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Lecture Comments (3)

0 answers

Post by Todd McKinney on December 9, 2014

You make it a lot easier to understand.
Thanks

1 answer

Last reply by: Mary Pyo
Sat Feb 4, 2012 12:48 AM

Post by Sanjay Nenawati on October 27, 2011

a-s-a is does have a angle side relationship, because 180 - a -a= other angle. then you will have a side angle side relationship.

Related Articles:

Law of Cosines

  • Law of Cosines: Let ΔABC be any triangle with a, b, and c representing the measures sides opposite angles with measures A, B, and C, respectively. Then,
  • Use the Law of Cosines when:
    • The triangle is a non-right triangle
    • The measures of an angle and the side opposite are not given

Law of Cosines

Determine whether the following statement is true or false.

In ∆ABC, a2 = b2 + c2 − 2bc cosA.
True.
Determine whether the following statement is true or false.

Using Law of Cosines is the only way to find m∠A.
False.
Determine which one should be filled in the blank.

Use the Law of Cosines when _____ of the following statements are true.
1. The triangle is a non - right triangle.
2. The measures of an angle and the side opposite are not given.

A. One of Them
B. Neither
C. Both
C

AB = 21, BC = 18, m∠B = 60, find AC.
  • AC2 = AB2 + BC2 − 2AB*BC*cosB
  • AC2 = 212 + 182 − 2*21*18*cos60
  • AC2 = 387
AC = 16.7.

Determine whether the Law of Cosines need to be used to find AC.
No.

Use the Law of Cosines to find m∠A.
  • 252 = 102 + 202 − 2*10*20*cosA
  • 625 = 500 − 400cosA
  • cosA = − [5/16]
m∠A = 108.2o
Determine whether the following statement is true or false.
In ∆ABC, if ∠A is an obtuse angle, then cosA is always negative.
True.
Determine which one should be used to find the missing measurement, Law of Sines or Law of Cosines.
Law of Sines.
Determine whether the following statement is true or false.
In triangle ABC, a and b are the opposite sides of ∠A and ∠B, if cosA > cosB, then a > b.
False.

Find CD.
  • AC2 = 142 + 102 − 2*14*10*cos110
  • AC2 = 392
  • AC = 19.8.
  • [(sin∠ACB)/14] = [sin110/19.8]
  • sin∠ACB = 0.664
  • m∠ACB = 41.6o
  • m∠CAD = m∠ACB = 41.6o
  • CD2 = 182 + 19.82 − 2*18*19.8*cos41.6
  • CD2 = 183
CD = 13.5

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Law of Cosines

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Law of Cosines 0:35
    • Law of Cosines
  • Law of Cosines 6:22
    • Use the Law of Cosines When Both are True
  • Law of Cosines 8:35
    • Example: Law of Cosines
  • Extra Example 1: Law of Sines or Law of Cosines? 13:35
  • Extra Example 2: Use the Law of Cosines to Find the Missing Measure 17:02
  • Extra Example 3: Solve the Triangle 30:49
  • Extra Example 4: Find the Measure of Each Diagonal of the Parallelogram 41:39

Transcription: Law of Cosines

Welcome back to Educator.com.0000

We are going to continue this unit with the Law of Cosines.0002

We are still on triangles; now, using Law of Cosines is very similar to when we used Law of Sines.0007

We are using both of those for non-right triangles; it is very important to remember that there are different things0016

that you can use for right triangles, and then there are things that you can use for non-right triangles.0024

So, Law of Sines and Law of Cosines are used for non-right triangles.0030

The Law of Cosines consists of three different formulas, depending on what you need--what measure you have to find.0038

If you have a triangle ABC with sides a, b, c, and angles A, B, C, angles are capital, and the sides opposite them are the lowercase of that letter.0048

So, if this is angle A, then this would be side a; if this is angle B, this is side b, and this is side c.0063

