For more information, please see full course syllabus of Trigonometry

For more information, please see full course syllabus of Trigonometry

### Related Articles:

### Word Problems and Applications of Trigonometry

**Main formulas**:

- Master formula for
*right*triangles: SOHCAHTOA! - The Law of Sines (in
*any*triangle) - The Law of Cosines (in
*any*triangle) - Heron's Formula (in
*any*triangle)

**Example 1**:

^{°}angle with the ground, how tall is the pole?

**Example 2**:

^{°}, then how long will the bridge be?

**Example 3**:

**Example 4**:

^{°}angle with the ground, how high is the kite?

**Example 5**:

^{°}and 110

^{°}with the coast. How far is it to the town along each of the roads?

### Word Problems and Applications of Trigonometry

^{°}above the horizon. Find the length of a shadow cast by a flagpole that is 35 feet tall.

- Start by drawing a picture using the information you are given
- Use SOHCAHTOA to find the length of the shadow
- tan40
^{°}= [35/x] ⇒ x = [35/(tan40^{°})]

^{°}.

- Start by drawing a picture using the information you are given
- Use SOHCAHTOA to find the height of the ladder to the ground
- sin53
^{°}= [h/20] ⇒ h = 20sin53^{°}

^{°}. What is the approximate height of the house?

- Start by drawing a picture using the information you are given
- Use SOHCAHTOA to find the height of the house
- tan48
^{°}= [h/185] ⇒ h = 185tan48^{°}

^{°}angle with the ground. How high is the kite?

- Start by drawing a picture using the information you are given
- Use SOHCAHTOA to find the height of the kite
- sin35
^{°}= [h/300] ⇒ h = 300sin35^{°}

- Start by drawing a picture using the information you are given
- Use SOHCAHTOA to find the angle of elevation
- tanθ = [85/60] ⇒ θ = arctan([85/60])

^{°}

- Start by drawing a picture using the information you are given
- Use SOHCAHTOA to find the length of the shadow
- sinθ = [2237/3500] ⇒ θ = arcsin([2237/3500])

^{°}

^{°}W. From point C, 150 meters from B, the bearings to B and A are S67

^{°}E and S18

^{°}E respectively. Find the distance from A to B.

- Start by drawing a picture using the information you are given
- C = 67
^{o}− 18^{o}= 49^{o} - B = 180
^{o}− 67^{o}− 33^{o}= 80^{o} - A = 180
^{o}− 49^{o}− 80^{o}= 51^{o} - Use Law of Sines
- [(sin51
^{°})/150] = [(sin49^{°})/c] ⇒ c sin51^{°}= 150sin49^{°}⇒ c = [(150sin49^{°})/(sin51^{°})]

^{°}from the vertical. At a point 45 meters from the tree, the angle of elevation is 32

^{°}. Find the height of the tree.

- Start by drawing a picture using the information you are given
- θ = 180
^{°}− 32^{°}− 96^{°}= 52^{°} - Use Law of Sines
- [(sin32
^{°})/h] = [(sin52^{°})/45] ⇒ h sin52^{°}= 45sin32^{°}⇒ c = [(45sin32^{°})/(sin52^{°})]

- Start by drawing a picture using the information you are given
- Use the Law of Cosines
- cosC = [(a
^{2}+ b^{2}− c^{2})/2ab] ⇒ cosC = [(620^{2}+ 560^{2}− 730^{2})/2(620)(560)] ⇒ C = arccos([165100/694400])

^{°}

- Start by drawing a picture using the information you are given
- Use Heron's Formula to find the area: A = √{s(s − a)(s − b)(s − c)} ⇒ where s = [1/2](a + b + c)
- s = [1/2](480 + 540 + 670) ⇒ s = 845
- A = √{845(845 − 480)(845 − 540)(845 − 670)} ⇒ A = √{845(365)(305)(175)}

*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

### Word Problems and Applications of Trigonometry

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
- Formulas to Remember 0:11
- SOHCAHTOA
- Law of Sines
- Law of Cosines
- Heron's Formula
- Example 1: Telephone Pole Height 4:01
- Example 2: Bridge Length 7:48
- Example 3: Area of Triangular Field 14:20
- Extra Example 1: Kite Height
- Extra Example 2: Roads to a Town

