SAT 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 SAT 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. $\frac{10}{26}$ (correct answer)
2. $\frac{24}{26}$
3. $\frac{26}{10}$
4. $\frac{24}{10}$

Explanation: We need to find cos(A) in a right triangle with AC = 10 and BC = 24. First, find the hypotenuse using the Pythagorean theorem: AB = √(10² + 24²) = √(100 + 576) = √676 = 26. For angle A, AC is the adjacent side and AB is the hypotenuse, so cos(A) = adjacent/hypotenuse = 10/26. A common error is using BC (the opposite side) instead of AC (the adjacent side) in the cosine ratio. Always identify sides relative to the specific angle.

Question 2

1. 64 m
2. 102 m (correct answer)
3. 120 m
4. 71 m

Explanation: A drone travels 120 m along a path at 58° to the ground, and we need its height. The 120 m is the hypotenuse (distance along the path), and height is the opposite side to the 58° angle, so we use sine: sin(58°) = height/120. Solving: height = 120 × sin(58°) = 120 × 0.848 ≈ 101.8 m, which rounds to 102 m. A common mistake is treating the 120 m as horizontal distance rather than the actual path length. In ascent problems, the path length is typically the hypotenuse.

Question 3

1. $28^\circ$
2. $16^\circ$ (correct answer)
3. $45^\circ$
4. $74^\circ$

Explanation: We need to find angle θ opposite a leg of length 7 in a right triangle with hypotenuse 25. Since we know the opposite side (7) and hypotenuse (25), we use sine: sin(θ) = 7/25 = 0.28. Taking the inverse sine: θ = sin⁻¹(0.28) ≈ 16.26°, which rounds to 16°. A common error is using the wrong inverse function or forgetting to switch your calculator to degree mode. When finding angles from trig ratios, always double-check your calculator settings.

Question 4

1. $\frac{12}{13}$
2. $\frac{5}{13}$
3. $\frac{5}{12}$
4. $\frac{12}{5}$ (correct answer)

Explanation: We need to find $\tan(J)$ in a right triangle where JL = 5 and KL = 12, with the right angle at L. For angle J, KL is the opposite side and JL is the adjacent side, so $\tan(J) = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{5}$. A common mistake is inverting the ratio or misidentifying which side is opposite versus adjacent to angle J. Remember that tangent is always opposite over adjacent (TOA in SOH-CAH-TOA).

Question 5

1. 20.1
2. 22.3 (correct answer)
3. 30.0
4. 19.9

Explanation: We need the side opposite angle M in a right triangle where angle M is 48° and hypotenuse MN is 30. Since we want the opposite side and have the hypotenuse, we use sine: sin(48°) = opposite/30. Solving: opposite = 30 × sin(48°) = 30 × 0.743 ≈ 22.3. The key is recognizing that the side opposite angle M doesn't touch point M. Always draw and label your triangle to avoid confusion about which side is which.

Question 6

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

Explanation: Given $\sin(\theta) = \frac{3}{5}$, we need to find $\cos(\theta)$ using the Pythagorean identity. From $\sin^2(\theta) + \cos^2(\theta) = 1$, we get $\cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$. Therefore, $\cos(\theta) = \frac{4}{5}$ (positive for acute angles). A common error is forgetting to simplify the fraction or making arithmetic mistakes with fractions. The 3-4-5 triangle is a classic Pythagorean triple that appears frequently on tests.

Question 7

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

Explanation: We need $\tan(\theta)$ where $\theta$ is opposite the leg of length 8 in a right triangle with legs 8 and 15. For angle $\theta$, the opposite side is 8 and the adjacent side is 15, so $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{8}{15}$. A common mistake is confusing which leg is opposite versus adjacent to the angle in question. When given two legs, carefully identify which angle you're working with before setting up the ratio.