It is really important to remember that the angle and the side opposite go together, like a pair: this angle and this side opposite, and so on.0073

If we are looking for, let's say, A, you can use this formula.0086

If you are given, let's say, a, b, and c, and you are looking for angle B, well, then, you have to use this one.0094

Or if you are, let's say, looking for b, and you are given sides c and a, and then you are given the angle C, then you would use this one.0105

You would only use one of these, depending on what you need.0117

That is a2 = b2 + c2 minus 2 times bc times cos(A).0121

If you look at this, this looks kind of like the Pythagorean theorem: a2 = b2 + c2, the other two sides squared.0132

But it is minus 2 times bc, and the cosine of A; so then, that is how it is going to work for all of them.0141

b2 = a2 + c2 - 2 times a and c, cosine of angle B.0149

c2 = a2 + b2 minus 2 times a times b, times the cosine of this angle C.0160

And that is the Law of Cosines.0170

Now, when you are given a non-right triangle, and you have to determine whether to use Law of Sines or Law of Cosines--0173

well, the Law of Sines would probably be the easier option, so if you can, you would use the Law of Sines.0183

But remember: the Law of Sines was sin(A) over side a = sin(B)/b and sin(C)/c.0191

So, in order for you to be able to use the Law of Sines, you would have to have an angle-side pair.0205

If you are given an angle-side pair, then you are always going to be given three measures.0215

Out of the six total, the three angles and three sides, you are going to be given three.0219

In order to use the Law of Sines for non-right triangles, out of those three, two of them would have to be an angle-side pair.0225

When you don't have an angle-side pair, and it is a non-right triangle, then you would have to use the Law of Cosines.0235

Again, for non-right triangles, you can use the Law of Sines or the Law of Cosines.0242

If you can, you will look to see if you have an angle-side pair; if you do, then you would use the Law of Sines.0250

But if all three measures that are given...none of them is an angle-side pair, then you would have to use the Law of Cosines.0258

It is either one or the other.0267

For example, let's say that the measure of angle B is 27 degrees, and let's say this over here is 5, and this over here is 10.0271

Well, here we have no angle-side pair, because angle C is not given; so then, we don't have that angle-side pair;0288

we have the measure of angle B, but no side b, so the angle-side pair is not given; and the same thing happens for this.0299

So, since we have no angle-side pair, we would have to use the Law of Cosines.0305

And see how we are given a and c, and we have angle B; so which one would we use?0310

Well, it depends on what we are looking for; if we are looking for this one right here--let's say we are looking for b--0321

that went first; then we would have to use this one right here, because this is what you are looking for.0331

You are given this; you are given this; and you are given the measure of angle B; so you would have to use this one.0337

Let's say we are looking for, let's see...once we find that, if we want to find, let's say...the measure of angle A;0344

then after we find B, we will have an angle-side pair; we will have the measure of angle B, and we will have side b.0359

So, from there, you can go ahead and use the Law of Sines, using this one and this one that we found, and then sin(A)/a.0366

Let's just do a few problems and see if we can get more familiar with this.0379

Use the Law of Cosines when both are true: the triangle is a non-right triangle (we know this one),0384

and then the measures of an angle and the side opposite are not given, meaning that there is no angle-side pair.0392

Here, if I draw this out, and this is a; this is b; and this is c; we know that, for us to use the Law of Cosines, these have to be true.0401

If these are true, then we have to use the Law of Cosines.0418

And from the three measures that they give you, there are going to be three different types of problems0421

where you are going to use the Law of Cosines, where both of these would be true.0428

If I am given a side, a side, and a side--the measure of all three sides of a non-right triangle--then that would be the Law of Cosines.0432

Also, if you have side-angle-side (with the one that we just did, that was side-angle side, because that was side b,0443

and then angle C, and then side a), that doesn't give you any angle-side pairs.0453

So, side-angle-side would be another one; and then, the third one is going to be angle-side-angle.0458

