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### 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

In ∆ABC, a

^{2}= b

^{2}+ c

^{2}− 2bc cosA.

Using Law of Cosines is the only way to find m∠A.

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

AB = 21, BC = 18, m∠B = 60, find AC.

- AC
^{2}= AB^{2}+ BC^{2}− 2AB*BC*cosB - AC
^{2}= 21^{2}+ 18^{2}− 2*21*18*cos60 - AC
^{2}= 387

Determine whether the Law of Cosines need to be used to find AC.

Use the Law of Cosines to find m∠A.

- 25
^{2}= 10^{2}+ 20^{2}− 2*10*20*cosA - 625 = 500 − 400cosA
- cosA = − [5/16]

^{o}

In ∆ABC, if ∠A is an obtuse angle, then cosA is always negative.

In triangle ABC, a and b are the opposite sides of ∠A and ∠B, if cosA > cosB, then a > b.

Find CD.

- AC
^{2}= 14^{2}+ 10^{2}− 2*14*10*cos110 - AC
^{2}= 392 - AC = 19.8.
- [(sin∠ACB)/14] = [sin110/19.8]
- sin∠ACB = 0.664
- m∠ACB = 41.6
^{o} - m∠CAD = m∠ACB = 41.6
^{o} - CD
^{2}= 18^{2}+ 19.8^{2}− 2*18*19.8*cos41.6 - CD
^{2}= 183

*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

### Geometry Online Course

### 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 things*0016

*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 a ^{2} = b^{2} + c^{2} minus 2 times bc times cos(A).*0121

*If you look at this, this looks kind of like the Pythagorean theorem: a ^{2} = b^{2} + c^{2}, 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

*b ^{2} = a^{2} + c^{2} - 2 times a and c, cosine of angle B.*0149

*c ^{2} = a^{2} + b^{2} 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 problems*0421

*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 a ^{2} = b^{2} + c^{2} - 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 b ^{2} = a^{2} + c^{2} (it is the other two sides), minus 2 times ac cosine of B.*0569

*And then, side c ^{2} = a^{2} + b^{2} - 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 a ^{2} = b^{2}...*0618

*our b is 21, so 21 ^{2}, 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: 21 ^{2} is 441,*0725

*plus...14 ^{2} 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 a ^{2} 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 2 ^{nd} 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 a ^{2} = b^{2} + c^{2} - 2bc cosine of angle A.*1048

*b ^{2} = a^{2} + c^{2} - 2ac cos(B).*1061

*c ^{2} = a^{2} + b^{2} - 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 b ^{2}, which is 5^{2}, equals a^{2} (is 13 squared), plus 12^{2}, 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 think*1387

*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 2^{nd} 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 a ^{2} = b^{2} + c^{2} - 2 times b times c times the cosine of angle A.*1895

*Then, b ^{2} = a^{2} + c^{2} - 2 times a times c times the cosine of angle B.*1909

*c ^{2} = a^{2} + b^{2} - 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 a ^{2} = 20^{2} + 24^{2} - 2(b)(c)(cos(47)).*1971

*a ^{2} = 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)/a*2150

*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 2 ^{nd} 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 a ^{2} = b^{2} + c^{2} - 2 bc cosine of angle A.*2581

*Here, a ^{2} 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, a ^{2} = b^{2}...we are going to say 8^{2}...*2616

*+ c ^{2}, just the third side, 11^{2}, minus 2 times those two sides cosine of angle A.*2624

*It would be the angle with that side.*2638

*8 ^{2} is 64, plus 11^{2} is 121, minus...this would be 176cos(A).*2646

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

*a ^{2} = 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...a ^{2} = 185 - -60.196.*2705

*a ^{2} = 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 a ^{2} would be b^{2}, which is 11^{2}, plus 8^{2}, minus 2(11)(8) cosine of angle A.*2993

*So, a ^{2} = 121 +...8^{2} is 64, minus 176 cosine of 70 degrees.*3013

*a ^{2} = 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; a ^{2} = 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

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.