TF-1 Right Triangle Trigonometry

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Section TF-1 Right Triangle Trigonometry

Worksheet Right Triangle Trigonometry

For this outcome Specifically, for this outcome, students should be able to:

Convert between degrees and radians
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Evaluate trigonometric functions of special angles (multiples of 30° and 45°)
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Evaluate reference angles and coterminal angles.
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Evaluate the six trigonometric functions of an angle in a right triangle
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Use the triangle and unit circle definitions of the six trigonometric functions
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Given a point in the plane, find the six trigonometric functions of the angle in standard position.
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Given the value of a trigonometric function of an angle, find the other five trigonometric functions of that angle.
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Set up and solve right triangle applications.
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Angles and their Measures

An angle is formed by rotating a ray about its endpoint. The initial side of an angle is the starting position of the ray, and the terminal side is the final position of the ray. An angle in standard position has its vertex at the origin and its initial side along the positive x-axis.

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There are two common units for measuring angles: degrees and radians. Degrees are based on dividing a circle into 360 equal parts, while radians are based on the radius of a circle. One radian is the angle that intercepts an arc equal in length to the radius of the circle.

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To convert between degrees and radians, we can use the following formulas:

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Degrees to Radians: \(radians = degrees \cdot \frac{\pi}{180}\)

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Radians to Degrees: \(degrees = radians \cdot \frac{180}{\pi}\)

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Right Triangle Trigonometry

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A right triangle is a triangle with one angle measuring 90 degrees. The side opposite the right angle is called the hypotenuse, and the other two sides are called legs. The hypotenuse is always the longest side of a right triangle. The other angles in a right triangle are acute angles (less than 90 degrees).

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Right triangle trigonometry involves the relationships between the sides and angles of a right triangle. The six trigonometric functions are defined in terms of the sides of a right triangle:

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\(\displaystyle \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\)

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\(\displaystyle \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\)

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\(\displaystyle \tan \theta = \frac{\text{opposite}}{\text{adjacent}}\)

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\(\displaystyle \csc \theta = \frac{\text{hypotenuse}}{\text{opposite}}\)

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\(\displaystyle \sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}}\)

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\(\displaystyle \cot \theta = \frac{\text{adjacent}}{\text{opposite}}\)

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These functions can be used to find missing sides or angles in a right triangle, as well as to solve real-world problems involving right triangles.

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Trigonometric Functions of Acute Angles

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The trigonometric functions of acute angles are defined in terms of the sides of a right triangle. For an acute angle \(\theta\text{,}\) the trigonometric functions are:

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\(\displaystyle \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\)

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\(\displaystyle \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\)

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\(\displaystyle \tan \theta = \frac{\text{opposite}}{\text{adjacent}}\)

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\(\displaystyle \csc \theta = \frac{\text{hypotenuse}}{\text{opposite}}\)

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\(\displaystyle \sec \theta = \frac{\text{hypotenuse}}{\text{adjacent}}\)

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\(\displaystyle \cot \theta = \frac{\text{adjacent}}{\text{opposite}}\)

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For any acute angle \(\theta\text{,}\) the trigonometric functions are positive. The signs of the trigonometric functions depend on the quadrant in which the angle lies. The pythagorean theorem states that for a right triangle with legs of length \(a\) and \(b\text{,}\) and hypotenuse of length \(c\text{,}\) we have \(a^2 + b^2 = c^2\text{.}\) We can use this theorem to find missing sides of right triangles. This correlates to our first trigonometric identity which is \(\sin^2 \theta + \cos^2 \theta = 1\text{.}\) This will be made more clear in the next section.

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Trigonometric Functions of Any Angle

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For any angle in standard position, we can define the trigonometric functions using the unit circle. The unit circle is a circle with center at the origin and radius 1. The point on the unit circle corresponding to an angle \(\theta\) has coordinates \((\cos \theta, \sin \theta)\text{.}\) This allows us to define trigonometric functions for any angle in terms of the coordinates of the point on the unit circle. The signs of the trigonometric functions depend on the quadrant in which the angle lies. The circular functions are:

\(\displaystyle \sin \theta = y\text{-coordinate of the point on the unit circle}\)
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\(\displaystyle \cos \theta = x\text{-coordinate of the point on the unit circle}\)
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\(\displaystyle \tan \theta = \frac{y\text{-coordinate}}{x\text{-coordinate}}\)
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\(\displaystyle \csc \theta = \frac{1}{y\text{-coordinate}} = \frac{1}{\sin \theta}\)
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\(\displaystyle \sec \theta = \frac{1}{x\text{-coordinate}} = \frac{1}{\cos \theta}\)
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\(\displaystyle \cot \theta = \frac{1}{\tan \theta} = \frac{x\text{-coordinate}}{y\text{-coordinate}}\)
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Worksheet TF-1 Right Triangle Trigonometry

Trigonometry Fundamentals

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Right Triangle Trigonometry

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Triangle and Unit Circle Definitions

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Section TF-1 Rubric

Worksheet Worksheet

Before attempting the TF-1 outcome, you should review the following rubric. This shows the criteria you will be graded on and the expectations to earn each mastery level.

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Table 22. TF-1 Rubric

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Criteria	M	P	R	N
Simplifying Algebraic Expressions	All answers are fully simplified. Exact answers are given, not approximations. Radical and exponential expressions are used appropriately and are simplified in the correct manner.	No algebraic errors but some expressions are not simplified.	There are some algebraic errors and some expressions were not simplified.	There are some algebraic errors and some expressions were not simplified.
Basics of Graphing	Graph is labeled with units for each axis. The graph includes all of the important features (intercepts, asymptotes, deleted points, etc). The graph is appropriately drawn to accommodate all features.	Graph is labeled and has most of the required features.	Graph has most of the required features.	No graph was given.
Problem Structure and Mathematical Communication	Problems have a clear beginning, middle and end. Work progresses clearly from one step to the next. The work provided uses appropriate methods and provides a clear solution.	Problems have a clear beginning, middle and end. Work progresses clearly from one step to the next. The work provided uses algebraic methods and provides a solution.	Parts of the mathematical structure are missing or steps in the progress of the solution are missing. The work provided uses algebraic methods and provides a solution.	There is not a clear structure to the solution and it is unclear what methods are being used.
Trigonometric Ratios from a right triangle	Given two sides of a right triangle, find the missing side and find the six trigonometric function ratios based on the stated angle. The functions have correct notation and simplified values.	Given two sides of a right triangle, find the missing side with minor errors and find the six trigonometric function ratios based on the stated angle. The functions have correct notation and simplified values.	Found at least two of the trigonometric ratios with correct notation and simplified values.	Not enough evidence shown to determine the understanding of trigonometric ratios based on a right triangle.
Sketch an angle and the reference triangle	Given a point in the plane, graph, and label the correct angle in standard position. Identify the reference angle and correctly create and label the reference triangle.	Given a point in the plane, graph, an angle in standard position. Identify the reference angle and create the reference triangle.	Given a point in the plane, create a triangle based on the coordinates.	Not enough evidence to determine understanding of angles in standard position or reference triangles.

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Worksheet TF-1 Right Triangle Trigonometry

This outcome covers using right triangles to solve problems involving trigonometric ratios. You should be able to use the definitions of the six trigonometric functions to find missing sides and angles in right triangles. You should also be able to apply these skills to solve trigonometric functions of any point in the coordinate plane.

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For the quiz on this outcome you would be given 30 minutes to complete 2 problems. Here are samples of the types of problems you will encounter:

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Objectives

Simplifying Expressions: Simplify algebraic expressions by combining like terms and applying the distributive property. Answers should be completely simplified.
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Basics of Graphing: Give a graph with appropriate scales and labels. The line should have the correct intercepts and general behavior.
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Problem structure: Solution should be organized and easy to follow, with clear, complete steps that follow progressively from one to the next.
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Mathematical Communication: Solutions should be presented in a clear and logical manner, with appropriate use of mathematical notation and terminology. Equations should have at most one equal sign and expressions on both sides.
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Solve a right triangle: Given two sides of a right triangle, find the missing side and use the trigonometric definitions to find the values of the six trigonometric functions.
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Find the trigonometric functions of a point: Given a point in the coordinate plane, draw the angle in standard position; draw and label the reference triangle. Find the values of the six trigonometric functions using the definitions of the trigonometric functions and the signs based on the quadrants in the coordinate plane.
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1.

Given a right triangle RST with legs s=9 and t=12:

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(a)

Find the values of the six trigonometric functions of angle T.

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Hint.

