ACT Math study guide · Chapter 26
Trigonometry Beyond Right Triangles
Three trigonometric identities let one ratio be rewritten in terms of another, without needing a calculator or a specific angle. The Pythagorean identity, sin²θ + cos²θ = 1, connects sine and cosine directly. The reciprocal identities, cscθ = 1/sinθ, secθ = 1/cosθ, and cotθ = 1/tanθ, invert the three basic ratios. The quotient identity, tanθ = sinθ/cosθ, expresses tangent in terms of sine and cosine.
A related identity, 1 + tan²θ = sec²θ, follows directly from the Pythagorean identity by dividing every term by cos²θ. Recognizing that these identities are all connected, instead of memorizing each in isolation, makes simplifying an unfamiliar trig expression far more manageable.
The Law of Sines and Law of Cosines extend right-triangle trigonometry to triangles with no right angle. The Law of Sines states that a/sin A = b/sin B = c/sin C, relating each side to the sine of its opposite angle; it applies when two angles and a side are known, or two sides and a non-included angle. The Law of Cosines, c² = a² + b² − 2ab·cos C, applies when two sides and the included angle are known, or when all three sides are known and an angle is needed.
When the givens are two sides and a non-included angle (SSA), check for the ambiguous case. The sine equation can produce a second possible angle, so the information may describe zero, one, or two triangles. Any second angle must still leave the triangle with a positive third angle.
Before picking up a calculator, identify which of three routes the triangle actually calls for.
| What the triangle gives you | Route to use |
|---|---|
| A right angle, with two known side or angle relationships | Sine, cosine, tangent, or the Pythagorean theorem |
| No right angle, but a side is paired with its opposite angle, plus one more angle or side | Law of Sines is usually the shortest route |
| Two sides and the included angle, or all three sides | Law of Cosines |
If none of these patterns is visible in the given information, that's a signal to re-read the figure and identify what is actually paired, not to start entering numbers into a calculator.
Where this appears as a model: two paths leaving a shared starting point at an angle to each other. Two ships leaving the same port, one traveling 30 miles and the other 40 miles along paths that form a 70° angle, end up about 40.98 miles apart, found with the Law of Cosines once the triangle is recognized as having no right angle.
▸ Using an identity to simplify
Worked Example If sin θ = 5/13 and θ is in the first quadrant, find cos θ.
Apply the Pythagorean identity.
cos²θ = 1 − 25/169 = 144/169
cosθ = 12/13 (positive in the first quadrant)
▸ The Law of Sines
Worked Example In triangle ABC, angle A = 55°, angle B = 65°, and side a = 9. Find side b, rounded to the nearest tenth.
Set up the Law of Sines using the known angle-side pair.
9/sin(55°) = b/sin(65°)
b = 9 · sin(65°)/sin(55°) ≈ 10.0
▸ The Law of Cosines
Worked Example A triangle has sides of length 6 and 8, with an included angle of 50°. Find the length of the third side, rounded to the nearest hundredth.
Use the Law of Cosines with the two known sides and their included angle.
c² = 6² + 8² − 2(6)(8)cos(50°)
c² = 100 − 96cos(50°) ≈ 38.29
c ≈ 6.19
Worked Example Simplify (1 − cos²θ)/(sinθ · cosθ).
Recognize. The numerator matches the Pythagorean identity: 1 − cos²θ = sin²θ. Rewrite it before simplifying, instead of canceling anything directly.
Set up.
(1 − cos²θ)/(sinθ·cosθ) = sin²θ/(sinθ·cosθ)
Solve.
= sinθ/cosθ = tanθ
Check. Testing θ = 45°: the original expression gives (1−0.5)/(0.7071×0.7071) = 0.5/0.5 = 1, and tan(45°) = 1, which match.
Best route. Hand simplification using the identity. A calculator can spot-check one specific angle, but it does not simplify the general expression.
Common error. Canceling cos²θ directly against the cosθ in the denominator without first rewriting 1 − cos²θ as sin²θ.