Question 8

1. $\sin^{-1}\!\left(\dfrac{10}{26}\right)\approx 22^\circ$ (correct answer)
2. $\cos^{-1}\!\left(\dfrac{10}{26}\right)\approx 68^\circ$
3. $\tan^{-1}\!\left(\dfrac{26}{10}\right)\approx 69^\circ$
4. $\sin^{-1}\!\left(\dfrac{26}{10}\right)\approx 90^\circ$

Explanation: We have a right triangle with hypotenuse 26 and one leg 10, and θ is the angle opposite the leg of length 10. To find θ, we use sine since we have the opposite side and hypotenuse: $sin(θ) = opposite/hypotenuse = 10/26$. Therefore, $θ = sin^{-1}(10/26)$ ≈ $sin^{-1}(0.385)$ ≈ $22^\circ$. Be careful not to use the wrong inverse function; since we're using opposite/hypotenuse, we need inverse sine, not inverse cosine or tangent.

Question 9

1. $\dfrac{12}{\cos 35^\circ}$
2. $12\cos 35^\circ$ (correct answer)
3. $12\sin 35^\circ$
4. $12\tan 35^\circ$

Explanation: We need to find the length of the leg adjacent to angle A in a right triangle. Since we have the hypotenuse (12 cm) and need the adjacent side to the 35° angle, we use cosine: cos(35°) = adjacent/hypotenuse. Setting up the equation: cos(35°) = AC/12, so AC = 12cos(35°). A common error is using sine instead of cosine, which would give the opposite side rather than the adjacent side. Remember: CAH in SOH-CAH-TOA means Cosine = Adjacent/Hypotenuse.

Question 10

1. $\dfrac{120}{\tan 31^\circ}\approx 200$
2. $120\sin 31^\circ\approx 62$
3. $120\tan 31^\circ\approx 72$ (correct answer)
4. $\dfrac{120}{\sin 31^\circ}\approx 233$

Explanation: The surveyor is 120 m from the building's base, and the angle of elevation is 31°. We need the building's height, which is the side opposite to the 31° angle, while 120 m is the adjacent side. Using tangent: tan(31°) = opposite/adjacent = height/120. Therefore, height = 120tan(31°) ≈ 120(0.601) ≈ 72 m. A common mistake is using sine or cosine, but remember that when you have adjacent and need opposite (or vice versa), use tangent.

Question 11

1. $\frac{15}{12}$
2. $\frac{9}{12}$
3. $\frac{12}{15}$ (correct answer)
4. $\frac{9}{15}$

Explanation: We need to find $sin(\theta)$ where $\theta$ is angle P in a right triangle with legs $PQ = 9$ and $QR = 12$. First, we find the hypotenuse using the Pythagorean theorem: $PR = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$. Since QR is opposite to angle P and PR is the hypotenuse, $sin(P) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{15}$. A common error is confusing which side is opposite to angle P; remember that the opposite side doesn't touch the angle. Always label your triangle clearly to avoid mixing up sides.

Question 12

1. 14.7 ft (correct answer)
2. 6.6 ft
3. 24.0 ft
4. 13.5 ft

Explanation: We need to find the length of a ramp that rises 6 ft at a 24° angle with the ground. The 6 ft height is the side opposite the 24° angle, and the ramp length is the hypotenuse, so we use sine: sin(24°) = 6/ramp length. Solving: ramp length = 6/sin(24°) = 6/0.407 ≈ 14.7 ft. A common mistake is using cosine or tangent instead of sine when the opposite side and hypotenuse are involved. Remember that sine relates the opposite side to the hypotenuse in SOH-CAH-TOA.

Question 13

1. $\frac{2}{7}$
2. $\frac{7}{2}$ (correct answer)
3. $\frac{9}{7}$
4. $\frac{7}{9}$

Explanation: This problem uses the complementary angle relationship for tangent. If $tan(\theta) = \frac{2}{7}$, then $tan(90^\circ - \theta)$ equals the reciprocal: $tan(90^\circ - \theta) = \frac{7}{2}$. This follows from the cofunction identity: $tan(90^\circ - \theta) = cot(\theta) = \frac{1}{tan(\theta)}$. Since $tan(\theta) = \frac{2}{7}$, we have $tan(90^\circ - \theta) = \frac{1}{\frac{2}{7}} = \frac{7}{2}$. A common mistake is thinking $tan(90^\circ - \theta) = -tan(\theta)$ or forgetting the reciprocal relationship. Remember that complementary angles in a right triangle swap the roles of opposite and adjacent sides.