So, if you have an angle, a side, and an angle, that is also a case where you are not going to have any angle-side pairs.0468

These three are the types of problems, given the measures, where you are going to have to use the Law of Cosines: side-side-side...0479

and don't get this confused with the side-side-side postulates and the congruence theorems,0491

because that is when you are using two triangles, and you are trying to make them congruent.0499

With this, I am just doing this so that you can see that, if you are given the measure of a side, the measure of a side,0503

the measure of a side, of one triangle, then you are going to have to use the Law of Cosines, and so on.0509

OK, for this triangle ABC, we are going to look for side a, which is this side right here, the side opposite angle A.0516

I know that I have no angle-side pair; I have a side, an angle, and this side.0527

And so, using one of the formulas (I know that I have to use the Law of Cosines, so), here are the three formulas.0533

It is a2 = b2 + c2 - 2bc times the cosine of angle A.0543

Remember: it has to be the cosine, because it is the Law of Cosines.0555

And remember that all lowercase letters are the sides, and the capital letters are angles.0559

The second one is b2 = a2 + c2 (it is the other two sides), minus 2 times ac cosine of B.0569

And then, side c2 = a2 + b2 - 2ab cosine of angle C.0582

Those are your formulas; now, we are going to see which one we need to use.0597

I have side c, side b, and angle A; if I am looking for side a, well, I am given b, c, and angle A, and this is what I am looking for.0602

So, I have to use the first one: that is going to be a2 = b2...0618

our b is 21, so 212, plus...c is 14, squared, minus 2 times 21 times 14, cosine of 60.0629

From here, it is all calculator math.0646

Now, if your calculator is the type that doesn't perform everything in one screen, then just go ahead and solve it in chunks.0652

But keep in mind--remember the order of operations: we have to multiply all of those together.0663

And this right here is the cosine of 60, so you would have to punch in cos(60) on your calculator to find that number.0672

You cannot separate the cosine and the angle measure.0682

It is going to be 2 times 21 times 14 times whatever this number is, so it is four numbers; and cos(60) is actually 1/2,0686

so it will be those four, multiplied; you are going to square this; you can add these together,0694

because here, multiplying this, this is far away from this right here; so you can solve this, and they won't affect each other.0702

Just remember that you have to multiply the 2 by all of those numbers before subtracting the 2.0709

So, be careful not to just solve it all in the order of this plus this minus 2; it has to be minus the whole thing.0716

I am going to just go ahead and do 21; and I do have a calculator on my screen: 212 is 441,0725

plus...142 is 196, minus...and then, let's do all of that, too.0735

2 times 21 times 14 times cos(60), .5, equals 294, so a2 is going to be 441 + 196 - 294, so 343.0750

Then, the square root of that is 18.5, so that is the measure of side a, 18.5.0795

All right, we are going to work on some more examples.0817

Determine whether the Law of Sines or the Law of Cosines should be used to solve each triangle.0820

We are not going to actually solve them; we are just going to determine if we can use the Law of Sines or the Law of Cosines.0826

Remember that both the Law of Sines and the Law of Cosines are used for non-right triangles.0835

And you want to look to see if you have an angle-side pair.0841

If you do, then you can go ahead and use the Law of Sines, because the Law of Sines is a lot easier.0847

If you don't, then you would have to use the Law of Cosines.0852

Now, if they say to solve the triangle, we went over that before; solving the triangle just means looking for all of the unknown measures.0856

So, in this case, they have given you this side, this side, and this angle.0866

To solve the triangle, we would have to solve for the measure of angle B, the measure of angle C, and AB, because those are all unknown measures.0870

The first one: we have a non-right triangle, and let's see how we have the measure of angle A, and we have side a, and then we have side b.0885

So, we do have an angle-side pair; since we have an angle-side pair...0900

now, I am not going to actually solve it, but I just want you to see that the Law of Sines is this.0904

Remember: you don't use all three of these ratios--you would just use two of them to create a proportion, whichever two you need.0917

