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Using Trigonometry

GK-12 Lesson/Unit Template

Title:  Using Trig.                       

Submitted by:        Brenda Rios and Jenna Pike                                             

Short description:   The students are presented with real life situations where trigonometry is used.

Duration of lesson(s): 1 lesson 

Grade level(s) and/or target gro          up(s):  Students studying trigonometry

Subject(s):  Trigonometry                                     

Technologies used:  When these problems were presented, a smart board was used to display the diagrams as well as make notations as we progressed working on the project.

 

 

 

 

 

 

Objectives:   Student will be able to use trigonometry and other knowledge of mathematics to solve application problems.

 

Key Questions/Driving Questions:  How can trigonometry be used?

Prerequisites and Sequence of Lessons in Unit: 

Students should be familiar with calculating the sine, cosine, and tangent of a given angle.  They should also be able to calculate the missing side of a right triangle given a side and an angle.

         

Lesson Introduction: 

Do-now :

Students are to find the length of a missing side of a right triangle given an angle and a side. 

And

Students are to find the measure of an angle given two sides of a right triangle.  ( inverse trig functions)

Lesson Core:  Students apply the trigonometric functions in real world problems.  The problems are worked on by the students in groups while teacher facilitates the inquiry of questions. 

See attached :  Three Examples of Practical Applications of Trigonometry

Lesson Closure:  Selected or volunteer students are to present their detailed explanations of the problems.

Three Examples of Practical Applications of Trigonometry

 

1. Staircase construction

Jorge’s studio has a steep staircase made out of a ladder that he wants to replace with a real staircase.  It needs to be less steep and has to leave room for the lower bedroom door to open.  Using trigonometry, how can you design a better staircase?

 

Figure 1.  Schematic diagram of the original staircase and bedroom area.

 

The length of the wall is 25 feet.  The height of the original staircase is 12 feet.  The doors are 2.75 feet wide.  The lower bedroom door is 10 feet from the front edge of the bedroom area.

One solution is to build a landing outside the upper bedroom and then extend the staircase, as in Figure 2.

Figure 2. Proposed position of new staircase.

 

 

What is the length of the new staircase?

 

What is the height and width of each new stair?

 

What is the difference between the angle made by the new staircase and the old staircase?


2. Sarah owns a piece of property that she needs to know the square feet of assessment purposes.  The figure that the city surveyors came up with is 7012.5 ft2, and she thinks it is wrong, and will raise her properties taxes unnecessarily.  They didn’t subtract the circular portion of the cul-de-sac that the city owns.  Her lot is 55 ft wide. One side is 108.96 feet; the other side is 146.04 feet.  There is a circular area removed from the property, with an arc length of 78.21 feet and radius 40 feet.

Figure 3. Schematic diagram of Sarah’s property.

The length of the arc from A to B (78.21 feet) is given by

78.21 = R θ

R is the radius of the circle (40 feet) and θ is the angle measurement in radians.


Now find the length of the line AB (the base of the triangle ACB).  A diagram helps.  In the new diagram, we create a new point D midway between A and B.

                   D

          C        B

                                              (33.1669 feet)

2BD = AB                                            (66.3338 feet)

The height of the triangle ACB is

                                                          (22.3957 feet)

Thus, the area of the triangle ACB is            

                                                               

The area of the sector of the circle bounded by the lines of AC and CB, and the arc from A to B is

                                                    square feet

So the area of the crescent shaped region bounded by the line AB and the arc of the circle is

1564.2 - 741.603 = 822.597 square feet.

Hence the area of the lot is

   square feet

3. X-ray diffraction in materials science

Materials scientists use trigonometry to learn specific details about the materials they are studying.  We use a technique called x-ray diffraction, which shoots x-rays at a material, then we study the ways that the x-rays bounce back out of the sample.  Each material has it’s own pattern, like a fingerprint.


            This is an example of an x-ray diffraction pattern.  Different materials have peaks at different positions on the x-axis because of the positions of the atoms within the material.  We use Bragg’s Law to understand this property.

Bragg’s Law: 

If we know that λ = 1.54, we can find d that corresponds to each peak.  There are major peaks at 44°, 50.5°, and 74.5°.  Find d for each of these peaks.

What does this mean?  This means that there are rows of atoms d distance apart from each other in this material.

 

 

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