CK-12 Trigonometry

by CK-12 Foundation

  Lesson Summary

  Scalar multiplication with vectors involves distributing the scalar to both coordinates of the vector. If the scalar is positive, the direction is the same. If the scalar is negative, the direction is opposite

  Component vectors are helpful because we position them at right angles with one another. This allows us to use trigonometric ratios and the Theorem of Pythagoras to solve problems.

  A vector has a horizontal and vertical component. If we know the magnitude and direction of the vector, we can find the horizontal and vertical components.

  In order to find the resultant of the sum of two vectors that are not perpendicular, we need to use the parallelogram method. This allows use to utilize the Law of Sines and the Law of Cosines to find the magnitude and direction of the resultant.

  Review Questions

  Find the resulting ordered pair that represents in each equation if you are given and and .

  Find the magnitude of the horizontal and vertical components of the following vectors given that the coordinates of their initial and terminal points.

  Find the magnitude of the horizontal and vertical components if the resultant vector’s magnitude and direction are given.

  Two forces of and act on an object at right angles. Find the magnitude of the resultant and the angle that it makes with the smaller force.

  Forces of and act on an object. The angle between the forces is . Find the magnitude of the resultant and the angle it makes with the larger force.

  Find the resultant and the direction made with vector if the magnitudes of vectors and and the angle between them is given. Make a rough sketch of the vectors, and then make a drawing of the resultant. Check your answers with the drawing to see if it makes sense.

  An incline ramp is long and forms an angle of with the ground. Find the horizontal and vertical components of the ramp.

  An airplane is traveling at a speed of . It needs to head in a direction of while there is a wind from . What should the airplane’s heading be?

  A speedboat is capable of traveling at , but is in a river that has a current of . In order to cross the river at right angle, in what direction should the boat be heading?

  If is any vector, what is ?

  Review Answers

  to

  to

  to

  to

  from the horizontal

  counterclockwise from the smaller force

  from the horizontal

  from the horizontal

  from the horizontal

  The airplane is traveling at at a heading of .

  against the current.

  Vocabulary

  component vectors

  Two or more vectors whose vector sum, the resultant, is the given vector. Components can be on axes or more generally in space.

  scalar

  A real number by which a vector can be multiplied. The magnitude of a vector is always a scalar.

  unit vector

  A vector that has a magnitude of one unit. These generally point on coordinate axes.

  Real-World Triangle Problem Solving

  Learning Objectives

  A student will be able to:

  Represent situations using right and oblique triangles and label the information given.

  Formulate a problem-solving plan to find the unknowns.

  Choose the appropriate tool to solve the problem from the Law of Sines, the Law of Cosines, the Theorem of Pythagoras, trigonometric ratios, vectors, and area formulas.

  Confirm your findings using another method than the one you originally chose.

  Introduction

  In this section, we will look at three different real-world applications. The situations are given below:

  A mountain climber is getting ready to scale a climbing wall that is high. If the angle of elevation from where he’s standing to the top of the wall is and the wall is perpendicular with the ground, how far way from the wall is he? How much rope will he need to get to the top if his partner will be standing in his current position?

  An engineer needs to solve a storage problem. The engineer must design a way to keep three long cylinders together in a stack, until they are used. One cylinder has a radius of , the second has a radius of , and the third has a radius of . The center of each cylinder has an axel projecting outwards. What is the length of a steel cable needed to hold the cylinders together? Consider both ends in your answer. What are the angles that the cable will make?

  Two tractors are being used to pull down the framework of an old building. One tractor is pulling on the frame with a force of and is headed directly north. The second tractor is pulling on the frame with a force of and is headed north of east. What is the magnitude of the resultant force on the building? What is the direction of the result force?

  Each of these situations is different and requires a different method for solving them. Up until this point, we have been learning many different tools to solve various types of problems. In this section, we will explore each of these problems, develop a problem-solving plan, find a solution, and check that solution to see if it is correct. We will be utilizing all of the tools we have learned in this chapter, as well as some tools learned in previous chapters.

  Represent Problem Situations as Triangle(s)

  Each of the situations above presents a unique problem. In order to form a plan for solving the problem, we must first know what information we have and what information we still need to find. Our first step in solving all of these problems will be to represent the situations using a triangle or multiple triangles. We must then label each triangle with the information we are given. Once we have a visual image of our problem, we can figure out what we need to find and which tool(s) would be most helpful in finding that information.

  First, let’s look at situation #1.

