Calculus ETF 6e
Solve real-life problems involving right triangles
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MA.912.G.5.4 Archived Standard
Clarifications
- Item Type(s): This benchmark may be assessed using: MC , FR item(s)
- Clarification : Students will apply properties of right triangles to solve real-world problems.
Items may require students to apply the Pythagorean theorem, special right triangle relationships, and/or characteristics of triangles resulting from the altitude of a right triangle drawn from the right angle to the hypotenuse.
Items may include the application of the geometric mean.
Items assessing MA.912.G.5.2 may be set in either mathematical or real-world contexts. All other items must be set in real-world context.
Any radical expressions in the item stem must be in simplified or rationalized form.
Graphics should be used in most of these items, as appropriate.
- Response Attributes : Any radical expressions in multiple-choice options will be provided in simplified or rationalized form.
- Test Item #: Sample Item 1
- Difficulty: N/A
- Type: MC: Multiple Choice
- Test Item #: Sample Item 2
Nara created two right triangles. She started with LJKL and drew an altitude from point K to side JL. The diagram below shows LJKL and some of its measurements, in centimeters (cm).
Based on the information in the diagram, what is the measure of x to the nearest tenth of a centimeter?
- Type: FR: Fill-in Response
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Right-Triangle Word Problems
What is a right-triangle word problem.
A right-triangle word problem is one in which you are given a situation (like measuring something's height) that can be modelled by a right triangle. You will draw the triangle, label it, and then solve it; finally, you interpret this solution within the context of the original exercise.
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Right Triangle Word Problems
Once you've learned about trigonometric ratios (and their inverses), you can solve triangles. Naturally, many of these triangles will be presented in the context of word problems. A good first step, after reading the entire exercise, is to draw a right triangle and try to figure out how to label it. Once you've got a helpful diagram, the math is usually pretty straightforward.
- A six-meter-long ladder leans against a building. If the ladder makes an angle of 60° with the ground, how far up the wall does the ladder reach? How far from the wall is the base of the ladder? Round your answers to two decimal places, as needed.
First, I'll draw a picture. It doesn't have to be good or to scale; it just needs to be clear enough that I can keep track of what I'm doing. My picture is:
To figure out how high up the wall the top of the ladder is, I need to find the height h of my triangle.
Since they've given me an angle measure and "opposite" and the hypotenuse for this angle, I'll use the sine ratio for finding the height:
sin(60°) = h/6
6 sin(60°) = h = 3sqrt[3]
Plugging this into my calculator, I get an approximate value of 5.196152423 , which I'll need to remember to round when I give my final answer.
For the base, I'll use the cosine ratio:
cos(60°) = b/6
6×cos(60°) = b = 3
Nice! The answer is a whole number; no radicals involved. I won't need to round this value when I give my final answer. Checking the original exercise, I see that the units are "meters", so I'll include this unit on my numerical answers:
ladder top height: about 5.20 m
ladder base distance: 3 m
Note: Unless you are told to give your answer in decimal form, or to round, or in some other way not to give an "exact" answer, you should probably assume that the "exact" form is what they're wanting. For instance, if they hadn't told me to round my numbers in the exercise above, my value for the height would have been the value with the radical.
- A five-meter-long ladder leans against a wall, with the top of the ladder being four meters above the ground. What is the approximate angle that the ladder makes with the ground? Round to the nearest whole degree.
As usual, I'll start with a picture, using "alpha" to stand for the base angle:
They've given me the "opposite" and the hypotenuse, and asked me for the angle value. For this, I'll need to use inverse trig ratios.
sin(α) = 4/5
m(α) = sin −1 (4/5) = 53.13010235...
(Remember that m(α) means "the measure of the angle α".)
So I've got a value for the measure of the base angle. Checking the original exercise, I see that I am supposed to round to the nearest whole degree, so my answer is:
base angle: 53°
- You use a transit to measure the angle of the sun in the sky; the sun fills 34' of arc. Assuming the sun is 92,919,800 miles away, find the diameter of the sun. Round your answer to the nearest whole mile.
First, I'll draw a picture, labelling the angle on the Earth as being 34 minutes, where minutes are one-sixtieth of a degree. My drawing is *not* to scale!:
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Hmm... This "ice-cream cone" picture doesn't give me much to work with, and there's no right triangle.