### Trigonometry Online Course

I. Trigonometric Functions | ||
---|---|---|

Angles | 39:05 | |

Sine and Cosine Functions | 43:16 | |

Sine and Cosine Values of Special Angles | 33:05 | |

Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D | 52:03 | |

Tangent and Cotangent Functions | 36:04 | |

Secant and Cosecant Functions | 27:18 | |

Inverse Trigonometric Functions | 32:58 | |

Computations of Inverse Trigonometric Functions | 31:08 | |

II. Trigonometric Identities | ||

Pythagorean Identity | 19:11 | |

Identity Tan(squared)x+1=Sec(squared)x | 23:16 | |

Addition and Subtraction Formulas | 52:52 | |

Double Angle Formulas | 29:05 | |

Half-Angle Formulas | 43:55 | |

III. Applications of Trigonometry | ||

Trigonometry in Right Angles | 25:43 | |

Law of Sines | 56:40 | |

Law of Cosines | 49:05 | |

Finding the Area of a Triangle | 27:37 | |

Word Problems and Applications of Trigonometry | 34:25 | |

Vectors | 46:42 | |

IV. Complex Numbers and Polar Coordinates | ||

Polar Coordinates | 1:07:35 | |

Complex Numbers | 35:59 | |

Polar Form of Complex Numbers | 40:43 | |

DeMoivre's Theorem | 57:37 |

### Transcription: Word Problems and Applications of Trigonometry

*We are trying some more examples of applied trigonometry to solve word problems.*0000

*In this first one, we are given a child flying a kite on 200ft of string and the kite’s string makes an angle of 40 degrees with the ground.*0006

*We want to figure out how high is the kite of the ground.*0014

*Remember, whenever you have these word problems, the important thing is to draw a picture right away.*0019

*That helps you convert from lots of words into some triangles or geometric picture where you can invoke the equation that you know.*0023

*Let me draw this now, the child is flying a kite on 200ft of string and the kite’s string makes an angle of 40 degrees with the ground, that is a 40 degree angle.*0034

*I want to figure out how high the kite is off the ground so let me draw in something to mark the height here.*0050

*There is the kite up there and here is the child, we are not going to worry about how tall the child is because that really will not make a difference.*0066

*We are trying to find this distance right here, that is the height of the kite off the ground.*0077

*This is a right triangle so we can use our formulas for right triangles which is SOHCAHTOA.*0086

*Remember SOHCAHTOA only works for right triangles, if you do not have a right triangle then you will going use something like the law of sin or law of cos, or heron’s formula.*0092

*If you have a right triangle, SOHCAHTOA is usually faster.*0102

*Let us see what I know here, I do not really know anything except the hypotenuse of this triangle and one angle, and then this is the opposite side to the angle that I know.*0108

*I’m going to use this part of SOHCAHTOA, sin=opposite/hypotenuse, sin(theta)=opposite/hypotenuse that is because I know the hypotenuse and I’m looking for the opposite side there.*0122

*Sin(40)=opposite/200 and if I solve that by for the opposite side, I will get the opposite side as equal to (200) sin (40).*0140

*Let me check on my calculator what that is, since that is not a common value, that is not one that I just remember.*0162

*What I get is an approximate value of 128.6 and we are told that the unit of measurement here is feet, so, 128.6ft.*0171

*That one was a pretty quick one, the reason it was quick because we really had a right angle there so we are able to use SOHCAHTOA.*0199

*If we did not have a right angle and we had to use the law of cos or law of sin, it probably would have been longer.*0205

*Let us recap what it took to do that problem.*0212

*First of all, you read the word and try to make a sense of it then you draw a picture.*0215

*It is very important that you draw a picture to illustrate what is going on.*0219

*We drew a picture with a child here, the kite here, and we know that the child is flying a kite on 200ft of string.*0223

*We fill that in, we are told that the kite’s string makes a 40 degree angle with the ground and we draw that in.*0233

*We are trying to find how high is the kite? We draw in the height of the kite.*0240

*We have a right angle, definitely a candidate for SOHCAHTOA and the reason I picked sin is because I want to find the opposite side to the angle and I know the hypotenuse.*0247

*That is why I worked with sin, sin=opposite/hypotenuse.*0258

*I filled in the values that I know 40 and 200, and I just solved it for the opposite.*0263

*I got a number and then I checked back to see that we are using feet.*0268

*That tells me that my unit for measurement for the answer should be feet.*0274

*Our last example here is another pretty wordy one, two straight roads lead from different points along the coast to an in land town.*0000

*Surveyors working on the coast measure that the roads are 12 miles apart and make angles of 40 degrees and 110 degrees with the coast.*0009

