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Trigonometry : Solving Word Problems with Trigonometry
Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : solving word problems with trigonometry.
You can draw the following right triangle using the information given by the question:
Since you want to find the height of the platform, you will need to use tangent.
You can draw the following right triangle from the information given by the question.
In order to find the height of the flagpole, you will need to use tangent.
You can draw the following right triangle from the information given in the question:
In order to find out how far up the ladder goes, you will need to use sine.
Example Question #4 : Solving Word Problems With Trigonometry
In right triangle ABC, where angle A measures 90 degrees, side AB measures 15 and side AC measures 36, what is the length of side BC?
This triangle cannot exist.
Example Question #5 : Solving Word Problems With Trigonometry
A support wire is anchored 10 meters up from the base of a flagpole, and the wire makes a 25 o angle with the ground. How long is the wire, w? Round your answer to two decimal places.
To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25 o , the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w.
Now, we just need to solve for w using the information given in the diagram. We need to ask ourselves which parts of a triangle 10 and w are relative to our known angle of 25 o . 10 is opposite this angle, and w is the hypotenuse. Now, ask yourself which trig function(s) relate opposite and hypotenuse. There are two correct options: sine and cosecant. Using sine is probably the most common, but both options are detailed below.
We know that sine of a given angle is equal to the opposite divided by the hypotenuse, and cosecant of an angle is equal to the hypotenuse divided by the opposite (just the reciprocal of the sine function). Therefore:
To solve this problem instead using the cosecant function, we would get:
The reason that we got 23.7 here and 23.81 above is due to differences in rounding in the middle of the problem.
Example Question #6 : Solving Word Problems With Trigonometry
When the sun is 22 o above the horizon, how long is the shadow cast by a building that is 60 meters high?
To solve this problem, first set up a diagram that shows all of the info given in the problem.
Next, we need to interpret which side length corresponds to the shadow of the building, which is what the problem is asking us to find. Is it the hypotenuse, or the base of the triangle? Think about when you look at a shadow. When you see a shadow, you are seeing it on something else, like the ground, the sidewalk, or another object. We see the shadow on the ground, which corresponds to the base of our triangle, so that is what we'll be solving for. We'll call this base b.
Therefore the shadow cast by the building is 150 meters long.
If you got one of the incorrect answers, you may have used sine or cosine instead of tangent, or you may have used the tangent function but inverted the fraction (adjacent over opposite instead of opposite over adjacent.)
Example Question #7 : Solving Word Problems With Trigonometry
From the top of a lighthouse that sits 105 meters above the sea, the angle of depression of a boat is 19 o . How far from the boat is the top of the lighthouse?
To solve this problem, we need to create a diagram, but in order to create that diagram, we need to understand the vocabulary that is being used in this question. The following diagram clarifies the difference between an angle of depression (an angle that looks downward; relevant to our problem) and the angle of elevation (an angle that looks upward; relevant to other problems, but not this specific one.) Imagine that the top of the blue altitude line is the top of the lighthouse, the green line labelled GroundHorizon is sea level, and point B is where the boat is.
Merging together the given info and this diagram, we know that the angle of depression is 19 o and and the altitude (blue line) is 105 meters. While the blue line is drawn on the left hand side in the diagram, we can assume is it is the same as the right hand side. Next, we need to think of the trig function that relates the given angle, the given side, and the side we want to solve for. The altitude or blue line is opposite the known angle, and we want to find the distance between the boat (point B) and the top of the lighthouse. That means that we want to determine the length of the hypotenuse, or red line labelled SlantRange. The sine function relates opposite and hypotenuse, so we'll use that here. We get:
Example Question #8 : Solving Word Problems With Trigonometry
Angelina just got a new car, and she wants to ride it to the top of a mountain and visit a lookout point. If she drives 4000 meters along a road that is inclined 22 o to the horizontal, how high above her starting point is she when she arrives at the lookout?
As with other trig problems, begin with a sketch of a diagram of the given and sought after information.
Angelina and her car start at the bottom left of the diagram. The road she is driving on is the hypotenuse of our triangle, and the angle of the road relative to flat ground is 22 o . Because we want to find the change in height (also called elevation), we want to determine the difference between her ending and starting heights, which is labelled x in the diagram. Next, consider which trig function relates together an angle and the sides opposite and hypotenuse relative to it; the correct one is sine. Then, set up:
Therefore the change in height between Angelina's starting and ending points is 1480 meters.
Example Question #9 : Solving Word Problems With Trigonometry
Two buildings with flat roofs are 50 feet apart. The shorter building is 40 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 48 o . How high is the taller building?