The two lines along the side of my triangle measure the lines of sight from Earth to the sides of the Sun. What if I add another line, being the direct line from Earth to the center of the Sun?
Now that I've got this added line, I have a right triangle — two right triangles, actually — but I only need one. I'll use the triangle on the right.
(The angle measure , "thirty-four arc minutes", is equal to 34/60 degrees. Dividing this in half is how I got 17/60 of a degree for the smaller angle.)
I need to find the width of the Sun. That width will be twice the base of one of the right triangles. With respect to my angle, they've given me the "adjacent" and have asked for the "opposite", so I'll use the tangent ratio:
tan(17/60°) = b/92919800
92919800×tan(17/60°) = b = 459501.4065...
This is just half the width; carrying the calculations in my calculator (to minimize round-off error), I get a value of 919002.8129 . This is higher than the actual diameter, which is closer to 864,900 miles, but this value will suffice for the purposes of this exercise.
diameter: about 919,003 miles
- A private plane flies 1.3 hours at 110 mph on a bearing of 40°. Then it turns and continues another 1.5 hours at the same speed, but on a bearing of 130°. At the end of this time, how far is the plane from its starting point? What is its bearing from that starting point? Round your answers to whole numbers.
The bearings tell me the angles from "due north", in a clockwise direction. Since 130 − 40 = 90 , these two bearings create a right angle where the plane turns. From the times and rates, I can find the distances travelled in each part of the trip:
1.3 × 110 = 143 1.5 × 110 = 165
Now that I have the lengths of the two legs, I can set up a triangle:
(The angle θ is the bearing, from the starting point, of the plane's location at the ending point of the exercise.)
I can find the distance between the starting and ending points by using the Pythagorean Theorem :
143 2 + 165 2 = c 2 20449 + 27225 = c 2 47674 = c 2 c = 218.3437657...
The 165 is opposite the unknown angle, and the 143 is adjacent, so I'll use the inverse of the tangent ratio to find the angle's measure:
165/143 = tan(θ)
tan −1 (165/143) = θ = 49.08561678...
But this angle measure is not the "bearing" for which they've asked me, because the bearing is the angle with respect to due north. To get the measure they're wanting, I need to add back in the original forty-degree angle:
distance: 218 miles
bearing: 89°
Related: Another major class of right-triangle word problems you will likely encounter is angles of elevation and declination .
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Real Life Applications Of Right Angle Triangle
Geometry provides us with an elementary introduction to shapes around us. This branch of maths is useful in mastering shape reasoning. If you ever wonder how math is going to be used in real life, geometry offers the answer most readily. We know that circles, triangles, and rectangles are the 2D figures that we see around most commonly. Another 2D figure, which is a ramification of the triangle is a right triangle. It is one of the most stable structures and has its utility in the fields of carpentry, architecture, etc.
How right triangles are used in real-life?
What exactly is a right triangle? If you visualize elaborately, a right triangle is two line segments placed at 90 o angle to each other and their free ends are joined by a diagonally oriented segment. This is a shape that we may locate quite effortlessly around us. Staircases when observed from the side may give you an impression of several right angles juxtaposed to each other. Apart from this common example locatable in homes, listed here are the other real-life uses of the right triangles.
- Shadow cast by a tall object: A tree forms right angle to the land. Its top and the open end of the shadow form the hypotenuse allowing you to visualize the right triangle structure.
- A climbing ladder leaning to the wall: A slanting ladder to the wall forms right triangle shape where hypotenuse corresponds to the ladder.
- Angles for placing a slab: Wall mounted shelves are placed on right triangular holding points.
- A person standing at a tower and locating a ball on the land : The gaze of a person standing on the tower forms the hypotenuse of the right triangle shape whose two vertical and horizontal arms are height of the tower and distance between the tower and ball.
Solving right triangle and its application in various professions
Right triangles fit various spheres of our lives like a glove. These geometrical figures have an entire branch of math dedicated to it. We all know it by the name of ‘trigonometry’. Various relationships between the two from the three sides of the right triangle (trigonometric functions) help solve real-life problems various professionals face [ 1 ] . Let’s take a look at those.