*I want to find out how far is it to the town along each one of the roads?*0019

*This is a pretty complicated one, when you get like one like this with lots of words, it is absolutely essential that the first thing you do is draw a picture.*0025

*Let me draw a picture, that is supposed to be the coastline there.*0034

*Those are supposed to be fish swimming in the ocean and we have got two different points along the coast.*0047

*Two roads leads from these points to an in land town and these roads make angles of 40 degrees and 110 degrees with the coast.*0055

*That is about 110 and that is about 40, we know that straight roads go off and they go to some town somewhere in land.*0069

*The last thing we are told is that the roads are 12 miles apart, I’m going to write in my third side as 12 miles there.*0086

*We want to figure out how far it is to the town along in each of the roads.*0099

*First of all, let us look at this triangle we got here.*0104

*What I have been given is two angles of the triangle and the side in between them, so we have been given an (angle, side, angle) situation.*0108

*When you are given an (angle, side, angle) situation, the thing you want to check is whether those angles are really legitimate.*0120

*In other words, whether they add up to less than 180 degrees.*0127

*The angles sum up to, in this case (110+40=150) which is less than 180, which is good, there is a unique solution.*0132

*I’m going to label everything I know here, I’m going to label that unknown angle as (C) and the others are (A and B).*0160

*I will label my sides (a, b – remember you put the sides opposite the angles of the same letter and that side that we know is 12 is (c).*0172

*Now I have labeled everything I know and I want to find the length of the roads, that is (a and b), so how can I find those?*0189

*This is an (angle, side, angle) situation, (angle, side, angle) if you remember that is the one you want to solve using the law of sin.*0202

*That one you want to solve with the law of sin, it does not work very well with the law of cos.*0218

*Certainly, it does not work with SOHCAHTOA because we do not have a right angle here.*0223

*We are going to use the law of sin, let me remind you what that is, that says sin(A)/a=sin(B)/b=sin(C)/c.*0227

*That is the law of sin and we want to use that to solve for (a) and (b).*0245

*Let me work on (a), I will write down sin(A)/(a)=sin(C)/c, the reason I’m going to that is because I know what (c) is and I do not know what any of the others are.*0251

*I can find (C) very easily, (C) is 180-(A)-(B), that is because the three angles of a triangle add up to 180.*0271

*That is 180-110-40, 180-150=30, so (C) is 30 degrees, that is not my full answer to the problem but that is going to be useful to me.*0284

*Sin(A)=sin(110), sin(a)= I do not know yet, sin(C)=sin(30), sin(c)=12.*0307

*I’m going to cross multiply to solve for (a), 12 sin(110)=(a)sin(30), if I solve that for (a), (a)=(12)sin (110)/sin(30).*0325

*I’m going to reduce that using my calculator, remember to set your calculators on degree mode if that is what you are using.*0349

*If you use radian mode, your calculator will interpret this as a 110 radians and 30 radians and will give you answers that do not make sense.*0355

*I’m going to simplify (12)sin(110)/sin(30), it tells me that it is approximately 22.6.*0364

*The unit of measurement here is miles be we were told here that the measurement on the coast was given as 12 miles.*0388

*That tells us how long the A road is, 22.6 miles.*0398

*Now let us figure out how long the B road is, again we are using the law of sin.*0405

*Sin(B)/(b)=sin(C)/sin(c), I fill in what I know there.*0409

*I know that (B) is 40 degrees, I do not know (b), that is what I’m solving for, sin(C) is sin(30), (c) is (12).*0421

*I will cross multiply there, I get (12)sin(40)=(B)sin(30), I want to solve for (b), (b)=(12)sin(40)/sin(30).*0441

*I will plug that into my calculator.*0470

*Make sure that it is in degree mode and I get an approximate answer of 15.4 miles for (b).*0477

*Now I have the lengths of those two roads to the town in land.*0493

*Let us recap what we did there, we were graded with this long problem and lots of words.*0498

*The first thing to do is draw a picture and try to identify everything you are being given in the problem.*0502

*We have different points along the coast, we have an in land town, we measure the roads are 12 miles apart.*0509

*I filled that 12 miles into my distance along the coast, I filled in the two angles, and that was really all the problem gave me.*0517

*But that was enough to set up a triangle and to notice that I have (angle, side, angle).*0529

*Once I know that I have (angle, side, angle) I can the measure of the angles, make sure it is less than 180 and that tells me that it has a unique solution.*0535

*Now with (angle, side, angle) it is really not a good one to work with cos, certainly it does not work with SOHCAHTOA because we do not have a right angle.*0545

