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Could you advise me where to find real- life or word problems using trigonometry, involving solving triangles, suitable for high-school students? Most of the problems i found were pseudo-real problems with simplified information and pre-drawn models. I need problems with more irrelevant information, which could actually helps students with mathematical modelling competence.

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    $\begingroup$ Welcome to the site! Could you clarify what you mean by "solving triangles"? Do you mean computing certain information about a triangle when other information is given? $\endgroup$ Commented Jul 7 at 13:01
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    $\begingroup$ @JochenGlueck yes, the purpose is to find the missing sides and angles of the triangles, but it will be put in a real context, triangles are just models that students should draw themselves in order to solve the real-life problems. $\endgroup$
    – Linda
    Commented Jul 7 at 16:36
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    $\begingroup$ Measuring the height of a tree you are about to cut down: Tree Height Measurement. $\endgroup$ Commented Jul 8 at 0:41
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    $\begingroup$ Some land surveying perhaps? They made maps of entire countries using sticks and trigonometry in the 1700s which you could overlap with Google Earth and have a near perfect match. maps.arcanum.com/hu/map/firstsurvey-hungary $\endgroup$
    – vsz
    Commented Jul 8 at 5:11
  • $\begingroup$ Perhaps computing time dilation using Einstein's "mirror on a train" gedankenexperiment (e.g. phys.libretexts.org/Bookshelves/Modern_Physics/…). Not sure it fits "real life ;-). $\endgroup$ Commented 2 days ago

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There are many real-life problems involving solving triangles in "Trigonometry for the Practical Man" by James Edgar Thompson. Here are some from Chapter 5:

Problem 1

From the top of a mountain three miles above sea level, the angle of depression of the ocean horizon is found to be 2° 13' 50". Find the radius of the earth.

Answer: 3960 miles

Problem 2

An observer sights a telescope on the sun, then turns the telescope straight up to the zenith, measuring an angle of 30° 0' 13.2". At this same moment (by previous agreement), another observer 8280 miles away reports the sun is just setting. Find the distance from the center of the earth to the center of the sun. (You will need the earth’s radius found in problem 1.)

Answer: 92,800,000 miles

Problem 3

From a point on the surface of the earth, the sun is sighted with a transit, first on one edge and then on the other, and the angle between the lines of sight is found to be 32' 4". Find the radius of the sun. (You will need the answers to problems 1 and 2.)

Answer: 433,000 miles

Problem 4

On January 1, an astronomer measures the sun-earth-star angle of the star Proxima Centauri, and finds it to be 95° 27' 45", and six months later measures the angle and finds it has changed to 84° 32' 13.6". Find the distance from the earth to the star on January 1. (You will need to use the earth-sun distance found in problem 2.)

Answer: 4.63 light years

Problem 6 of Chapter 5: On a level plain at some distance from the mountain, the elevation of the summit was read and found to be 30 degrees, and at a point one mile farther from the mountain the elevation was again read and found to be 25 degrees. What is the height of the mountain? Answer: 12,790 feet.

There are also very nice surveying problems in this book.

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    $\begingroup$ I'm left wondering how they knew the height of the mountain in problem 1. $\endgroup$ Commented Jul 7 at 21:03
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    $\begingroup$ @DanielR.Collins More trigonometry I imagine? e.g. Given two observers at equal altitude, diametrically opposite the peak of the mountain, measure the angle of inclination of the peak, and the distance as the crow flies between the two observers. If calculating that distance is hard, I think you can also do it with 3 nearby observers on the same side of the mountain, where the ground might be flat. Or even simpler, look at the mountain from a long (mostly known) distance away and measure a nearby height with the same angle of inclination. $\endgroup$ Commented Jul 7 at 22:13
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    $\begingroup$ @DanielR.Collins Previous surveys? Maybe there was a nice marker already there. Or they threw the most useless team member off and timed his fall. Either way works. $\endgroup$
    – Thierry
    Commented Jul 7 at 22:57
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    $\begingroup$ @JochenGlueck: Altitude is insignificant at the scale this problem is operating on. Longitude is not required in order to draw a quadrilateral whose vertices are the two observers, the center of the sun, and the center of the Earth, and then use trigonometry to solve for the missing pieces with the angles we are given. Note that the distance between observers is too great to be a chord of the Earth, so it must instead represent an arc length over the surface of the Earth, and therefore we have to do more trigonometry to get the straight line distance between them. $\endgroup$
    – Kevin
    Commented 2 days ago
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    $\begingroup$ @Kevin: To rephrase my issue without referring to longitude and latitude: If we choose the coordinate system such that the center of the earth is located at the origin, the center of the sun is located on the (positive) $x_1$-axis and the second observer is located on the $x_3$-axis, then the position of the first observer is located on a (semi-)circle that is aligned in parallel to the $x_1x_2$-plane. Depending on the first observer's position within this semi-circle, the measured angle will give a different value for the distance to the sun. So we need to know one more parameter. $\endgroup$ Commented 2 days ago
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I’ve got one very specific answer to this question, because I had to dust off my old trigonometry knowledge just yesterday. I tell it as a real-world use of what I had to learn in school out of left field.

Assume you want to sew a quilt with Ohio stars. They are composed of squares, some built with four triangles that meet in the middle.

You end up with a lot of triangles that need to be sewn together. You know the final size of the (isosceles right) triangles. Let’s say you want the hypotenuse to have 10 inch (resulting in 10x10 inch squares on the finished quilt).