To solve this problem, let's start by drawing a diagram of the two buildings, the distance in between them, and the angle between the tops of the two buildings. Then, label in the given lengths and angle.
Example Question #10 : Solving Word Problems With Trigonometry
Two buildings with flat roofs are 80 feet apart. The shorter building is 55 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 32 o . How high is the taller building?
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Home > Math Worksheets > Word Problems > Trigonometry
These types of problems focus on the lengths of sides and angles within a triangle. After all that is what trigonometry is all about. Start to approach these problems by focusing on understanding the figures that are involved and model this by drawing diagrams. On the diagram highlight all the information that we know about each stage of the diagram. Make sure that if a unit of measure is attached to that information that you show that as well. We then start to analyze the system and identify if Pythagorean Theorem comes into play or if we can apply various trigonometric functions to determine the angle of elevation or depression.
In these worksheets, your students will solve word problems involving trigonometry. Diagrams are included to accompany some problems. These are moderately complex problems and a sound understanding of trigonometry is required in order for students to be successful with these worksheets. There are six worksheets in this set. This set of worksheets contains lessons, step-by-step solutions to sample problems, and both simple and more complex problems. It also includes ample worksheets for students to practice independently. Students may require extra paper on which to do their calculations. Most worksheets contain between eight and eleven problems. When finished with this set of worksheets, students will be able to solve word problems that involve trigonometry. These worksheets explain how to solve word problems by using trigonometry. Sample problems are solved and practice problems are provided.
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Trigonometry word problems worksheets, click the buttons to print each worksheet and answer key., trigonometric word problem lesson.
This worksheet explains how to solve word problems by using trigonometry. A sample problem is solved, and two practice problems are provided.
You are stationed at a radar base and you observe an unidentified plane at an altitude h = 2000 m flying towards your radar base at an angle of elevation = 30o. After exactly one minute, your radar sweep reveals that the plane is now at an angle of elevation = 60o maintaining the same altitude. What is the speed (in m/s) of the plane?
A fisherman drops a boat anchor and pays out 93 meter of anchor rope. The rope makes an angle of depression of 59.5 degrees with the horizontal. Assuming a level bottom, how deep in the water?
Review and Practice
What is the measure of the diagonal of the rectangle ABHG ?
A tree in a back yard is to be cut down. The base of the tree from the house is 23 meters away, and the angle of elevation from the house to the top of the tree is 45.5 degrees. Could the tree hit the house when it is cut down?
An aerial 30 meter tail is to be supported by 2 guy wires each 50 meter from its base. How long will each guy wire be?
Trigonometric Word Problems Worksheets
How to Solve Trigonometric Word Problems - Students often get to solving the Trigonometry Problems stage when they are in the ninth grade, and it can get tricky at times. In this topic, we will be covering a general or basic idea regarding Solving Trigonometry Problems along with some useful tips. In mathematics, it is essential to understand how you understand something rather than memorizing the steps. Trigonometry is the study of triangles. Let’s discuss some of the tips. 1. The first step involves remembering the formulas and definitions. Unless and until you are familiar with the identities and the background information of a trigonometric problem, till then, you cannot get better at Solving Trigonometry Problems. 2. The second tip is practice. The real reason why most students struggle with solving trigonometric problems is because of a lack of practice. Learning the formulas is the easier part; the bigger challenge is to maintain the continuous practice of every single formula and learning variations of problems. 3. Practice your way into difficulty. If you are getting too comfortable with a particular level of difficulty, then it is recommended you increase the level and do more difficult ones.
Guides students solving various trigonometry based word problems. What is formula to find x?
Demonstrates how to solve complex absolute value problems. A point on the ground 125 feet from the foot of a tree, the angle of elevation of the top of the tree is 32 degrees. Find to the nearest foot, the height of the tree.
Independent Practice 1
A really great activity for allowing students to understand the concept of trigonometric word problems. A ladder leans on a building. The foot of ladder is 6 feet from the building and reaches a height of 14 feet on the building. Find the angle the ladder makes with the ground?
Independent Practice 2
Students find the final value of trigonometric word problems in assorted problems. The answers can be found below. Calculate the height of a tree, the angle of elevation from a point A is 70 degrees. The distance is 18 m from the base of the tree. How high is the tree?
Students are provided with problems to achieve the concepts of trigonometric word problems.
This tests the students ability to evaluate trigonometric word problems. Example: A kite is 33 m above the ground. The kite string makes an angle of 38 degrees with the ground. Assuming that the string is taut, how much string is out?
