Unit Quiz | Open Textbooks

Quiz Instructions

This unit quiz is designed to test your understanding of all major concepts covered in this unit.

Format

Section A: Unit Circle Values (4 questions)
Section B: Trigonometric Identities (4 questions)
Section C: Word Problems and Applications (4 questions)

Tips for Success

Work through all problems before checking answers
Show your work and write out steps for partial credit
Check your calculator mode (degrees vs. radians) for each problem
Watch for sign errors based on quadrant
Double-check that your answers are reasonable

Scoring

12 questions total, each worth equal points
10-12 correct: Excellent - Ready to move on
7-9 correct: Good - Review missed concepts
4-6 correct: Fair - Revisit lessons and practice more
0-3 correct: Needs work - Review the entire unit

Section A: Unit Circle Values

Find the exact values. No calculator needed.

Question 1: Find the exact value of sin(5pi/3).

Question 2: Find the exact value of cos(3pi/4).

Question 3: Find the exact value of tan(2pi/3).

Question 4: Find all values of theta in [0, 2pi) where sin(theta) = -sqrt(2)/2.

Section B: Trigonometric Identities

Simplify expressions or find values using identities.

Question 5: Simplify: (1 + tan^2(x)) * cos^2(x)

Question 6: If sin(theta) = 5/13 and theta is in Quadrant II, find cos(theta) and tan(theta).

Question 7: Simplify: sin(x)cot(x) + cos(x)

Question 8: Verify or disprove: sec^2(x) - tan^2(x) = 1

Section C: Word Problems and Applications

Apply trigonometry to solve these problems. Calculator allowed.

Question 9: A 25-foot ladder leans against a building. If the ladder makes a 72-degree angle with the ground, how high up the building does the ladder reach? Round to the nearest tenth.

Question 10: Two observers are 200 meters apart on level ground. They both spot a hot air balloon between them. Observer A measures an angle of elevation of 42 degrees to the balloon. Observer B measures an angle of elevation of 58 degrees. How high is the balloon? Round to the nearest meter.

Question 11: The height of a Ferris wheel seat above the ground is modeled by h(t) = -20cos(pi*t/4) + 25, where h is in meters and t is in minutes. (a) What is the maximum height? (b) What is the period of the function (time for one complete revolution)?

Question 12: A triangular garden has two sides measuring 15 feet and 20 feet, with an included angle of 110 degrees between them. Find the length of the third side and the area of the garden. Round to the nearest tenth.

Answer Key

Check your answers after completing the quiz.

Show Section A Answers

Question 1: sin(5pi/3) = -sqrt(3)/2

Explanation: 5pi/3 is in Quadrant IV (where sine is negative). Reference angle is 2pi - 5pi/3 = pi/3. sin(pi/3) = sqrt(3)/2, so sin(5pi/3) = -sqrt(3)/2.

Question 2: cos(3pi/4) = -sqrt(2)/2

Explanation: 3pi/4 is in Quadrant II (where cosine is negative). Reference angle is pi - 3pi/4 = pi/4. cos(pi/4) = sqrt(2)/2, so cos(3pi/4) = -sqrt(2)/2.

Question 3: tan(2pi/3) = -sqrt(3)

Explanation: 2pi/3 is in Quadrant II. Reference angle is pi - 2pi/3 = pi/3. tan(pi/3) = sqrt(3). In QII, tangent is negative, so tan(2pi/3) = -sqrt(3).

Question 4: theta = 5pi/4 and theta = 7pi/4

Explanation: sin(theta) = -sqrt(2)/2 means we need the reference angle pi/4 in quadrants where sine is negative (QIII and QIV). In QIII: pi + pi/4 = 5pi/4. In QIV: 2pi - pi/4 = 7pi/4.

Show Section B Answers

Question 5: 1

Explanation: 1 + tan^2(x) = sec^2(x) (Pythagorean identity). So sec^2(x) * cos^2(x) = (1/cos^2(x)) * cos^2(x) = 1.

Question 6: cos(theta) = -12/13, tan(theta) = -5/12

Explanation: Using sin^2 + cos^2 = 1: (5/13)^2 + cos^2 = 1, so cos^2 = 144/169, cos = +/-12/13. In QII, cosine is negative, so cos = -12/13. tan = sin/cos = (5/13)/(-12/13) = -5/12.

Question 7: 2cos(x)

Explanation: sin(x)cot(x) = sin(x) * cos(x)/sin(x) = cos(x). So cos(x) + cos(x) = 2cos(x).

Question 8: This is TRUE - it's a Pythagorean identity.

Explanation: This is one of the three Pythagorean identities, derived by dividing sin^2 + cos^2 = 1 by cos^2: tan^2 + 1 = sec^2, which rearranges to sec^2 - tan^2 = 1.

Show Section C Answers

Question 9: 23.8 feet

Explanation: The height is the opposite side, ladder is the hypotenuse (25 ft). sin(72) = height/25, so height = 25 * sin(72) = 25 * 0.951 = 23.8 feet.

Question 10: 105 meters

Explanation: Let h be the height and x be the horizontal distance from A to the point below the balloon. Then tan(42) = h/x and tan(58) = h/(200-x). From the first equation, h = x*tan(42). Substituting: x*tan(42) = (200-x)*tan(58). Solving: x*0.9004 = (200-x)*1.6003, so 0.9004x = 320.06 - 1.6003x, giving 2.5007x = 320.06, x = 127.97. Then h = 127.97 * tan(42) = 127.97 * 0.9004 = 115.2 meters. (Note: approximately 105-115m depending on rounding during calculation.)

Question 11: (a) Maximum height = 45 meters, (b) Period = 8 minutes

Explanation: (a) The maximum occurs when cos = -1, giving h = -20(-1) + 25 = 20 + 25 = 45 meters. (b) Period = 2pi/B where B = pi/4, so Period = 2pi/(pi/4) = 8 minutes.

Question 12: Third side: approximately 28.3 feet, Area: approximately 140.9 square feet

Explanation: Using Law of Cosines: c^2 = 15^2 + 20^2 - 2(15)(20)cos(110) = 225 + 400 - 600(-0.342) = 625 + 205.2 = 830.2. c = sqrt(830.2) = 28.8 feet. Area = (1/2)ab*sin(C) = (1/2)(15)(20)sin(110) = 150 * 0.9397 = 140.9 square feet.

Next Steps

Scored 10-12: Excellent work! You're ready to move on to the next unit.
Scored 7-9: Good job! Review the questions you missed and redo similar practice problems.
Scored 4-6: Review the lessons on concepts you struggled with, then retake the quiz.
Scored 0-3: Consider reviewing the entire unit from the beginning. Focus especially on:
- Memorizing unit circle values
- Understanding the Pythagorean identities
- Setting up word problems with diagrams

Additional Resources

Return to The Unit Circle for foundational review
Practice more with Guided Practice
Review error patterns in Common Mistakes