Trigonometric applications outside the classroom

Purpose

This unit follows on from others that introduce students to sine, cosine and tangent. In this unit students will gain practical experience in using trigonometry in practical situations; mainly outside the classroom.

Achievement Objectives
GM5-10: Apply trigonometric ratios and Pythagoras' theorem in two dimensions.
Specific Learning Outcomes
  • describe and demonstrate how trigonometry can be used to find the height of a tall building or tree
  • describe and demonstrate how trigonometry can be used to find the height of a high hill, or other high object where one cannot stand directly beneath the highest part
  • describe in broad terms how trigonometry might be used to find the distance between the earth and the moon
Description of Mathematics

This unit is a ‘hands on’ unit. The students are not learning new mathematical concepts, but they are learning to use concepts they have already come across, but now, in practical situations. The aim of the unit is to both convince the students that trigonometry can be used for practical purposes, and consolidate the understandings they already have.

Required Resource Materials
  • Clear plastic protractors
  • Clear plastic rulers
  • Clear tape
  • Cotton thread
  • Materials for constructing one clinometer for each group of three students
  • Small weights
  • Drill and small bit
  • Long tape measure
  • World atlas
Activity

Session 1

In this session the students are grouped in threes. Each group constructs a clinometer, uses it to measure angles, and prepares for a practical application of trigonometry to be undertaken in Session 2.

  1. Issue each group with a clear plastic ruler, a clear plastic protractor, clear tape, cotton and a small weight. Before the session it is suggested that the teacher drill a small hole in the centre of the protractors large enough to pass the cotton through. Each group makes a clinometer. See diagram below.

clinometer.

  1. The next task is to gain some facility in using it to measure angles. Explain how it is done by using the diagram below.

diagram.

  1. The clinometer allows the students to measure the angle ABC. They are then able to calculate the angle BAC. The students can then be sent to a few places around the school to practise measuring such angles. Usually it is best for one person to hold the clinometer in such a way that his or her eye looks along the ruler to the top of the object concerned, and a team member reads off the angle. Students should take turns doing this. The aim, here, is to become proficient at taking angles so that this is not an issue during Session 2. For each measurement, the students should sketch a diagram like Diagram 1.2 and calculate the angle BAC.

  2. The above may be completed before the session has concluded. If this happens, the time could be spent usefully revising some of the material from trigonometry units studied earlier.

Session 2

In this session the students learn how to measure the height of one or more suitable objects, carry out the activity, and write up a poster explaining how this procedure is carried out.

  1. Each student will need to measure the distance between his or her feet and eyes. Imagine the tall object to be measured is a tree. The diagram below indicates how this is done.

diagram.

 

  1. Students measure the distance d from the base of the tree. They use the clinometer to find the angle θ. Using tan they can find h (because h = d x tan(θ)). And, of course, they need to add on the distance between their feet and eyes at the end.

  2. Before you send the groups out, go through two practice examples using imaginary figures. Once you are sure the groups understand the process get each group to take the necessary measurements and calculate the heights concerned. Each group should make tree measurements of the angle concerned (each at a different distance); one for each member of the group.

  3. On their return to class, lest the activity be quickly forgotten, get each student to make a small poster that fully describes how a clinometer is made and how it was used for making the relevant measurements just completed.

Session 3

What if you cannot get to the base of the tall object to be measured? What if the object was, not a tree, but a hill? Or a tree surrounded by a lake? In this session we measure slightly more inaccessible objects.

  1. The method here is slightly more complicated, but trigonometry can still enable us to find such a height. See diagram below.

diagram.

  1. The method here involves taking a reading from the point A, then moving a fixed distance towards the hill to a new point B—in this case exactly 100m—and then taking another reading. Perhaps surprisingly, by using a combination of algebra and trigonometry, we can find the height of the hill.

  2. Go through the example in the diagram above. The algebra is a bit complicated. Work to 3 decimal places accuracy. Don’t pay attention to rounding errors because this will only confuse matters unnecessarily. The aim is for the students to understand the method. Greater accuracy can be used later once the method is understood.

  3. Let BC = k, and CD = h. Thus AC = 100 + k. Using triangle BCD, h = k x tan(46), ie h = 1.036k. Call this equation 1. Using triangle ACD, h = (100 + k) x tan(42), ie h = 0.900(100 + k), ie h = 90 + 0.900k. Call this equation 2.
    Equations 1 and 2 each have unknowns h and k. Our aim is to find h.

  4. At this point it will be helpful to the students to look at some simpler simultaneous equations of the same sort. For example: h = 2k and h = 10 + 0.5k. In this case h can be found as follows. If h = 2k then k = 0.5h. Substitute this in the second equation gives: h = 10 + 0.25h. Thus 0.75h = 10, and h = 40/3. Give the students more examples of this sort so that they become familiar with the algebra needed. Once they are comfortable, return to the main problem.

  5. We have equation 1:  h = 1.036k. And equation 2: h = 90 + 0.900k. Rearranging equation 1 gives: k = 1/1.036 x h, ie k = 0.965h. Substitute this in equation 2. Thus h = 90 + 0.900 x 0.965h. Ie, h = 90 + 0.869h. Ie, h – 0.869h = 90. Ie, h(1 – 0.869) = 90. Ie, h = 90/0.132. Ie, h = 682m. Of course the student height needs to be added on at the end.

Session 4

In this session the students measure the height of a nearby hill in the way outlined in Session 3. Once this is done each student prepares a poster that details the method and the calculations done in the practicum.

Session 5

In this session the students are given an insight into how trigonometry might be used, under the appropriate circumstances, to measure the height of the moon. But first the students need to be reminded what ‘latitude’ means, and what the moon rising or setting over the horizon means.

  1. Get the students to draw a circle representing the earth. Get them to mark a point P on the earth and imagine they are standing at this point. Now they draw a tangent through that point. See diagram below.

  1. The tangent is basically the horizon at P. When the moon is above this line it is above the horizon and visible from P. When it crosses the line it is either rising or setting. When it is below the line it is not visible from P.

  2. Next revise the meaning of latitude. See diagram below.

diagram.

  1. Get the student to find, using an atlas, four countries at a latitude of 20 degrees, two in the southern hemisphere and two in the northern. If you are at a latitude 60 degrees, where might you be? What is the weather likely to be at a latitude 0 degrees? Why? If you are at a latitude of 90 degrees, where are you?

  2. Now for the problem. See diagram below.

diagram.

 

  1. A friend of yours is at the equator, at E. You are at P, at latitude 89 degrees. Your friend and you are in phone contact. Accordingly you are able to establish that, just as the moon is directly overhead for your friend, it is just setting for you. You know that the radius of the earth is approximately 6000km. What is the distance, d, from the earth to the moon?

  2. Students make their own calculation.
    (The answer is: d x cos(89) = 6000. Thus, d = 343, 792km.)

  3. Here are some relevant facts: The earth radius is more correctly 6378km. The mean distance between the earth and the moon is 384,400km. The radius of the moon is 1738km. Its mass is 0.012 x mass of earth. The moon’s mean velocity in orbit is 1km/sec. Find other interesting relevant facts from the atlas.

  4. Students complete this session by making a poster explaining what it means when the moon passes over the horizon, what latitude means with some examples, how the calculation was made, and some interesting facts associated with this problem.


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