Homework and Quiz Answer Key
Answers for the homework and quiz.
Answers for the lesson and practice sheets.
Here's a great excuse to try the next time you forget your trigonometry homework: " I locked the paper in my trunk but a four-dimensional dog got in and ate it."
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Home » Mathematics » Pre Calculus » Word Problems and Applications of Trigonometry
Word Problems and Applications of Trigonometry
Table of contents, pre calculus word problems and applications of trigonometry.
Section 8: Applications of Trigonometry: Lecture 5 | 34:25 min
In these word problems and applications of trigonometry, you'll use the knowledge of the previous lectures: The SOHCAHTOA, The Law of Sines, The Law of Cosines, and Heron's Formula. Remember that the SOHCAHTOA is used in right triangles only, while the other rules can be used in any triangle. In this lecture, you'll learn how to find the height of a telephone pole if you are given the length of its shadow and the angle that the sun's rays make with the ground. In other examples, you'll work with bridge length, roads to a town and some other real life situations.
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Working on the solution...
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 :
Example 2 :
Example 3 :
Example 4 :
Example 5 :
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.
Download All Slides
- Formulas to Remember 0:11
- 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
Precalculus with Limits Online Course
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
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Unit 1: Right triangles & trigonometry
About this unit, ratios in right triangles.
- Getting ready for right triangles and trigonometry (Opens a modal)
- Hypotenuse, opposite, and adjacent (Opens a modal)
- Side ratios in right triangles as a function of the angles (Opens a modal)
- Using similarity to estimate ratio between side lengths (Opens a modal)
- Using right triangle ratios to approximate angle measure (Opens a modal)
- Right triangles & trigonometry: FAQ (Opens a modal)
- Use ratios in right triangles Get 3 of 4 questions to level up!
Introduction to the trigonometric ratios
- Triangle similarity & the trigonometric ratios (Opens a modal)
- Trigonometric ratios in right triangles (Opens a modal)
- Trigonometric ratios in right triangles Get 3 of 4 questions to level up!
Solving for a side in a right triangle using the trigonometric ratios
- Solving for a side in right triangles with trigonometry (Opens a modal)
- Solve for a side in right triangles Get 3 of 4 questions to level up!
Solving for an angle in a right triangle using the trigonometric ratios
- Intro to inverse trig functions (Opens a modal)
- Solve for an angle in right triangles Get 3 of 4 questions to level up!
Sine and cosine of complementary angles
- Intro to the Pythagorean trig identity (Opens a modal)
- Sine & cosine of complementary angles (Opens a modal)
- Using complementary angles (Opens a modal)
- Trig word problem: complementary angles (Opens a modal)
- Trig challenge problem: trig values & side ratios (Opens a modal)
- Trig ratios of special triangles (Opens a modal)
- Relate ratios in right triangles Get 3 of 4 questions to level up!
Modeling with right triangles
- Right triangle word problem (Opens a modal)
- Angles of elevation and depression (Opens a modal)
- Right triangle trigonometry review (Opens a modal)
- Right triangle trigonometry word problems Get 3 of 4 questions to level up!
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TRIGONOMETRY WORD PROBLEMS WITH SOLUTIONS
Problem 1 :
The angle of elevation of the top of the building at a distance of 50 m from its foot on a horizontal plane is found to be 60 °. Find the height of the building.
Draw a sketch.
Here, AB represents height of the building, BC represents distance of the building from the point of observation.
In the right triangle ABC, the side which is opposite to the angle 60° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and the remaining side is called adjacent side (BC).
Now we need to find the length of the side AB.
tanθ = Opposite side/Adjacent side
tan60° = AB/BC
√3 x 50 = AB
Approximate value of √3 is 1.732
AB = 50 (1.732)
AB = 86.6 m
So, the height of the building is 86.6 m.
Problem 2 :
A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the ladder is inclined at an angle of 60 ° . Find how far the ladder is from the foot of the wall.
Here AB represents height of the wall, BC represents the distance between the wall and the foot of the ladder and AC represents the length of the ladder.
In the right triangle ABC, the side which is opposite to angle 60° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).
Now, we need to find the distance between foot of the ladder and the wall. That is, we have to find the length of BC.
tanθ = opposite side/adjacent side
BC = (6/√3) x (√3/√3)
BC = (6√3)/3
Approximate value of √3 is 1.732.
BC = 2 (1.732)
BC = 3.464 m
So, the distance between foot of the ladder and the wall is 3.464 m.