▸ Practice
1. If sin θ = 3/5 and θ is in the first quadrant, find cos θ using the Pythagorean identity.
A) 16/25 B) 4/5 C) 3/5 D) 5/4
2. Simplify sin θ · csc θ.
F) 0 G) 1 H) 2 sin θ J) sin²θ
3. If cos θ = 5/13 and θ is in the first quadrant, find sin θ.
A) 12/13 B) 8/13 C) 5/13 D) 13/12
4. Simplify tan θ · cos θ.
F) tan θ G) 1 H) sin θ J) cos θ
5. If tan θ = 4/3 and θ is in the first quadrant, find sec θ.
A) 5/3 B) 4/3 C) 25/9 D) 3/5
6. Simplify (1 − cos²θ) / sin²θ.
F) cos²θ G) 1 H) 0 J) sin²θ
7. If sin θ = 7/25 and θ is in the first quadrant, find cot θ.
A) 7/24 B) 24/7 C) 25/7 D) 24/25
8. Simplify sec θ · cos θ.
F) 0 G) sec²θ H) 1 J) 2 cos θ
9. Find sin²(30°) + cos²(30°).
A) 3/4 B) 1/2 C) 1 D) 1/4
10. Simplify csc θ · tan θ.
F) cos θ G) sec θ H) 1 J) csc θ
11. In triangle ABC, a = 10, angle A = 30°, and angle B = 45°. Find side b.
A) 10√3 B) 5√2 C) 5 D) 10√2
12. In triangle ABC, angle A = 40°, angle B = 60°, and side a = 8. Find side b, rounded to the nearest tenth.
F) 10.8 G) 11.5 H) 5.9 J) 9.2
13. In triangle ABC, a = 7, b = 9, and angle C = 60°. Find side c, rounded to the nearest hundredth.
A) 6.24 B) 9.85 C) 11.40 D) 8.19
14. In triangle ABC, a = 5, b = 6, and c = 7. Find angle C, rounded to the nearest tenth of a degree.
F) 78.5° G) 82.3° H) 66.4° J) 73.1°
15. In triangle ABC, angle A = 50°, angle C = 70°, and side c = 12. Find side a, rounded to the nearest tenth.
A) 8.4 B) 9.8 C) 14.7 D) 10.6
16. Two sides of a triangle measure 8 and 10, with an included angle of 45°. Find the length of the third side, rounded to the nearest hundredth.
F) 6.02 G) 12.81 H) 8.94 J) 7.13
17. A triangle has sides 5, 7, and 10. Find the measure of the largest angle, rounded to the nearest tenth of a degree.
A) 104.6° B) 40.5° C) 111.8° D) 97.2°
18. Simplify sin²θ − 1.
F) 1 − sin²θ G) cos²θ H) −sin²θ J) −cos²θ
19. In triangle ABC, angle A = 35°, angle B = 85°, and side a = 6. Find side c, rounded to the nearest tenth.
A) 9.1 B) 7.6 C) 4.0 D) 10.3
20. If cos θ = −1/2 and θ is between 90° and 180°, find sin θ.
F) √3/2 G) −1/2 H) −√3/2 J) 1/2
▸ Answers and Explanations
1: B. The Pythagorean identity is sin²θ + cos²θ = 1: cos²θ = 1 − 9/25 = 16/25, so cosθ = 4/5 (positive in the first quadrant). Choice A reports cos²θ directly without taking the square root.
2: G. cscθ = 1/sinθ, so sinθ · cscθ = sinθ · (1/sinθ) = 1. Choice F treats the product as if the two terms canceled to zero instead of to one.
3: A. sin²θ = 1 − cos²θ = 1 − 25/169 = 144/169, so sinθ = 12/13 (positive in the first quadrant). Choice B reports the given cosθ value instead of solving for sinθ.
4: H. tanθ = sinθ/cosθ, so tanθ · cosθ = (sinθ/cosθ) · cosθ = sinθ. Choice G reports 1, mistaking this product for secθ·cosθ instead of tanθ·cosθ.