*Remember that SOHCAHTOA only works with right angles.*0554

*We are stuck using the law of sin, which actually works very well for (angle, side, angle).*0558

*First thing we had to do is find that third angle, the way we did that was by noticing that the angles add up to 180 degrees.*0564

*We used that to find the value of the third angle, that the value of angle C is 30 degrees.*0572

*We were able to plug that in to these incarnations of the law of sines, sin(A)/(a)=sin(C)/(c).*0578

*We plug in everything we know and the only thing missing is this (a).*0588

*We were able to solve this down and get (a) equal to 22.6 and we figured out that the unit of measurement is miles.*0594

*That is why we said that the answer was in term of miles.*0603

*We used the law of sines again, sin(B)/(b)=sin(C)/(c) and fill in everything we know, reduced it down, solve for (b) and give our answer in terms of miles.*0607

*That tells us the length of those two roads that finishes off this problem.*0620

*That also finishes the lecture on word problems and applications of triangle trigonometry.*0624

*Thanks for watching this trigonometry lectures on www.educator.com.*0631

*Hi, these are the trigonometry lectures for educator.com, and today we're going to look at some word problems and some applications of triangle trigonometry.*0000

*We're going to be using all the major formulas that we've learned in the previous lectures.*0009

*I hope you remember those very well.*0014

*The master formula which works for right triangles is SOH CAH TOA.*0016

*You can remember that as Some Old Horse Caught Another Horse Taking Oats Away.*0021

*Remember, that only works in a right triangle.*0025

*If you have an angle θ, that relates the sine, cosine and tangent of θ to the hypotenuse, the side opposite θ, and the side adjacent to θ.*0030

*You interpret that as the sin(θ) is equal to opposite over hypotenuse, the cos(θ) is equal to adjacent over hypotenuse, and the tan(θ) is equal to opposite over adjacent.*0043

*The law of sines works in any triangle.*0056

*Let me draw a generic triangle here, doesn't have to be a right triangle for the law of sines to work, so a, b, and c ...*0060

*Generally, you'd label the angles with capital letters, and label the sides with lowercase letters opposite the corner with the same letter.*0071

*That makes this a, this is b, and this is c.*0082

*The law of sines says that sin(A)/a=sin(B)/b=sin(C)/c.*0086

*That's the law of sines.*0106

*Law of cosines also works in any triangle.*0108

*Let me remind you what that one is.*0112

*We had a whole lecture on it earlier, but just to remind you quickly, it says that c ^{2}=a^{2}+b^{2}-2abcos(C).*0114

*That's useful when you know all three sides, you can figure out an angle very quickly using the law of cosines.*0129

*Or if you know two sides and the angle in between them, you can figure out that third side using the law of cosines.*0134

*Remember, law of cosines works in any triangle, doesn't have to be a right triangle.*0142

*It still works in right triangles.*0147

*Of course, in right triangles, if C is the right angle, then cos(C) is 0, so the law of cosines just boils down to the Pythagorean formula.*0149

*You can think of the law of cosines as kind of a generalization of the Pythagorean theorem to any triangle, doesn't have to be a right triangle anymore.*0158

*Finally, Heron's formula.*0166

*Heron's formula tells you the area of a triangle when you know the lengths of the three sides.*0173

*Heron says that the area is equal to the square root of s×(s-a)×(s-b)×(s-c).*0178

*The a, b, and c are the lengths of the sides, you're supposed to know what those are before you go into Heron's formula.*0193

*This s I need to explain is the semi-perimeter.*0199

*You add a, b, and c, you get the perimeter of the triangle, then you divide by 2 to get the semi-perimeter.*0205

*That tells you what the s is, then you can drop that into Heron's formula and find the area of the triangle.*0212

*We'll be using all of those, and sometimes it's a little tricky to interpret the words of a problem and figure out which formula you use.*0218

*The real crucial step there is as soon as you get the problem, you want to draw a picture of the triangle involved, and then see which formula works.*0225

*Let's try that out on a few examples and you'll get the hang of it.*0237

*The first example here is a telephone pole that casts a shadow 20 feet long.*0241

*We're told the sun's rays make a 60-degree angle with the ground.*0247

*We're asked how tall is the pole.*0250

*Let me draw that.*0252

*Here's the telephone pole, and we know it casts a shadow, and we know that that shadow is 20 feet long.*0254

*That's a right angle.*0262

*The reasons it casts a shadow is because of these rays coming from the sun.*0265