Problem 3 :
A string of a kite is 100 meters long and t he inclination of the string with the ground is 60°. Find the height of the kite, assuming that there is no slack in the string.
Here AB represents height of kite from the ground, BC represents the distance of kite from the point of observation.
In the right triangle ABC the side which is opposite to angle 60° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).
Now we need to find the height of the side AB.
sinθ = opposite side/hypotenuse side
sinθ = AB/AC
sin60° = AB/100
√3/2 = AB/100
(√3/2) x 100 = AB
AB = 50√3 m
So, the height of kite from the ground 50√3 m.
Problem 4 :
From the top of the tower 30 m height a man is observing the base of a tree at an angle of depression measuring 30 ° . Find the distance between the tree and the tower.
Here AB represents height of the tower, BC represents the distance between foot of the tower and the foot of the tree.
Now we need to find the distance between foot of the tower and the foot of the tree (BC).
tan30° = AB/BC
1/√3 = 30/BC
BC = 30(1.732)
BC = 51.96 m
So, the distance between the tree and the tower is 51.96 m.
Problem 5 :
A man wants to determine the height of a light house. He measured the angle at A and found that tan A = 3/4. What is the height of the light house if A is 40m from the base?
Here BC represents height of the light house, AB represents the distance between the light house from the point of observation.
In the right triangle ABC the side which is opposite to the angle A is known as opposite side (BC), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (AB).
Now we need to find the height of the light house (BC).
tanA = opposite side/adjacent side
tanA = BC/AB
Given : tanA = 3/4.
3/4 = BC/40
Multiply each side by 40.
So, the height of the light house is 30 m.
Problem 6 :
A ladder is leaning against a vertical wall makes an angle of 20° with the ground. The foot of the ladder is 3 m from the wall. Find the length of ladder.
Here AB represents height of the wall, BC represents the distance of the wall from the foot of the ladder.
In the right triangle ABC, the side which is opposite to the angle 20° is known as opposite side (AB),the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).
Now we need to find the length of the ladder (AC).
cosθ = adjacent side/hypotenuse side
Cosθ = BC/AC
Cos 20° = 3/AC
0.9397 = 3/AC
AC = 3/0.9396
So, the length of the ladder is about 3.193 m.
Problem 7 :
A kite is flying at a height of 65 m attached to a string. If the inclination of the string with the ground is 31°, find the length of string.
Here AB represents height of the kite. In the right triangle ABC the side which is opposite to angle 31° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and the remaining side is called adjacent side (BC).
Now we need to find the length of the string AC.
sin31° = AB/AC
0.5150 = 65/AC
AC = 65/0.5150
AC = 126.2 m
Hence, the length of the string is 126.2 m.
Problem 8 :
The length of a string between a kite and a point on the ground is 90 m. If the string makes an angle θ with the ground level such that tan θ = 15/8, how high will the kite be ?
Here AB represents height of the balloon from the ground. In the right triangle ABC the side which is opposite to angle θ is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).
tanθ = 15/8 ----> cotθ = 8/15
cscθ = √(1+ cot 2 θ)
cscθ = √(1 + 64/225)
cscθ = √(225 + 64)/225
cscθ = √289/225
cscθ = 17/15 ----> sinθ = 15/17
But, sinθ = opposite side/hypotenuse side = AB/AC.
AB/AC = 15/17
AB/90 = 15/17
So, the height of the tower is 79.41 m.
Problem 9 :
An airplane is observed to be approaching a point that is at a distance of 12 km from the point of observation and makes an angle of elevation of 50°. Find the height of the airplane above the ground.
Here AB represents height of the airplane from the ground. In the right triangle ABC the side which is opposite to angle 50° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse side (AC) and remaining side is called adjacent side (BC).
From the figure given above, AB stands for the height of the airplane above the ground.
sin50° = AB/AC
0.7660 = h/12
0.7660 x 12 = h
So, the height of the airplane above the ground is 9.192 km.
Problem 10 :
A balloon is connected to a meteorological station by a cable of length 200 m inclined at 60 ° angle with the ground. Find the height of the balloon from the ground. (Imagine that there is no slack in the cable)
Here AB represents height of the balloon from the ground. In the right triangle ABC the side which is opposite to angle 60° is known as opposite side (AB), the side which is opposite to 90° is called hypotenuse (AC) and the remaining side is called as adjacent side (BC).
From the figure given above, AB stands for the height of the balloon above the ground.
sin60° = AB/200
√3/2 = AB/200
AB = (√3/2) x 200
AB = 100(1.732)
AB = 173.2 m
So, the height of the balloon from the ground is 173.2 m.
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