Question 14

1. $37^\circ$
2. $60^\circ$
3. $45^\circ$
4. $53^\circ$ (correct answer)

Explanation: Points A(0,0), B(6,0), and C(6,8) form a right triangle with the right angle at B. To find angle A, we note that AB = 6 (horizontal) and BC = 8 (vertical), so for angle A, the opposite side is BC = 8 and the adjacent side is AB = 6. Using tangent: tan(A) = 8/6 = 4/3. Taking inverse tangent: angle A = tan⁻¹(4/3) ≈ 53.13°, which rounds to 53°. Coordinate geometry problems often simplify to basic right triangle trigonometry once you identify the sides.

Question 15

1. 34 m
2. 62 m (correct answer)
3. 190 m
4. 211 m

Explanation: A drone travels 200 m along a straight path at an 18° angle of elevation, and we need its height above ground. The 200 m is the hypotenuse of the right triangle, and the height is the opposite side to the 18° angle, so we use sine: sin(18°) = height/200. Solving: height = 200 × sin(18°) = 200 × 0.309 ≈ 61.8 m, which rounds to 62 m. A common mistake is treating the 200 m as the horizontal distance and using tangent, which would give a much larger height. In elevation problems, the distance traveled along the incline is the hypotenuse.

Question 16

1. $37^\circ$
2. $42^\circ$
3. $53^\circ$ (correct answer)
4. $58^\circ$

Explanation: In right triangle XYZ with right angle at Y, XY = 9 and YZ = 12, we need angle X. Since YZ is opposite to angle X and XY is adjacent to angle X, we use tangent: tan(X) = opposite/adjacent = 12/9 = 4/3. Taking the inverse: angle X = arctan(4/3) ≈ 53.13°, which rounds to 53°. A common error is finding angle Z instead, which would be arctan(9/12) = arctan(3/4) ≈ 37°. Always verify which angle you're finding by checking which sides are opposite and adjacent to it.

Question 17

1. $\frac{24}{26}$
2. $\frac{5}{13}$ (correct answer)
3. $\frac{10}{26}$
4. $\frac{12}{13}$

Explanation: We need $cos(\angle D)$ where the hypotenuse $DE = 26$ and the adjacent side $DF = 10$. Using the cosine ratio: $cos(\angle D) = \text{adjacent}/\text{hypotenuse} = DF/DE = 10/26 = 5/13$. The fraction 10/26 simplifies to 5/13 by dividing both numerator and denominator by 2. A common error is using the wrong side or not simplifying the fraction. Since F is the right angle and DF touches angle D, DF is adjacent to angle D. Always simplify fractions in your final answer.

Question 18

1. $10\sin 62^\circ$
2. $\dfrac{10}{\sin 62^\circ}$ (correct answer)
3. $\dfrac{10}{\cos 62^\circ}$
4. $10\cos 62^\circ$

Explanation: We need the hypotenuse AB when angle A = 62°, angle C = 90°, and BC = 10 (opposite to angle A). Using sine: sin(62°) = opposite/hypotenuse = 10/AB. Solving for AB: AB = 10/sin(62°). A common error is multiplying instead of dividing, or using the wrong trigonometric ratio. Remember SOH: when you know the opposite side and need the hypotenuse, use sine and solve by dividing.

Question 19

1. $\frac{9}{41}$
2. $\frac{40}{41}$ (correct answer)
3. $\frac{41}{40}$
4. $\frac{40}{9}$

Explanation: With legs of 9 and 40, we first find the hypotenuse: $c^2 = 9^2 + 40^2 = 81 + 1600 = 1681$, so $c = 41$. For angle $\theta$ opposite the leg of length 9, the adjacent leg is 40. Therefore, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{40}{41}$. This is another Pythagorean triple (9-40-41) that's useful to recognize. The key is correctly identifying which leg is adjacent to $\theta$ - it's the one that isn't opposite to $\theta$.

Question 20

1. $40$ m (correct answer)
2. $51$ m
3. $25$ m
4. $81$ m

Explanation: The kite string is 65 m (hypotenuse) making a 38° angle with the ground, and we need the height (opposite side). Using sine: sin(38°) = height/65. Solving: height = 65 × sin(38°) = 65 × 0.616 ≈ 40.0 m. The key is recognizing that the string is the hypotenuse and the height is opposite to the ground angle. A common error is using cosine, which would give the horizontal distance instead of the vertical height.