*There's the sun casting the shadow.*0274

*We want to figure out how tall is the pole.*0278

*We're told that the sun's rays make a 60-degree angle with the ground.*0281

*That means that angle right there is 60 degrees.*0284

*We want to solve for the height of the telephone pole, that's the quantity we want to solve for.*0288

*That's the side opposite the angle that we know.*0295

*We also are given the side adjacent to the angle we know.*0303

*I see opposite and adjacent, and I see a right angle.*0305

*I'm going to use SOH CAH TOA here.*0309

*I know the adjacent side, I'm looking for the opposite side.*0315

*It seems like I should use the tangent formula here.*0325

*Tan(60) is equal to opposite over adjacent.*0330

*Tan(60), 60 is one of those common values, that's π/3.*0339

*I know what the tan(60) is, I've got that memorized and hopefully you do too, square root of 3 is the tan(60).*0346

*If you didn't remember that, I at least hoped that you remember the sin(60) and the cos(60), that tan=sin/cos.*0353

*You can always work out the tangent if you don't remember exactly what the tan(60) is.*0364

*The sin(60) is root 3 over 2, the cos(60) is 1/2, that simplifies down to square root of 3.*0368

*That's what the tan(60) is.*0377

*The opposite, I don't know what that is, I'll just leave that as opposite, but the adjacent side I was given is 20.*0380

*I'll solve this opposite is equal to 20 square root of 3.*0387

*Since this is an applied problem, I'll convert that into a decimal, 20 square root of 3 is approximately 34.6.*0394

*The unit of measurement here is feet, I'll give my answers in terms of feet.*0411

*That tells me that the telephone pole is 34.6 feet tall.*0420

*Let's recap there.*0428

*We were given a word problem, I don't know at first exactly what it's talking about.*0430

*First thing I do is I draw a picture, so I drew a picture of my telephone pole, I drew a picture of the shadow then I noticed that's a right triangle, I could use SOH CAH TOA to solve it.*0433

*I tried to figure out which quantities do I know, which quantities do I not know.*0447

*I knew the adjacent side, I knew the angle, but I didn't know the opposite side and that seemed like it was going to work well with the tangent formula.*0451

*I plugged in what I knew, I solved it down using the common value that I knew, and I got the answer.*0461

*In the next one, we're trying to build a bridge across a lake but we can't figure out how wide the lake is, because we can't just walk across the lake to measure it.*0469

*It says that these engineers measure from a point on land that is 280 feet from one end of the bridge, 160 feet from the other.*0481

*The angle between these two lines is 80 degrees.*0490

*From that, we're supposed to figure out what the bridge will be, or how long the bridge will be.*0494

*Lots of words here, it's a little confusing when you first encounter this because there's just so many words here and there's no picture at well.*0498

*Obviously, the first thing we need to do is to draw a picture.*0504

*I had no idea what shape this lake is really but I'm just going to draw a picture like that.*0513

*I know that these engineers are trying to build a bridge across it.*0523

*Let's say that that's one end of the bridge right there and that's the other end.*0529

*They measure from a point on land that is 280 feet from one end of the bridge and 160 feet from the other.*0532

*It says the angle between these two lines measures 80 degrees.*0551

*If you look at this, what I have is a triangle, and more than that I have two sides and the included angle of a triangle.*0557

*I have a side angle side situation, and I know that that gives me a unique solution as long as my angle is less than 180 degrees.*0566

*Of course, 80 is less than 180 degrees.*0576

*I know I have a unique solution.*0580

*I'm trying to find the length of that third side.*0585

*If you have side angle side and you need the third side, that's definitely the law of cosines.*0590

*I'm going to label that third side little c, and call this capital C, label the other sides a and b.*0595

*Now, the law of cosines is my friend here, c ^{2}=a^{2}+b^{2}-2abcos(C).*0605

*We know everything there except for little c.*0623

*I'm just going to plug in the quantities that I know and reduce down and solve for little c.*0626

*Let me plug in c ^{2}, I don't know that yet, a^{2} is 160^{2}, plus b^{2} is 280^{2}, minus 2×160×280×cos(C), the angle is 80 degrees.*0634

*I don't know exactly what that is but I can find that on my calculator.*0658

*160 ^{2}=25600, 280^{2}=78400, 2×160×280, I worked that out as 89600, the cos(80), I'll do that on my calculator ...*0664

*Remember to put your calculator in degree mode if that's what you're using here.*0692

