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Math 20-1

Math 20-1 · Worksheets

Trigonometry

Ten questions of mixed difficulty, covering Trigonometry. Print it, or work through it on screen — the answer key starts on its own page.

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Trigonometry

Math 20-1 · maddyhelps.com

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Circle the best answer for each question. Show your work in the space provided.

  1. For 0° ≤ θ ≤ 360°, the solutions to sin θ = 0.5 are

    1. a) 30° only
    2. b) 30° and 210°
    3. c) 60° and 120°
    4. d) 30° and 150°

  2. The exact value of cos 120° is

    1. a) −½
    2. b) √3/2
    3. c) −√3/2
    4. d) ½

  3. For 0° ≤ θ ≤ 360°, the solutions to tan θ = −1 are

    1. a) 135° only
    2. b) 45° and 315°
    3. c) 135° and 315°
    4. d) 45° and 225°

  4. The point P(−3, 4) lies on the terminal arm of θ. Then sin θ is

    1. a) −4/3
    2. b) −3/5
    3. c) 4/5
    4. d) 3/5

  5. In triangle ABC, a = 6, b = 9 and ∠A = 30°. The number of possible triangles is

    1. a) none
    2. b) three
    3. c) one
    4. d) two

  6. In triangle ABC, a = 5, b = 7 and ∠C = 60°. The length of c is about

    1. a) 7.9
    2. b) 4.4
    3. c) 8.6
    4. d) 6.2

  7. If sin θ > 0 and cos θ < 0, then θ is in

    1. a) quadrant III
    2. b) quadrant IV
    3. c) quadrant II
    4. d) quadrant I

  8. An angle in standard position has

    1. a) its vertex at the origin and its initial arm on the positive y-axis
    2. b) both arms on the x-axis
    3. c) its vertex anywhere and its initial arm horizontal
    4. d) its vertex at the origin and its initial arm on the positive x-axis

  9. An angle of 150° in standard position lies in

    1. a) quadrant I
    2. b) quadrant II
    3. c) quadrant IV
    4. d) quadrant III

  10. The reference angle for 150° is

    1. a) 210°
    2. b) 60°
    3. c) 150°
    4. d) 30°

Answer key · Trigonometry

Math 20-1 · maddyhelps.com

  1. d) 30° and 150° — The reference angle is 30°, and sine is positive in quadrants I and II, giving 30° and 180° − 30° = 150°. A calculator returns only the first one, so the second has to be reasoned out.
  2. a) −½ — The reference angle is 60°, and cos 60° = ½. In quadrant II the x-coordinate is negative, so cosine is negative: −½. Getting the size right and the sign wrong is the most common way to lose this mark.
  3. c) 135° and 315° — The reference angle is 45°, and tangent is negative where sine and cosine have opposite signs — quadrants II and IV. That gives 180° − 45° = 135° and 360° − 45° = 315°.
  4. c) 4/5 — First find r = √(9 + 16) = 5, always taken as positive. Then sin θ = y/r = 4/5. The x-coordinate is negative, so cos θ = −3/5, but sine only uses y.
  5. d) two — This is the ambiguous case: two sides and an angle not between them. The height from C is b sin A = 4.5. Since 4.5 < a = 6 < b = 9, the side a is long enough to reach the base but short enough to reach it twice, so two different triangles fit.
  6. d) 6.2 — Two sides and the angle between them means the cosine law: c² = 25 + 49 − 2(5)(7)cos 60° = 74 − 35 = 39, so c ≈ 6.2. The sine law cannot start here, because no side is paired with a known opposite angle.
  7. c) quadrant II — Sine follows the sign of y and cosine the sign of x. Positive y with negative x is up and to the left, which is quadrant II. This is the whole content of the CAST rule, worked out rather than memorised.
  8. d) its vertex at the origin and its initial arm on the positive x-axis — Standard position fixes two things so that every angle is described the same way: the vertex sits at the origin and the initial arm runs along the positive x-axis. Only the terminal arm moves, counterclockwise for a positive angle.
  9. b) quadrant II — Quadrant I runs to 90°, quadrant II to 180°, so 150° is in quadrant II. Its terminal arm is above the x-axis and to the left of the y-axis.
  10. d) 30° — The reference angle is the acute angle between the terminal arm and the x-axis — never the y-axis. In quadrant II that is 180° − 150° = 30°.