ACT Math Quiz: Trigonometry

What this quiz covers

This quiz focuses on Trigonometry, giving you a quick way to practice the rules, question types, and explanations that matter most for ACT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\frac{\pi}{2}$
2. $2$
3. $3$
4. $4$ (correct answer)

Explanation: The correct answer is D (4). The period of f(x) = A sin(bx) + c is given by 2π/b. Here b = π/2. Period = 2π ÷ (π/2) = 2π × (2/π) = 4. A (π/2) reports the b-value itself as the period rather than computing 2π/b. B (2) results from computing 2π/b with b = π (misreading the coefficient as π instead of π/2): 2π/π = 2. C (3) reports the amplitude coefficient rather than the period — confusing the A and b parameters. Pro tip: the period formula is 2π/b where b is the coefficient multiplying x, not x itself.

Question 2

1. 4
2. 5
3. 7 (correct answer)
4. 10

Explanation: This is a trigonometric graph analysis question testing how a negative coefficient interacts with the vertical shift to determine maximum value. Choice C (7) is correct — identify the parameters of g(x) = −3cos(πx/2) + k. Amplitude = 3, vertical shift = k. The minimum of −3cos(πx/2) occurs when cos = +1 (giving −3), so the minimum of g is −3 + k = 1 → k = 4. The maximum occurs when cos = −1 (giving +3), so the maximum of g = 3 + k = 3 + 4 = 7. Choice A (4) reports k itself — finding the vertical shift but confusing it with the maximum value. Choice B (5) adds the amplitude to the minimum: 1 + |−3| = 1 + 4 = 5, incorrectly treating the amplitude as k. Actually B = min + amplitude = 1 + 4 = 5 if student thinks range = 2×amplitude centered at min. Choice D (10) adds k and the amplitude twice: 4 + 3 + 3 = 10 or similar double-counting. Pro tip: When a cosine function has a NEGATIVE leading coefficient, the function reaches its maximum when cosine is at its MINIMUM (−1), not its maximum. Always think: "What value of the trig function makes the WHOLE expression as large as possible?" Here, −3(−1) = +3 is the largest the trig part can be, giving max = 3 + k.

Question 3

1. $0$
2. $1$ (correct answer)
3. $\sqrt{3}$
4. $\frac{\sqrt{3}}{2}$

Explanation: In the unit circle, $\pi/4$ radians equals $45^\circ$. Using SOH-CAH-TOA, tangent represents opposite over adjacent. For the special angle $\pi/4$ ($45^\circ$), $\tan(\pi/4) = 1$. Choice C ($\sqrt{3}$) is actually $\tan(\pi/3)$ or $\tan(60^\circ)$.

Question 4

1. $\frac{\sqrt{3}}{2}$
2. $\frac{1}{2}$ (correct answer)
3. $\frac{\sqrt{2}}{2}$
4. $0$

Explanation: The angle 30° is a special angle on the unit circle. sin(30°) = 1/2, which is a standard value to memorize. Choice A shows √3/2, which is actually cos(30°), not sin(30°).

Question 5

1. 12/7
2. 4/3
3. 3/4 (correct answer)
4. 12/9

Explanation: By SOH-CAH-TOA, the tangent of an angle is the opposite side over the adjacent side. Here the opposite side is 9 and the adjacent side is 12, so $\tan(\theta) = 9/12$, which reduces to 3/4. The value 4/3 is the reciprocal, produced by dividing adjacent by opposite, and 12/9 is that same reversed ratio left unsimplified. The value 12/7 pairs the adjacent side with a number that is not a side of this triangle at all, so it does not come from either trigonometric ratio. Write the ratio in words before plugging in numbers, then reduce it, since an unsimplified or flipped fraction is the most common way tangent problems go wrong.

Question 6

1. 3/5 (correct answer)
2. 4/5
3. 5/6
4. 6/5

Explanation: For angle θ, the opposite side is 6 and the hypotenuse is 10. Using SOH-CAH-TOA, sin(θ) = opposite/hypotenuse = 6/10. This simplifies to 3/5. Choice B (4/5) would be the cosine if the adjacent side were 8.

Question 7

1. √3
2. 1
3. 0 (correct answer)
4. √3/2

Explanation: tan(0°) is a standard unit circle value. Using SOH-CAH-TOA, tangent represents the opposite side over the adjacent side. For the angle 0°, tan(0°) = 0. Choice B (1) is actually tan(45°), not tan(0°).

Question 8

1. $30^\circ$
2. $45^\circ$ (correct answer)
3. $60^\circ$
4. $90^\circ$

Explanation: We need to find which common acute angle has sin(θ) = √2/2. From the unit circle and special triangles, sin(45°) = √2/2. This is the fundamental value for a 45° angle in a 45-45-90 triangle. Choice C gives 60°, but sin(60°) = √3/2, not √2/2.