*A lot of people have their calculator set in radian mode, and then that gives you strange answers, because your calculator would be interpreting that as 80 radians.*0694

*It's very important to set your calculator to 80 degrees.*0704

*Cos(80)=0.174, let's see, 25600+78400=104000, 89600×0.174=15559, that's approximate of course, if we simplify that, we get 88441.*0710

*That's c ^{2}, I'll take the square root of that, c is approximately equal to 297.4.*0761

*Our unit of measurement here is feet, so I'll give my answer in terms of feet.*0778

*Let's recap what made that one work.*0791

*We're given this kind of long paragraph full of words and it's a little hard to discern what we're supposed to be doing.*0794

*First thing we see is, okay, it's a lake, so I drew just a random lake, it's a bridge across a lake, so I drew a picture.*0800

*That's really the key ideas to draw a picture.*0808

*I drew my bridge across the lake here.*0810

*That's the bridge right there.*0813

*It says we measure from a point that is 280 feet from one end of the bridge, and 160 feet from the other.*0817

*I drew that point and I filled in the 160 and the 280.*0825

*Then it gave me the angle between those two lines, so I filled that in.*0831

*All of a sudden, I've got a standard triangle problem, and moreover, I've got a triangle problem where I know two sides and the angle between them, and what I want to find is the third side of the triangle.*0835

*That's definitely a law of cosines problem.*0847

*I write down my law of cosines, I filled in all the quantities that I know then I simplified down and I solved for the answer on that.*0851

*We'll try another example here.*0862

*This time, a farmer measures the fences along the edges of a triangular field as 160, 240 and 380 feet.*0864

*The farmer wants to know what the area of the field is.*0873

*Just like all the others, I'll start out right away by drawing a picture.*0876

*It's a triangular field, my picture's probably not scaled, that doesn't really matter, 160, 240 and 380.*0885

*I have a triangle and I want to figure out what the area is.*0895

*If you have three sides of a triangle and you want to find the area that's pretty much a dead give-away that you want to use Heron's formula.*0900

*Let me remind you what Heron's formula is.*0907

*Heron's formula says the area is equal to the square root of s×(s-a)×(s-b)×(s-c).*0910

*The a, b, and c here just the lengths of the three sides but this s is the semi-perimeter of the triangle.*0929

*That's 1/2 of the perimeter a+b+c.*0939

*The perimeter is the distance around if you kind of walk all the way around this triangle.*0945

*That's (1/2)(160+240+300), 160+240=400, plus 300 is 700, 1/2 of that is 350, so s is 350.*0949

*I just drop the s and the three sides into Heron's formula.*0968

*That's 350×(350-160)×(350-240)×(350-300), (350-160)=190, (350-240)=110, and (350-300)=50.*0976

*I can pull 100 out of that immediately, so this is 100 times the square root of 35×19×11×5.*1014

*I'm going to go to my calculator for that, 35×19×11×5=36575, square root of that is 191, times 100 is 19.1, rounds to 25.*1026

*This is the area of a field, the units are squared here, and we were talking in terms of feet, so this must be square feet.*1060

*Let's recap how you approach that problem.*1081

*First of all, you read the words and right away you go to draw a picture, I see that I have a triangular field, the edges are 160, 240 and 300 feet.*1086

*I'm asked for the area of the field.*1099

*Now, once you have the three sides and you want the area, there's pretty much no question that you want to use Heron's formula on that.*1101

*That's the formula that tells you the area based on the three sides very quickly.*1109

*You write down Heron's formula, that's got a, b and c there.*1115

*It's also got this s, the s is the semi-perimeter.*1118

*You figure that out from the three sides.*1122

*You plug that into Heron's formula and you plug in the three sides.*1126

*The a, b and c are 160, 240 and 300.*1133

*Then, it's just a matter of simplifying down the numbers until you get an answer and figuring out that the units have to be square feet, because the original measurements in the problem were in terms of feet, and we're describing an area now, it must be square feet.*1137

*We'll try some more examples later.*1153

1 answer

Last reply by: Dr. William Murray

Tue Jan 7, 2014 11:55 AM

Post by Anwar Alasmari on January 1, 2014

Hello Doctor,

in the example III, why the result of the root of 350(190)(110)(50) is equals 100 times root of 35(19)(11)(5)?

3 answers

Last reply by: Dr. William Murray

Tue Aug 13, 2013 5:17 PM

Post by Taylor Wright on July 19, 2013

I thought that the Law of Sines doesn't work for AAA, because there would be infinite solutions.