We know that we are going to use the angle A and side a; it is going to be sin(30)/7 =...0925

and then, this is side b; so then, we would be looking for angle B; it would be sin(B)/11.0937

And then, you would solve it that way; you would have to solve sin(30) on your calculator; you can multiply that by 11, cross-multiplying.0948

And then, you are going to divide the 7.0958

And then, when you solve sin(B), remember: you have to use the inverse sine, the sin-1.0963

It is probably the 2nd key, and then "sin" on your calculator; use the inverse sine to find the measure of angle B.0969

That is the Law of Sines for this one; and then, for this one, we are given all of the side measures, but no angles.0981

That means that we have no angle-side pairs, so we would have to use the Law of Cosines--the Law of Sines here, and the Law of Cosines here.0994

And since you have three formulas to work with, you would have to choose whichever one you need,1008

depending on what angle or what measure...what side...whatever it is that you are looking for.1014

That is Example 1; for the next example, we are going to use more Law of Cosines to find missing measures.1021

For the first one, I want to look for the measure of angle B.1032

I am going to label that as x, just so that you know that that is the missing measure.1036

And then, before we start, let's write out all of the formulas again.1042

That is a2 = b2 + c2 - 2bc cosine of angle A.1048

b2 = a2 + c2 - 2ac cos(B).1061

c2 = a2 + b2 - 2ab cos(C); there are the formulas.1072

And then, to find the measure of angle B, given all three sides, I want to see what I want to use--which formula I would use.1084

So, for this one right here, I would have to use the second formula, because that is the only one that gives me angle B.1094

If you are looking for an angle measure, then you would have to pick and choose your formula, depending on this right here.1108

These are the only parts of the formulas that give you the angle measures, so it has to be the second one.1115

It is going to be b2, which is 52, equals a2 (is 13 squared), plus 122, minus 2ac cosine of angle B.1127

This is going to be 25 = 169 + 144 - 2...here, let's solve this part out on the calculator.1154

Remember that you can square this; you can square that; exponents come first, always, always, by the order of operations.1171

And once you do that, make sure that you do not subtract 2; it is not this number, subtract 2, because 2 is multiplied to all of this.1180

So, it would have to be 2 times 13 times 12...and then, for this one, cos(B)...1193

we are looking for angle B, so this is our variable; but we cannot separate the cosine and the B, because they have to go together.1199

It has to be the cosine of an angle measure.1208

Think of this whole thing as the variable; that is what you are solving for, so think of that whole thing as x.1211

Here, it would be minus 2 times 13 times 12; these three numbers, multiplied together, is going to be the coefficient of this variable.1222

2 times 13 times 12 is 312, cosine of B; I will circle this again.1235

Then, this is going to be 25; it equals this plus this; it is going to be 313, minus 312, cosine of B.1254

Be careful, again: do not subtract these two numbers; this is not 25 = 1cos(B); do not subtract them; it is this minus this whole thing.1269

But what I can do is subtract this over to the other side.1280

If I subtract 313, I am going to get...that is going to be -288, which is equal to negative (don't forget the negative sign) 312 cosine of B.1288

Now, from here, again, if I circle this, and that is the variable, then this is the coefficient;1310

I would have to divide this number to both sides, so that I can solve for this variable.1317

Divide this by -312; divide this by -312 (I'll write that over).1326

Your number is going to be -288, divided by -312: 0.923 = cos(B).1340

Now, remember, again: we are solving for angle B; that means we have to find the inverse cosine,1363

because here, if we have this angle measure, then we can go ahead and punch in the cosine of that angle measure.1372

But since that is the actual number that we are looking for, you would have to use the inverse cosine, don't forget--1381

because when you punch in a number after sin, cos, or tan, then the calculator is going to think1387

that that number is the angle measure, because it is always the cosine, sine, or tangent of an angle measure.1395