Question 9

1. $\frac{8}{6}$
2. $\frac{4}{3}$
3. $\frac{6}{8}$
4. $\frac{3}{4}$ (correct answer)

Explanation: For angle θ, the opposite side is 6 and the adjacent side is 8. Using SOH-CAH-TOA, $tan(\theta) = opposite/adjacent$. Therefore, $tan(\theta) = 6/8 = 3/4$. Choice B ($\frac{4}{3}$) would be adjacent/opposite, which is the reciprocal relationship (cotangent).

Question 10

1. $1$
2. $0$ (correct answer)
3. $\frac{1}{2}$
4. $\frac{\sqrt{2}}{2}$

Explanation: The angle 0° represents a position on the positive x-axis of the unit circle. At this position, the y-coordinate is 0, so sin(0°) = 0. This is a fundamental trigonometric value where the angle has no vertical component. Choice A gives 1, which is actually cos(0°), not sin(0°).

Question 11

1. $\frac{\pi}{6}$
2. $\frac{\pi}{4}$
3. $\frac{\pi}{3}$ (correct answer)
4. $\frac{\pi}{2}$

Explanation: The correct answer is C (π/3). To convert degrees to radians, multiply by π/180: 60 × (π/180) = 60π/180 = π/3. A (π/6) corresponds to 30°, not 60° — the student may confuse the two or divide 60 by 360 instead of 180. B (π/4) corresponds to 45°. D (π/2) corresponds to 90°. The conversion factor π/180 is essential to memorize — or remember that 180° = π radians, so each degree is worth π/180 radians. Pro tip: memorize the common conversions: 30°=π/6, 45°=π/4, 60°=π/3, 90°=π/2, 180°=π.

Question 12

1. $\sqrt{2}/2$
2. $1$
3. $0$ (correct answer)
4. $1/2$

Explanation: $\sin(0^\circ)$ is a standard unit circle value. Using SOH-CAH-TOA, sine represents the y-coordinate (or opposite/hypotenuse). For the angle $0^\circ$, $\sin(0^\circ)$ = 0. Choice B (1) is actually $\cos(0^\circ)$, not $\sin(0^\circ)$.

Question 13

1. 1
2. 3
3. 4
4. 7 (correct answer)

Explanation: This is a trigonometric graphs question testing the effect of a negative amplitude on the maximum value. Choice D (7) is correct — the function y = −3sin(2x) + 4 has amplitude 3, a negative leading coefficient, and a vertical shift of +4. To MAXIMIZE y, we need −3sin(2x) to be as large as possible, which occurs when sin(2x) is at its MINIMUM value of −1: y = −3(−1) + 4 = 3 + 4 = 7. Choice A (1) comes from plugging in sin(2x) = +1 (the maximum of sine), which actually gives the MINIMUM of this function: −3(1) + 4 = 1. Choice B (3) reports the amplitude alone, ignoring the vertical shift. Choice C (4) identifies the vertical shift as the maximum — this would be correct only if the amplitude were 0. Pro tip: When the coefficient of sine is negative, the function reaches its maximum when sin is at its minimum (−1), not its maximum (+1). Always ask: what value of sin(2x) makes the entire expression as LARGE as possible?

Question 14

1. $\sqrt{39}$ (correct answer)
2. $\sqrt{74}$
3. $8$
4. $\sqrt{109}$

Explanation: This is a law of cosines question. Choice A (√39) is correct — the Law of Cosines: c² = a² + b² − 2ab·cos(C) = 5² + 7² − 2(5)(7)·cos(60°) = 25 + 49 − 70(0.5) = 74 − 35 = 39. Therefore c = √39. Choice B (√74) results from omitting the cosine term entirely: c² = 5² + 7² = 74 — applying the Pythagorean theorem as if the triangle were a right triangle. Choice C (8) results from rounding or estimating √39 ≈ 6.24... possibly from a computational error that produces c² = 64. Choice D (√109) results from adding the cosine term instead of subtracting: c² = 25 + 49 + 35 = 109 — a sign error on the formula. Pro tip: The Law of Cosines formula always subtracts the cosine term: c² = a² + b² − 2ab·cos(C). For acute angles (less than 90°), cos(C) is positive, so the term 2ab·cos(C) reduces c² below the Pythagorean sum. For obtuse angles, cos(C) is negative, and c² is larger than a² + b².

Question 15

1. $1$
2. $\dfrac{7}{5}$
3. $\dfrac{31}{20}$ (correct answer)
4. $2$

Explanation: This is a trigonometry question requiring derivation of multiple trig ratios from one. Choice C (31/20) is correct — sin(θ) = 3/5 means the triangle has opposite = 3 and hypotenuse = 5, so it's a 3-4-5 right triangle with adjacent = 4. Therefore: cos(θ) = 4/5 and tan(θ) = 3/4. Sum: 4/5 + 3/4 = 16/20 + 15/20 = 31/20. Choice A (1) likely invokes the Pythagorean identity sin²θ + cos²θ = 1, confusing that identity with the sum cos(θ) + tan(θ). Choice B (7/5) comes from computing 4/5 + 3/5 = 7/5 — correctly finding cos but computing tan as 3/5 (using the hypotenuse instead of the adjacent side in the denominator). Choice D (2) may come from estimating both ratios as approximately 1 and adding. Pro tip: If sin(θ) = a/c, immediately sketch a right triangle with opposite = a, hypotenuse = c, and use the Pythagorean theorem to find the adjacent side. From there, cos and tan follow directly.