  In this problem, we are told that the wall and the ground are perpendicular, which means we have a right triangle. The climber is standing away from the ground and looking up at the wall with an angle of elevation of . This means that the bottom of the wall, the climber, and the top of the wall make a right triangle. Below is a sketch of the current situation.

  In this figure, we have labeled what we know and what we still need to find. In this case, represents the distance the climber is from the wall. represents the length of the rope from the climber to the top of the wall. We will refer back to this diagram later on.

  Now, let’s look at situation #2:

  In our second problem, we need to begin by drawing three circular shapes to represent our cylinders and the axels in the middle of them. We then need to label each cylinder and its radius.

  Once we have done that, we see that our three axels form a triangle. We don’t know that any of the axels are perpendicular and therefore cannot assume that they form a right triangle. We need to find the perimeter of the triangle which we will call and we need to find each of the angles, which we will refer to as , , and .

  Finally, we will look at situation #3:

  In our last problem, we have two tractors with a different force and direction. We will use arrows to represent each tractor’s force and direction.

  We are asked to find the resultant force and direction, which means we are dealing with vectors. In order to complete our diagram, we will need to connect our two vectors and draw in our resultant. We will refer to the magnitude of our resultant as and the direction of our resultant as .

  Now that we have diagrams or visual representations of each of our problems, we will begin to formulate a problem-solving plan for each situation.

  Make a Problem-Solving Plan

  Once we have our visual aid and an understanding of what we know and what we need to find, we can formulate a problem-solving plan. Often, we will not be able to directly solve for what is asked for. Instead, we will have to find other pieces of information first before we can find our unknown. When coming up with a problem-solving plan, we need to ask:

  What do I
know?

  What am I trying to find?

  What other information do I need before I can find what I’m looking for?

  We will look at the three situations discussed earlier and formulate a problem-solving plan for each one using the questions above.

  Situation #1:

  In this triangle, we know two angles (right angle and ) and one side ( wall). The side we know is opposite of our angle.

  We are trying to find the other two sides of the triangle. Side is adjacent to our angle and side is the hypotenuse of our triangle.

  In order to solve this problem, we have all of the information that we need.

  Situation #2:

  In this triangle, we know the radius of all three cylinders, which happen to form the sides of our triangle.

  We are triangle to find p, the perimeter of our triangle, and all three angles in our triangle (, , and ).

  In order to find any of the angles in the triangle, we first need to know the lengths of the sides.

  Situation #3:

  In this situation, we know the magnitude and direction of each of our vectors.

  We need to find the magnitude and direction of our resultant.

  In order to solve this problem, we have all of the information we need.

  Choose Among All Available Tools

  Now that we know what information we have and what we still need to find, we can choose the best tool(s) to use. Below, we will again look at the three situations from earlier. We will discuss our plan of action as to how to solve for the unknowns in the problem. We will decide which of our tools would be most effective in finding what we are looking for,

  Situation #1: Since this problem involves a right triangle, we can choose from using the Theorem of Pythagoras or trigonometric ratios. We can only use the Theorem of Pythagoras if we know two sides of triangle, which we don’t. This means we will need to use trigonometric ratios to find and .

  To find :

  Our wall is opposite our angle and is adjacent. This means we will need to use the tangent function to solve for .

  Answer: The climber is away from the wall.

  To find :

  The side representing the rope is the hypotenuse of our right triangle. If we use the climbing wall again as our opposite side, we will need to use the sine function to find the length of the rope.

  Answer: The climber will need of rope.

  Situation #2: Since we are dealing with an oblique triangle, we will not be able to use the Theorem of Pythagoras or trigonometric functions. Our first step will be to find the perimeter of the triangle. Once we know the perimeter, we will know all three sides of the triangle. Knowing all three sides will allow us to use the Law of Cosines to find one of the angles. Then, we can use the Law of Sines and the Triangle Sum Theorem to find the other two angles.

  To find the perimeter:

  All of the radii of our triangles meet. This means we can figure out the length of each side of the triangle by adding together the two radii that form each side.

  Side a:

  Side b:

  Side c:

  Answer: The length of the cables needs to be for one side. This means we need a total of of cable, one for each end.

  To find all three angles:

  Now that we know the lengths of all three sides of our triangle, we can use the Law of Cosines to find one of the angles in the triangle. We will begin by finding angle A because it is across from our largest side. This helps us to avoid the ambiguous case when we use the Law of Sines later on.

