get cosine of theta is equal to... Let's see we could divide the numerator and the denominator by essentially negative 100. So these are both going to become positive. So cosine of theta is equal to 57 over 60. And actually that can be simplified even more. Three goes into 57, is that 19 times? Yep, so this is actually... This could be simplified. This is equal to 19 over 20. We actually didn't have to do that simplification step because we're about to use our calculators, but that makes the math a little more tractable. Right, 3 goes into 57, yeah, 19 times. And so now we can take the inverse cosine of both sides. So we could get theta is equal to the inverse cosine, or the arc cosine, of 19 over 20. So let's get our calculator out and see if we get something that makes sense. So we wanna do the inverse cosine of 19 over 20. And we deserve a drum roll. We get 18.19 degrees. And I already verified that my calculator is in degree mode. So it gets 18.19 degrees. So if we wanted to round, this is approximately equal to 18.2 degrees, if we wanna round to the nearest tenth. So that essentially gives us a sense of how steep this slope actually is.

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Solving for an angle with the law of cosines

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Voiceover:Let's say you're studying some type of a little hill or rock formation right over here. And you're able to figure out the dimensions. You know that from this point to this point along the base, straight along level ground, is 60 meters.

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- [Voiceover] Let's say that I've got a triangle, and this side has length b, which is equal to 12, 12 units or whatever units of measurement we're using. Let's say that this side right over here, this side right over here, has length c, and that happens to be equal to nine. And that we want to figure out the length of this side, and this side has length a, so we need to figure out what a is going to be equal to. Now, we won't be able to figure this out unless we also know the angle here, because you could bring the blue side and the green side close together, and then a would be small, but if this angle was larger than a would be larger. So we need to know what this angle is as well. So let's say that we know that this angle, which we will call theta, is equal to 87 degrees. So how can we figure out a? I encourage you to pause this and try this on your own. Well, lucky for us, we have the Law of Cosines, which gives us a way for determining a third side if we know two of the sides and the angle between them. The Law of Cosines tells us that a squared is going to be equal b squared plus c squared. Now, if we were dealing with a pure right triangle, if this was 90 degrees, then a would be the hypotenuse, and we would be done, this would be the Pythagorean Theorem. But the Law of Cosines gives us an adjustment to the Pythagorean Theorem, so that we can do this for any arbitrary angle. So Law of Cosines tell us a squared is going to be b squared plus c squared, minus two times bc, times the cosine of theta. And this theta is the angle that opens up to the side that we care about. So we can use theta because we're looking for a. If they gave us another angle right over here, that's not the angle that we would use. We care about the angle that opens up into the side that we are going to solve for. So now let's solve for a, because we know what bc and theta actually are. So a squared is going to be equal to b squared... so it's going to be equal to 144, plus c squared which is 81, so plus 81, minus two times b times c. So, it's minus two, I'll just write it out. Minus two times 12 times nine, times the cosine of 87 degrees. And this is going to be equal to, let's see, this is 225 minus, let's see, 12 times nine is 108. 108 times two is 216. Minus 216 times the cosine of 87 degrees. Now, let's get our calculator out in order to approximate this. And remember, this is a squared. Actually, before I get my calculator out, let's just solve for a. So a is just going to be the square root of this. So a is going to be equal to the square root... of all of this business, which I can just copy and paste. It's going to be equal to the square root of that. So let me copy and paste it. So a is going to be equal to the square root of that, which we can now use the calculator to figure out. Let me increase this radical a little bit, so that we make sure we're taking the square root of this whole thing. So let me get my calculator out. So I want to find that square root of 220. Actually, before I do that, let me just make sure I'm in degree mode, and I am in degree mode. Because we're evaluating a trig function in degrees here. So that's fine, so let me exit. So it's going to be 225 minus 216, times cosine of 87 degrees. Not 88 degrees, 87 degrees. And we deserve a drumroll now. This is going to be 14.61, or 14.618. If, say, we wanted to round to the nearest tenth, just to get an approximation, it would be approximately 14.6. So a is approximately equal to 14.6, whatever units we're using long.

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Law of cosines: solving for a side | Trigonometry (video) | Khan Academy

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Voiceover] Let's say that I've got a triangle, and this side has length b, which is equal to 12, 12 units or whatever units of measurement we're using. Let's say that this side right over here, this side right over here, has length c, and that happens to be equal to nine.