But you need to account for the seam allowance! From each of the three sides of the triangle you need to expand by half an inch.

How large do the triangles need to be cut, including the seam allowance?

This is a surprisingly tricky question. Making the hypotenuse simply 11 inches large is way too little. The result involves the geometric mean theorem and a fair share of Pythagoras and square roots of 2.

I can give the details if wanted. It took me (physicist by trade) almost a quarter hour to get to the correct result, but you need only high school maths for it, and it’s quite straight-forward as soon as you focus on the triangle heights.

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Triangulation (Wikipedia link) has a wide variety of real-world uses. As a wildlife ecologist, a particularly fun one I can think of is wildlife radio telemetry (Wikipedia link). In this case, a radio transmitter is attached to an animal, and then people go out with receivers and antennas to get bearings (a point in space and vector direction pointing towards the animal). With two bearings, you have all the data needed to formulate a triangle where the animal's location is one of the corners of the triangle. Add in a third bearing, and the animal's location becomes a triangular area (ideally, all three bearings would intersect at a single point, but in practice, there's usually error in the bearings, which is why extra bearings are useful).

enter image description here

Trigonometry is also used in trajectory/path analysis. In ecology/conservation, this is used to model animal movement. By tracking the position of an individual animal over time, you can connect all the points in order, calculate the "turning angles" at each point, and use those angles to make statistical inferences about the animal's movement state. For example, migration might be represented by a long string of points with a sequence of small turning angles, while feeding might appear as a cluster of points with a sequence of large turning angles. Here's a very crude example figure of two paths and a rose diagram of the turning angles:

enter image description here

While both of these examples might seem simple from the trigonometric perspective, what I'm describing is just the basic premises for more advanced mathematical analyses that are important for conservation decision-making.

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Squares and hexagons can be decomposed into triangles. This opens up a lot of practical problems that I encounter, such as:

  1. What is the largest hexagon that can fit into a circle? (you just need the radius of the circle)
  2. What is the largest circle that can fit in hexagon?
  3. What is the largest square that can fit into a circle?
  4. What is the largest circle that can fit into a square?
  5. If you are given the size of a hexagon from "flats to flats", what is the length of one of the edge surfaces?
  6. Given a circle with radius "r", what is the size of a triangle/hexagon/square that has the same area?

You can re-do these problems, but instead of using a square, assume a rectangle with a width equal to twice its height.

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A caveat: I work in computer vision, so my perspective my be a little biased by a desire to "evangelize" on a currently hot topic.

Students these days spend a lot of spare time looking at pictures captured with cellphones. So they may be stimulated by simple problems using the pinhole camera model. This could also link to whatever study of optics they may be doing in science class.

Sample questions - none involving detailed understanding of the camera model.

  1. You are at the park, there is a hot air balloon festival going on, and you take a photo of a particularly pretty one. You check the image info from your photos app, and it tells you your phone's camera has a field of view of 90deg, and the picture is 2048 pixel wide. The balloon in the picture looks like a circle at the center of the image, about 250 pixels in diameter. Looking at similar balloons on the ground nearby, you estimate that your balloon has a diameter of 25m. Estimate how far away from you was the balloon when you took the photo.

  2. Another balloon looks even more interesting. You have no idea how big it is, but luckily you took two photos of it. In the first one, it is centered in the image, in the second one it is about 200 pixels to the right. Your smartphone has GPS and orientation sensors, so the images' info from your photos app says confidently that you snapped both pictures with the camera pointed directly East, and the second photo was taken from a point 20m North of the first one. How far was this second balloon from the first camera location?

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Most college textbooks for trig have some good problems. (For example, see page 43 of this pdf: https://www.slcc.edu/math/docs/math-1060-oer-trigonometry-textbook-version-2.pdf)

(When you get to sine waves, I have some great problems to share.)

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    $\begingroup$ Most college textbooks for trig have some mediocre problems. It's rare to find something that isn't a rehash of the usual "find the height of the flagpole/tree/skyscraper". $\endgroup$
    – Mark
    Commented Jul 8 at 21:41
  • $\begingroup$ @Mark, if you downvoted, please do look at this particular textbook. I agree that there are plenty of mediocre problems in most textbooks, but most that I've seen also include some goodies. I wanted to provide a resource for the OP, and I do think this answer is useful. $\endgroup$
    – Sue VanHattum
    Commented Jul 8 at 21:52
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    $\begingroup$ Not the downvoter. I'm guessing the downvote came from this feeling like a "Google it" answer. $\endgroup$
    – Mark
    Commented Jul 8 at 22:05
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One of the classic problems is have them calculate the height of a building. This can be your school building if you want.

Given a tall enough building, or an irregularly shaped building, pulling a tape measure up to the roof is not generally practical. But measuring the distance on the ground from the building to an observer is often easy.

Once you have the distance and the angle the observer have to look up to look at the roof of the building you can calculate the height.

A simple experiment will usually give good enough results. However, there are several subtle issues that you can try to solve:

  • For an irregularly shaped building the tallest point is usually not aligned with the outside wall. You may need to measure the distance from the ground location of the tallest point to the observer.

  • Sometimes there is a wall between the ground location of the tallest point of the roof to the observer making measuring the distance directly impossible. You can apply more trigonometry to calculate the straight line distance by triangulating the distance of the two points to the nearest door.

Using the same technique and some google maps help you can also then calculate the height of the nearest mountain or skyscraper. I'd have them do the actual height of an actual building first to show that you don't need technology to help you do the calculations.

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