  Angle :

  Angle :

  Now to find angle , we will use the Law of Sines since it is much less computationally intense than the Law of Cosines. We will find angle first since it is the next largest angle.

  Angle :

  Since we now know two of our three angles, we can use the Triangle Sum Theorem to find our third angle. While we could use the Law of Sines or Law of Cosines again, the Triangle Sum Theorem is much quicker.

  Answer: Angle is , angle is , and angle is .

  Situation #3: This problem involves magnitude and direction, which means we will need to use vectors in order to solve it. When finding the resultant of two vectors, we can choose from either the triangle method or the parallelogram method. We will solve this problem using the parallelogram method.

  Looking at the diagram, we can see that the two vectors form an angle of , . This means that the angle opposite the angle formed by our two vectors is also . To find the other two angles in our parallelogram, we know that the sum of all the angles must add up to and that opposite angles must be congruent.

  Now, we can use two sides of our parallelogram and our resultant to form a triangle in which we know two sides and the included angle (SAS).

  This means that we can use the Law of Cosines to find the magnitude of the resultant.

  Answer: The magnitude of the resultant is . **There are four significant digits in the problem, so the answer should have only four digits

  To find the direction , we can use the Law of Sines since we now know an angle and the side opposite it.

  Now that we know , in order to find the angle of the resultant, we must add the from the axis to .

  Answer: The direction of the resultant is .

  Confirm with Alternate Methods

  Once we’ve solved our problem, we need to know if the answer we came up with is actually correct. In this section, we will look at ways to confirm our answer using different tools than what we used to solve each problem.

  Situation #1: In this situation we have found the twp missing sides in a right triangle. Now that we know all three sides, a simple way of checking our answer is to use the Theorem of Pythagoras.

  Situation #2: In order to double check that we found all three angles correctly, we can set up the Law of Sines for all three ratios and check to see if they are equal. If they are, we can assume that our angle measures are correct.

  Since all three ratios are equal, we can assume that our angle measures are correct.

  Another way to quickly check to see if our answer makes sense is to see if the largest angle is across from the largest side and the smallest angle is across from the smallest side. This doesn’t verify our answer, but it gives us a good idea as to whether or not we are on track.

  Situation #3: One way we can verify whether or not we found the correct magnitude for our resultant is to find the two component vectors that form it.

  Since our two component vectors form a right angle, we can use trigonometric ratios to find their lengths ( and ).

  Our horizontal component is . Next we will find our vertical component.

  Our vertical component is .

  Now that we know our horizontal component, our vertical component, and our resultant, we can use the Theorem of Pythagoras to verify our side lengths.

  Since our calculations using the Theorem of Pythagoras yield the same answer as the one we found for the magnitude of our resultant, we can assume our answer is correct.

  Points to Consider

  In the situations discussed in this section, are there alternate methods we could to verify our answers?

  Are there any situations where you might solve a problem and check your answer, but still get the problem wrong?

  Why might using the Law of Sines to check an answer be unreliable at times?

  In the above problems, are there other tools we could have used to initially solve the problem? If so, what are they?

  Lesson Summary

  We have many different methods to solve problems involving right and oblique triangles and vectors. These include: The Law of Sines

  The Law of Cosines

  The Theorem of Pythagoras

  Trigonometric rations

  Vectors

  It is important to begin solving a problem by drawing a diagram and labeling the given information.

  When creating a problem-solving plan, we must look at what we know, what we’re trying to find, and what othe
r information we need in order to find what we’re looking for.

  After we have a plan, we must choose the most appropriate tool given the type of triangle we have or what we are trying to find.

  Once we’ve arrived at an answer, we must check our work. We can do this by using another method than the one we used to find our answer.

  Review Questions

  For each question below, find a solution. When finding the solution, be sure to set up a diagram and label the known information.

  A soldier at a command post spots a helicopter that is high at an angle of elevation of . What is the horizontal distance from the command post to a point on the ground directly below the helicopter?

  A hiker is standing at the edge of a canyon, looking down at the base of the opposite canyon wall. The angle of depression from where he is standing to the base of the opposite canyon wall is . If he knows that the canyon wall on the opposite side is high, what is the distance across the canyon?

  Street A runs north and south and intersects with Street B, which runs east and west. Street C intersects both and , and it intersects Street A at a angle. There are stoplights at each of these intersections. If the distance between the two stoplights on Street C is , what is the distance between the two stoplights on Street A?