Just make sure you don't punch in the cosine of this number here, or else your calculator is going to think that that is the angle measure.1403

But instead, we are doing the cosine of the unknown angle measure; it equals...and that is the answer.1412

You would have to use the inverse cosine; that is cos-1; on your calculator, you might have to press the 2nd key and cosine.1420

It depends on your calculator; it might have its own separate button for it--just look for it.1436

See how you just punch that in with this number; and then it is going to be 22.6...let's see, let's just round it to hundredths.1445

So, that would be the measure of angle B, 22.6 degrees.1462

The next one: we are given angle, side, and angle; we have no angle-side pairs.1473

So, let's say that I want to look for two unknown measures for this one.1487

If I have this as the unknown measure--this is the first unknown measure--remember...1498

now, you can use the Law of Cosines, but it is a lot of work, and it does take quite some time.1504

Here, we actually don't have to use the Law of Cosines, because this is the third angle of a triangle.1511

So, try to remember that, if you are looking for the third angle of a triangle, then you can use the angle sum theorem,1518

where all three angles of a triangle add up to 180; so it is just going to be 180 = 40 degrees + 59 + x.1528

So, all three angles together add up to 180.1548

This is 99 + x; if we subtract it, that is going to be 81; it equals x, so the measure of angle C is 81 degrees.1550

Now, I also want to look for, let's say, side b.1580

So, in this case, we would have to use the Law of Cosines; which one would I use?1588

If I am looking for this here, for this one, do I have all the sides--what sides do I have?1598

I have this side right here; so here, I have 14; this is unknown; and then, for this one, let's see...1611

I have all three angles, so I don't have to worry about that; I have side c; I have this; I have this; the first one I have...1622

And then, this is what I am looking for; and then, I have the measure of angle A, and I have this, again.1638

I wouldn't be able to use that one, because I only have that side right there.1648

Now, if you notice that all three of these don't work, it is because here, now that I found this angle measure, I have an angle-side pair.1660

I would use Law of Sines, instead of Law of Cosines.1674

Let's save ourselves a lot of work; and then here, since the measure of angle C is now 81,1679

in order for me to look for side b, I can use the Law of Sines, because of this angle-side pair.1684

Here, I am going to do this with the sine of angle B.1694

If you have something like this, you wouldn't have to use the Law of Cosines, because you can just find the third angle measure.1709

Now, if you want, you could; but again, this would be a lot easier.1718

This here, the sine of C, the sine of 81, over 14, equals the sine of 59, over b.1725

So, the sine of 81...we are going to cross-multiply, so bsin(81) = 14sin(59).1738

I can divide the sine of 81 (remember: don't forget that you can't separate them)...so b would equal all of this, over sin(81).1757

I am going to put 12 over 0.988...so b is going to be 12.5.1785

Again, here we have a side and a side and a side given; and so, we are going to have to use this one, because the unknown measure would be angle B.1824

For this one, when you are given angle-side-angle, then just solve it by finding the third angle measure,1835

and then using the Law of Sines, using the angle-side pair.1844

For this next one, we are going to solve the triangle.1850

Remember: to solve the triangle means to solve for all unknown measures, angles and sides.1853

Here, I want to list out all of the different measures that I am going to look for: the measure of angle B, the measure of angle C, and then side a.1861

And that is just so that I don't forget to solve for something.1875

Now, we have no angle-side pairs; we have the measure of angle A, but we don't have side a...and here, and here.1880

So, there are no angle-side pairs; that means that I am going to have to use the Law of Cosines.1888

The three formulas, again, would be a2 = b2 + c2 - 2 times b times c times the cosine of angle A.1895

Then, b2 = a2 + c2 - 2 times a times c times the cosine of angle B.1909

c2 = a2 + b2 - 2 times a times b cosine of angle C.1921

First, in order for me to solve for this angle measure, I want this sine here.1935

If I want to solve for angle B, then I am going to have to know all of the measures of these sides.1941