Find the value of r then use the definitions of the trigonometric functions to find the values.

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Answer.

\(sin T = \frac{12}{15}\text{,}\) \(\cos T = \frac{9}{15}\text{,}\) \(\tan T = \frac{12}{9}\text{,}\) \(\csc T = \frac{15}{12}\text{,}\) \(\sec T = \frac{15}{9}\text{,}\) \(\cot T = \frac{9}{12}\)

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Solution.

First, we find the value of r using the Pythagorean theorem: \(r = \sqrt{s^2 + t^2}\) \(r = \sqrt{9^2 + 12^2}\) \(r= \sqrt{81 + 144} \) \(r= \sqrt{225} = 15\)

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Now we can use the definitions of the trigonometric functions:

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\(\sin T = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{15}\)

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\(\cos T = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{9}{15}\)

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\(\tan T = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{9}\)

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\(\csc T = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{15}{12}\)

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\(\sec T = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{15}{9}\)

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\(\cot T = \frac{\text{adjacent}}{\text{opposite}} = \frac{9}{12}\)

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2.

Given a right triangle UVW with leg u=8 and hypotenuse v=17:

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(a)

Find the values of the six trigonometric functions of angle U.

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Hint.

Find the value of w then use the definitions of the trigonometric functions to find the values.

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Answer.

\(\sin U = \frac{15}{17}\text{,}\) \(\cos U = \frac{8}{17}\text{,}\) \(\tan U = \frac{15}{8}\text{,}\) \(\csc U = \frac{17}{15}\text{,}\) \(\sec U = \frac{17}{8}\text{,}\) \(\cot U = \frac{8}{15}\)

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Solution.

First, we find the value of w using the Pythagorean theorem: \(w = \sqrt{v^2 - u^2}\) \(w = \sqrt{17^2 - 8^2}\) \(w= \sqrt{289 - 64} \) \(w= \sqrt{225} = 15\)

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Now we can use the definitions of the trigonometric functions:

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\(\sin U = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{15}{17}\)

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\(\cos U = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{8}{17}\)

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\(\tan U = \frac{\text{opposite}}{\text{adjacent}} = \frac{15}{8}\)

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\(\csc U = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{17}{15}\)

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\(\sec U = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{17}{8}\)

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\(\cot U = \frac{\text{adjacent}}{\text{opposite}} = \frac{8}{15}\)

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3.

The point (-4,6) is on the terminal side of angle θ in standard position:

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(a)

Sketch the angle θ in standard position. Sketch and label the reference triangle.

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Hint.

Use the given point to form a right triangle with the origin and the x-axis. Use the Pythagorean theorem to find the hypotenuse.

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Answer.

Diagram Exploration Keyboard Controls

Key	Action
Enter, A	Activate keyboard driven exploration
B	Activate menu driven exploration
Escape	Leave exploration mode
Cursor down	Explore next lower level
Cursor up	Explore next upper level
Cursor right	Explore next element on level
Cursor left	Explore previous element on level
X	Toggle expert mode
W	Extra details if available
Space	Repeat speech
M	Activate step magnification
Comma	Activate direct magnification
N	Deactivate magnification
Z	Toggle subtitles
C	Cycle contrast settings
T	Monochrome colours
L	Toggle language (if available)
K	Kill current sound
Y	Stop sound output
O	Start and stop sonification
P	Repeat sonification output

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(b)

Find the values of the six trigonometric functions of angle θ.

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Hint.

Find the value of r then use the definitions of the trigonometric functions to find the values.

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Answer.

\(\sin θ = \frac{6}{10}\text{,}\) \(\cos θ = \frac{-4}{10}\text{,}\) \(\tan θ = \frac{6}{-4}\text{,}\) \(\csc θ = \frac{10}{6}\text{,}\) \(\sec θ = \frac{10}{-4}\text{,}\) \(\cot θ = \frac{-4}{6}\)

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Solution.

First, we find the value of r using the distance formula: \(r = \sqrt{(-4)^2 + 6^2}\) \(r = \sqrt{16 + 36}\) \(r= \sqrt{52} = 2\sqrt{13}\)

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Now we can use the definitions of the trigonometric functions:

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\(\sin θ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6}{2\sqrt{13}}\)

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\(\cos θ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{-4}{2\sqrt{13}}\)

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\(\tan θ = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{-4}\)

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