ACT Math Trigonometry: 15 Practice Problems with Step-by-Step Explanations
Trigonometry on the ACT splits into two related but distinct skill sets: right triangle trigonometry, where SOH-CAH-TOA (sine = opposite/hypotenuse, cosine = adjacent/hypotenuse, tangent = opposite/adjacent) and the Pythagorean identity (sin2θ + cos2θ = 1) do most of the work, and periodic function trigonometry, where you're reading or building sine and cosine graphs and identifying amplitude, period, and midline. Both halves show up regularly, and a single harder problem will sometimes combine them — for example, giving you a triangle and asking for an inverse trig expression, or giving you one trig ratio and asking you to find another using the identity.
For right triangle problems, the key is correctly identifying which side is opposite, adjacent, or the hypotenuse relative to the angle you're working with — that identification, not the arithmetic, is where most errors happen. When a problem gives you one trig ratio (like sinθ) and asks for another (like cosθ or tanθ), it often helps to sketch the implied right triangle using the two numbers given as two of the three sides, then find the third side with the Pythagorean theorem before computing the requested ratio.
For graphing problems, y = A sin(Bx+C)+D (or cosine) has amplitude |A|, period 2π/|B|, and vertical shift (midline) D. Given specific points on a graph — like a maximum, a minimum, and where the curve crosses its midline — you can back out each of these values directly: amplitude is half the distance between the max and min, the midline is their average, and the period is the horizontal distance between two points where the same behavior repeats. Also watch for quadrant-based sign rules: when a problem restricts θ to a range like 90° to 180°, that tells you the sign of sine, cosine, or tangent before you even calculate. Work through the 15 problems below, including several that reference a figure, then click to reveal each step-by-step explanation.
On this page:
- Problem 1 — Sine from a triangle
- Problem 2 — Cosine from a triangle
- Problem 3 — Tangent from sine and cosine
- Problem 4 — Pythagorean identity
- Problem 5 — Solving a triangle side
- Problem 6 — Ladder word problem
- Problem 7 — Inverse trig from a triangle
- Problem 8 — Amplitude
- Problem 9 — Quadrant sign rules
- Problem 10 — Amplitude and period
- Problem 11 — Building a function from a graph
- Problem 12 — Quadrant sign rules
- Problem 13 — Pythagorean identity
- Problem 14 — Finding sine from tangent
- Problem 15 — Building ratios from given sides
Practice Problems
The side lengths of a triangle are given in the figure below. What is sin C?
- A) 12/6
- B) 6/9
- C) 9/6
- D) 6/12
Using SOH (sine = opposite/hypotenuse) relative to angle C: the side opposite C is AB = 6, and the hypotenuse (the side opposite the right angle) is BC = 12.
sin C = 6/12.
Two of the side lengths of a right triangle are given in the figure below. What is cos C?
- A) 3/5
- B) 5/3
- C) 5/4
- D) 4/5
Since this is a right triangle with legs 3 and 4, the hypotenuse is 5 (the classic 3-4-5 triple, confirmed by 32+42=52).
Using CAH (cosine = adjacent/hypotenuse) relative to angle C: the side adjacent to C is AC = 4, and the hypotenuse is 5. cos C = 4/5.
For an angle with measure θ, sinθ = 12/13 and cosθ = 5/13. What is the value of tanθ?
- A) 5/12
- B) 13/12
- C) 12/5
- D) 5/13
Tangent is the ratio of sine to cosine: tanθ = sinθ/cosθ = (12/13)/(5/13).
The 13's cancel, leaving tanθ = 12/5.
5sin2θ + 5cos2θ = x. For the function above, what is the value of x?
- A) −5
- B) 1
- C) 5
- D) 8
Factor out the common 5: 5(sin2θ + cos2θ) = x.
By the Pythagorean identity, sin2θ + cos2θ = 1 always, so x = 5(1) = 5.
If sinθ = 0.8 then in the triangle below, what is the value of x?
- A) 3.6
- B) 12.2
- C) 14.4
- D) 21.6
Relative to θ, the side x is opposite the angle, and 18 is the hypotenuse. Using SOH: sinθ = x/18.
Solve for x: x = 18 × sinθ = 18 × 0.8 = 14.4.
The figure below shows a 7-foot ladder leaning against a vertical wall. The ladder makes a 47° angle. Which of the following expressions gives the height where the top of the ladder hits the wall?