Since we are given the two sides, side c and side b, I am going to look for side a.1951

Then, since I have the measure of angle A, I want to use this first one, because that is what goes in there.1962

It is going to be a2 = 202 + 242 - 2(b)(c)(cos(47)).1971

a2 = 400 + 576 -...don't forget that you cannot subtract this number with this;1994

it is going to be 2 times 20 times 24 times the cosine of 47.2012

You are going to multiply all four of those numbers together...that number, and this...so this is 654.72.2017

And then, now that you have multiplied all of that, you are just going to take this, add that, and subtract that.2047

400...add the 576, and subtract 654.72; you are going to get 321.28.2061

And then, to find a, we are going to have to take the square root of that; you are going to get 17.92.2084

So, I am going to write that answer here, for this one...92; and then I need to solve for these two.2095

So then, the measure of angle B...now that I have this, if I write this in here...2104

Keep in mind that, for any given problem where you have to use the Law of Cosines, you only have to use it once--you don't have to keep using it.2115

So, here, if I am solving the triangle, then I have to find the measures of three unknowns.2122

I don't have to use the Law of Cosines for each one of these; I only have to use it once.2128

And then, once you use it once, you are going to end up with an angle-side pair.2136

So, since I solved for a, I have side a, and I have angle A; there is my angle-side pair.2141

If you have an angle-side pair, then you know to use the Law of Sines: so sin(A)/a2150

equals the sine of angle B, over side b, which is equal to sin(C) over side c.2158

Since we have the angle-side pair of angle A and side a, we are going to use this one.2168

And then, if we solve for angle B first, then I am going to use that one.2173

I will put it down here: sin(47)/17.92 = sin(B)/20.2184

So, you are going to cross-multiply; that will be 20sin(47) = 17.92sin(B).2203

This is going to be, using your calculator...remember that this is sine, not cosine...it is this number, and then divided by this number;2216

this is your variable that you are solving for; that is the coefficient; so I need to divide this number...17.92.2239

I get 0.816; I am just going to double-check that really quickly.2263

47...the sine of 47 degrees...multiply that by 20 (you can do 20 times sin(47); it depends on what your calculator can do); divide it by 17.92.2272

So, .816 = the sine of angle B; now again, here we are looking for the angle measure, so we are going to have to use the inverse sine.2294

That means that you are going to hit the 2nd button; and that is going to give us B; the measure of angle B is 54.71.2309

We still have one more to solve for, the measure of angle C.2335

Now, we want to look for the easiest possible way to solve for this.2344

If you want, you can go ahead and use the Law of Sines again, because you have two different angle-side pairs.2348

But for this one, since I have two angle measures, I just want to look for the third one using the angle sum theorem.2355

Keep that in mind: if you have two angles, that would be the easiest way to look for a third angle.2363

The measure of angle C is going to be 180 minus 47 added to 54.71.2370

180 minus...and then, what are these two?...that will be 101.71; that means that the measure of angle C is going to be 78.29.2390

180 minus 101.71...the measure of angle C is 78.29.2421

All three of these are your answers, because these are the unknown measures of the triangle.2436

When it says to solve the triangle, then you are going to be looking for all of the unknown measures.2444

The easiest way would be to just first write out what you have to find, and then do one at a time.2451

First, always look to see if you can use the Law of Sines or the Law of Cosines.2460

In our case, we had to use the Law of Cosines, because we had no angle-side pair.2464

We are going to look for that side; and then, from there, you only have to use Law of Cosines once in a single problem.2470

We use the Law of Cosines; and then, since that gave us an angle-side pair, we went ahead and used the Law of Sines to find one of the angle measures.2480

Then, since you are given two angle measures, you can solve for the third angle measure by the angle sum theorem--that is solving the triangle.2489

And the fourth example: In parallelogram ABCD, AD is 8; AB is 11; and the measure of angle A is 110; find the measure of each diagonal of the parallelogram.2501