1. Carpentry
Carpenters need to ascertain that the walls are perfectly straight and corners are squares to fit in the elements as a part of the interior designing process. They may apply the Pythagorean theorem that relates the sides of the right triangle to find the required measurements.
2. Finding height or length
In architecture, the experts solve right triangle problems to find the required height or length of any structure. This calculation is needed to assure compliance with the building plan.
3. Navigation
In navigation, finding the distance of the ship from the shore, or between two ships or other objects can help in finding the probable duration of the tour. It may also help find the safe speed and the suitable maneuvering angles to keep the vessel protected from hits or accidents.
4. Surveying
Surveyor uses the trigonometry functions or solves right triangle based problems to estimate the height of a building. Using a total station theodolite (TST), the surveyor finds the exact angle of elevation from a chosen distance.
5. Astronomy
Several astronomical calculations like estimating the radius of the Earth or the movement of certain objects in a particular period can be found using the principles and functions of trigonometry. The study of distances between the two astral bodies becomes possible to do by treating the distance as one arm of the right triangle and applying trigonometry principles.
6. Astronauts and space scientists
Astronauts, during their expedition to space, need to find the characteristics of weather, soil, air pressure, etc. of the places where they cannot visit physically. They use tools like robotic arms for such purposes. With the help of right triangles-based calculations, they can find the correct angle and orientation for maneuvering the robotic arm or other similar tools.
Teaching right triangles the activity way
All the examples mentioned above indicate the possibility of teaching right triangles by using activities. High school students have to hone their trigonometry basic skills. Just solving equations and mugging up values of cos, tan and sine may not produce the required outcomes. Better learning is achieved when they get to apply the concepts. Activities based on right triangles help learn by doing.
A few activities that involve solving right triangles’ variables are:
1. Estimating heights of objects using Lego Robot kit
Lego Robot Kit act as a portable counterpart of the things students may see around. They apply trigonometry principles to estimate the height of various objects given in the kit. On scaling up the measurements received, they can apply the learning to find the heights of buildings, towers, poles, etc.
2. Estimating boundaries of a plot
Prepare the model of a landscape with dummy hills, buildings, empty land, etc. Assign a few values to the objects’ distance from each other and contours of the land. By making hypothetical right triangles from the top of the dummy hill, building, etc., students can learn about how to find values of various variables like height, speed, distance, etc.
3. Parallax activity to guess the distance between stars and sun
Parallax means the difference in the position of an object when viewed from different points of view. This displacement can create an angle of view, and two arms that resemble a triangle. By using calculators and trigonometric functions’ tables, they can get the idea of the distance between astronomical objects . Moving forth, the distance between two landmarks can also be estimated.
Studying right triangles can help students understand how the geometric and trigonometric principles can be applied to solve real-life problems. It is this practical knowledge attained that helps blur the line between bookish knowledge and applied science. So, next time you are done mugging up the definitions and functions’ values, start making hypothetical models and apply various combinations of values to strengthen your learning. It brings fun into the learning hour and makes the study of mathematical concepts more interesting.
References:
- Libretexts. (2021, January 2). 1.3: Applications and Solving Right Triangles . Mathematics LibreTexts. https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Elementary_Trigonometry_(Corral)/01%3A_Right_Triangle_Trigonometry_Angles/1.03%3A_Applications_and_Solving_Right_Triangles
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High school geometry
Unit 5: lesson 9, right triangle word problem.
- Angles of elevation and depression
- Right triangle trigonometry review
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Solving Right Triangles
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Applications
The ability to solve right triangles has many applications in the real world. Many of these applications have to do with two-dimensional motion, while others concern stationary objects. We'll discuss both.
Two-Dimensional Motion
Two-dimensional motion can be represented by a vector. Every vector can be resolved into a vertical and a horizontal component. When a vector is combined with its vertical and horizontal component, a right triangle is formed.
Stationary Objects
Whenever you use a right triangle to model a real-life situation, it is immensely helpful to draw a picture or diagram of the situation. Then labeling the parts of the right triangle is easy and the problem can be simply solved.
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Introduction Objective: Solve real life situation problems using Right Triangle Trigonometry. Duration/Mode:45mins/Student-centered Instructions: -click.
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