- A) 7 cos 47°
- B) 7 tan 47°
- C) 7 sin 47°
- D) 7/cos 47°
The ladder is the hypotenuse (7 ft), and the height on the wall is the side opposite the 47° angle at the base. Using SOH: sin(47°) = height/7.
Solve for height: height = 7 sin(47°).
Which expression correctly solves for the measure of angle θ in the figure below?
- A) sin−1(17/15)
- B) sin−1(15/17)
- C) cos−1(17/8)
- D) tan−1(8/15)
The two legs shown are 8 and 15 — recognizing the 8-15-17 Pythagorean triple (82+152=172) gives the hypotenuse of 17 without needing to compute a square root.
Relative to θ, the opposite side is 15 and the hypotenuse is 17, so sinθ = 15/17. Taking the inverse sine of both sides: θ = sin−1(15/17).
For the function y = −5sin(2.5x) + 12, what is the amplitude?
- A) −5
- B) 1
- C) 2.5
- D) 5
In y = A sin(Bx+C)+D, the amplitude is |A| — the absolute value of the coefficient in front of sine, since amplitude describes a distance and can't be negative.
Here A = −5, so the amplitude is |−5| = 5.
If 0° < θ < 180° and tanθ = 5/12, then sinθ = ?
- A) −5/13
- B) −12/13
- C) 5/13
- D) 12/13
Tangent is positive in the given range (0° to 180°) only in the first quadrant (0° to 90°), so θ is acute. Since tanθ = 5/12, sketch a right triangle with opposite=5, adjacent=12; by the Pythagorean theorem, the hypotenuse is 13 (a 5-12-13 triple).
sinθ = opposite/hypotenuse = 5/13.
What is the sum of the amplitude and period of the function y = (5/2)sin(x/2 + 10) + 1?
- A) 5/2 + 4π
- B) 5/2 + π/2
- C) 10 + 4π
- D) 10 + π/2
The amplitude is the coefficient in front of sine: 5/2. The period is 2π divided by the coefficient of x inside the sine function: here that coefficient is 1/2, so the period is 2π/(1/2) = 4π.
Sum: 5/2 + 4π.
The function f(x) and three points on the function are shown below. Which of the following equations correctly describes the function f(x)?
f(x) with points (0, 8), (2π, −4), and (4π, 8)
- A) y = 12sin(2x) + 2
- B) y = 6sin(x/2) + 2
- C) y = 12cos(2x) + 2
- D) y = 6cos(x/2) + 2
Since f(0) = 8 is the maximum value and f(2π) = −4 is the minimum, the amplitude is half their difference: (8−(−4))/2 = 6. The midline (vertical shift) is their average: (8+(−4))/2 = 2.
Since f(0)=8 is a maximum (the curve starts at its peak, matching how cosine behaves at x=0), a cosine function fits, not sine (which starts at the midline). The period is 4π (the distance from one maximum back to the next), so the coefficient of x is 2π/4π = 1/2.
Combined: y = 6cos(x/2) + 2. Checking: at x=2π, y=6cos(π)+2=−6+2=−4 ✓.
If 90° < θ < 180° and sinθ = 8/17, then cosθ = ?
- A) −17/15
- B) −15/17
- C) −8/17
- D) 15/17
Since sinθ = 8/17, use the 8-15-17 Pythagorean triple to find the reference triangle's third side: 15. So the reference value for cosθ is 15/17.
The given range, 90° to 180°, is the second quadrant, where cosine is negative. So cosθ = −15/17.
If sin2θ = 9/11, what does cos2θ equal?
- A) −9/11
- B) 2/11
- C) 3/√11
- D) 6/11
By the Pythagorean identity, sin2θ + cos2θ = 1, so cos2θ = 1 − sin2θ.
Substitute: cos2θ = 1 − 9/11 = 2/11.
If tanθ = 4/7, which of the following is a possible value of sinθ?
- A) 7/√65
- B) 4/√65
- C) 3/7
- D) 33/49
Since tanθ = 4/7 = opposite/adjacent, sketch a right triangle with opposite=4 and adjacent=7. By the Pythagorean theorem, the hypotenuse is √(42+72) = √65.
sinθ = opposite/hypotenuse = 4/√65.
In a right triangle, sinθ = m/n and cotθ = p/m. secθ = ?
- A) m/p
- B) n/m
- C) p/n
- D) n/p
From sinθ = m/n: the opposite side is m, and the hypotenuse is n. Cotangent is adjacent/opposite, so cotθ = p/m means adjacent/m = p/m, giving adjacent = p.
Secant is hypotenuse/adjacent: secθ = n/p.
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