First, I am going to draw a parallelogram and label it ABCD.2515

AD is 8; AB is 11; the measure of angle A is 110; and then, find the measure of each diagonal.2529

That is the first diagonal; by drawing this diagonal in, we now have this triangle here.2548

If this is 11, and this is 8, then this is what we are looking for; this would be side a.2557

And then, in order to solve for that, I know it is a non-right triangle...do I have an angle-side pair?2566

No, I don't; so then, I have to use the Law of Cosines.2576

And remember: it is a2 = b2 + c2 - 2 bc cosine of angle A.2581

Here, a2 is what we are looking for; and even though it is not sides b and c, it would just be the other two sides.2602

There are three sides total; it is just all three, written out here.2612

So, a2 = b2...we are going to say 82...2616

+ c2, just the third side, 112, minus 2 times those two sides cosine of angle A.2624

It would be the angle with that side.2638

82 is 64, plus 112 is 121, minus...this would be 176cos(A).2646

Then, from here, remember that this is what we are solving for.2664

a2 = 185 - 176...OK, we are solving for side a; this would be 110; let me circle the variable right here;2671

then this is the cosine of 110; make sure you do not subtract these two numbers.2699

So here, we are going to just solve this out on the calculator...and then...a2 = 185 - -60.196.2705

a2 = 245.196; and then, a would be √245.196...15.67.2756

So, if I were to round to the hundredths, it would be 15.67; that is BD.2805

Now, they want us to find the measure of the other diagonal.2816

Now, this, obviously, doesn't look like it is 110; if you want to draw it so that it would fit better, you can draw it the other way, like this.2823

This was A, B, C, D; and it is the same thing, but just if you want to have it more scaled, this would be 110;2838

since this side is a little bit shorter...8...11...that is 110, and then we found that this right here is 15.67.2851

BD, one of the diagonals, is 15.67; now, if I erase that, and I look for the other diagonal right here,2861

well, I know that this is 11; and this is the side that I am looking for.2883

Here, this angle measure is...we have 8 and 11; and then, do we know any of the angle measures?2895

Well, what do we know about parallelograms?2909

If this is 110 (let me erase this diagonal, so that it is easier to see), we know that this angle right here,2914

with this angle, are supplementary, because if you were to draw these a little bit longer--2927

extend these sides and the side right here--then remember: we have same-side interior angles, or consecutive interior angles.2934

So then, with parallel lines and a transversal, if this is 110, then this angle right here would be that.2945

So, using that, let me erase this, and then redraw the diagonal right there.2953

I am going to be looking for the side AC; and it doesn't matter what we label it.2961

I can redraw just that triangle...70...and then I label this a, b, and c...8, 11, and this is what we are looking for.2971

Just for the sake of our formulas, I can use this same formula again, because this would be side a.2984

Then a2 would be b2, which is 112, plus 82, minus 2(11)(8) cosine of angle A.2993

So, a2 = 121 +...82 is 64, minus 176 cosine of 70 degrees.3013

a2 = 185 - 176cos(70); and then, you can solve this part right here.3032

Make sure that you multiply this number by this number on the calculator, instead of subtracting it first.3044

The cosine of 70, times 176, is minus 60.196; and then, take that and subtract the 180; a2 = 124.80; so a is going to equal 11.17.3052

That side is the other diagonal, and that is AC; that would be 11.17; those are the measures of the two diagonals of this parallelogram.3113

And remember how we initially drew this parallelogram.3128

Now, this is just so you have a visual; if you need to draw it more to scale, you can redraw it,3132

depending on what sides you have and what is bigger and what is smaller.3139

And then, we know that the measure of angle A was 110, so that would be an obtuse angle; so you can draw it in that way.3144

And here, I just redrew this triangle right here, so that it would be easier to see what we are dealing with, what we are using.3152

That is it for this lesson; thank you for watching Educator.com.3161