Question 16

1. $\frac{8}{17}$
2. $\frac{8}{15}$ (correct answer)
3. $\frac{15}{17}$
4. $\frac{15}{8}$

Explanation: This is a trigonometry question testing SOHCAHTOA applied to a labeled right triangle. Choice B (8/15) is correct — from angle C's perspective: the opposite side is AB = 8 and the adjacent side is BC = 15. tan C = opposite/adjacent = 8/15. (The hypotenuse = √(8² + 15²) = √289 = 17.) Choice A (8/17) gives sin C, not tan C — it correctly identifies opposite = 8 but uses the hypotenuse (17) as the denominator instead of the adjacent side. Choice C (15/17) gives cos C — adjacent over hypotenuse. Choice D (15/8) gives tan B, the other acute angle — it swaps opposite and adjacent, giving the tangent from the perspective of angle B rather than angle C. Pro tip: Before writing a trig ratio, identify which angle you're evaluating and label each side (opposite, adjacent, hypotenuse) relative to THAT angle. The same side can be "opposite" for one angle and "adjacent" for another.

Question 17

1. $5$
2. $5\sqrt{2}$
3. $5\sqrt{3}$ (correct answer)
4. $10$

Explanation: This is a Law of Sines question requiring careful identification of opposite sides and angles. Choice C (5√3) is correct — by the Law of Sines: q/sin Q = p/sin P → q/sin(120°) = 5/sin(30°). Since sin(30°) = 1/2 and sin(120°) = √3/2: q = 5 × (√3/2)/(1/2) = 5 × √3 = 5√3. Choice A (5) notes that ∠P = ∠R = 30°, making the triangle isosceles with p = r = 5, then incorrectly concludes q = 5 as well — forgetting that the largest angle (120°) is opposite the longest side. Choice B (5√2) uses sin(45°) = √2/2 instead of sin(120°) = √3/2: 5 × (√2/2)/(1/2) = 5√2. Choice D (10) uses sin(Q) = 1 (as if Q = 90°): q = 5 × 1/(1/2) = 10. Pro tip: The Law of Sines states a/sin A = b/sin B = c/sin C, where each side is paired with its OPPOSITE angle. In this triangle, side q is opposite ∠Q = 120°, and side p = 5 is opposite ∠P = 30°. Key values: sin(30°) = 1/2 and sin(120°) = sin(60°) = √3/2. Set up the ratio and cross-multiply carefully.

Question 18

1. $\frac{2\pi}{3}$
2. $\frac{3\pi}{4}$
3. $\frac{5\pi}{6}$ (correct answer)
4. $\frac{7\pi}{6}$

Explanation: This is a degrees-to-radians conversion question. Choice C (5π/6) is correct — multiply by the conversion factor π/180: 150 × π/180 = 150π/180 = 5π/6 (dividing numerator and denominator by 30). Choice A (2π/3) corresponds to 120°: 120 × π/180 = 2π/3. Off by 30°. Choice B (3π/4) corresponds to 135°: 135 × π/180 = 3π/4. Off by 15°. Choice D (7π/6) corresponds to 210°: 210 × π/180 = 7π/6. These are all benchmark radian values adjacent to 5π/6, making them plausible traps. Pro tip: Memorize the benchmark radian values: 30° = π/6, 45° = π/4, 60° = π/3, 90° = π/2, 120° = 2π/3, 135° = 3π/4, 150° = 5π/6, 180° = π. For 150°: it's 5 × 30° = 5 × (π/6) = 5π/6. Alternatively, 150° = 180° − 30° = π − π/6 = 5π/6.

Question 19

1. $\frac{3}{5}$ (correct answer)
2. $\frac{4}{5}$
3. $\frac{3}{7}$
4. $\frac{4}{3}$

Explanation: The correct answer is A (3/5). Given tan θ = 3/4, identify the sides: opposite = 3, adjacent = 4. Find the hypotenuse using the Pythagorean theorem: 3² + 4² = 9 + 16 = 25 → hypotenuse = 5. This is the 3-4-5 Pythagorean triple. sin θ = opposite/hypotenuse = 3/5. B (4/5) gives cos θ = adjacent/hypotenuse — the correct formula applied to the wrong ratio. C (3/7) adds the given numerator and denominator to make the denominator (3 + 4 = 7), which has no geometric basis. D (4/3) is the tangent of the complementary angle — the reciprocal of the given tan. Always draw the triangle and label all three sides before computing trig ratios.

Question 20

1. √3/2
2. 1
3. 1/2
4. 0 (correct answer)

Explanation: cos(90°) is a standard unit circle value. Using SOH-CAH-TOA, cosine represents the x-coordinate (or adjacent/hypotenuse). For the special angle 90°, cos(90°) = 0. Choice B (1) is actually sin(90°), not cos(90°).