5: A. The identity 1 + tan²θ = sec²θ gives sec²θ = 1 + 16/9 = 25/9, so secθ = 5/3 (positive in the first quadrant). Choice B reports the given tanθ value instead of solving for secθ.
6: G. Since 1 − cos²θ = sin²θ by the Pythagorean identity, the expression simplifies to sin²θ/sin²θ = 1. Choice H treats the numerator and denominator as canceling to zero instead of to one.
7: B. cos²θ = 1 − 49/625 = 576/625, so cosθ = 24/25 (positive in the first quadrant). cotθ = cosθ/sinθ = (24/25)/(7/25) = 24/7. Choice A inverts the ratio, reporting tanθ (7/24) instead of cotθ (24/7).
8: H. secθ = 1/cosθ, so secθ · cosθ = (1/cosθ) · cosθ = 1. Choice F treats the product as if the two terms canceled to zero instead of to one.
9: C. By the Pythagorean identity, sin²θ + cos²θ = 1 for any angle θ, including 30°. Choice B reports sin²(30°) alone, forgetting to add cos²(30°).
10: G. cscθ · tanθ = (1/sinθ) · (sinθ/cosθ) = 1/cosθ = secθ. Choice J reports cscθ alone, forgetting to complete the multiplication by tanθ.
11: D. By the Law of Sines, a/sin A = b/sin B: 10/sin(30°) = b/sin(45°), so b = 10 · sin(45°)/sin(30°) = 10√2. Choice B inverts the ratio, computing 10·sin(30°)/sin(45°) instead of 10·sin(45°)/sin(30°).
12: F. By the Law of Sines: 8/sin(40°) = b/sin(60°), so b = 8 · sin(60°)/sin(40°) ≈ 10.8. Choice H inverts the ratio, computing 8·sin(40°)/sin(60°) instead of 8·sin(60°)/sin(40°).
13: D. By the Law of Cosines, c² = a² + b² − 2ab·cos C: c² = 49 + 81 − 2(7)(9)cos(60°) = 130 − 63 = 67, so c ≈ 8.19. Choice A omits the −2ab·cos C term entirely, computing √(a²+b²) as if this were a right triangle.
14: F. By the Law of Cosines solved for the angle: cos C = (a²+b²−c²)/(2ab) = (25+36−49)/60 = 0.2, so C = cos⁻¹(0.2) ≈ 78.5°. Choice H divides by ab instead of 2ab, giving an incorrect cosine value and a wrong angle.
15: B. First find angle B = 180° − 50° − 70° = 60°. By the Law of Sines: a/sin(50°) = 12/sin(70°), so a = 12 · sin(50°)/sin(70°) ≈ 9.8. Choice C inverts the ratio, computing 12·sin(70°)/sin(50°) instead of 12·sin(50°)/sin(70°).
16: J. By the Law of Cosines: c² = 8² + 10² − 2(8)(10)cos(45°) = 164 − 160cos(45°) ≈ 50.86, so c ≈ 7.13. Choice G omits the −2ab·cos C term entirely, computing √(a²+b²) as if this were a right triangle.
17: C. The largest angle is opposite the longest side (10). By the Law of Cosines: cosθ = (5²+7²−10²)/(2·5·7) = −26/70 ≈ −0.371, so θ ≈ 111.8°. Choice B finds the angle opposite the side of length 7 instead of the side of length 10, computing the wrong angle entirely.
18: J. Since sin²θ = 1 − cos²θ by the Pythagorean identity, sin²θ − 1 = (1−cos²θ) − 1 = −cos²θ. Choice F leaves the expression in its unsimplified form instead of applying the Pythagorean identity to reach −cos²θ.
19: A. First find angle C = 180° − 35° − 85° = 60°. By the Law of Sines: c/sin(60°) = 6/sin(35°), so c = 6 · sin(60°)/sin(35°) ≈ 9.1. Choice C inverts the ratio, computing 6·sin(35°)/sin(60°) instead of 6·sin(60°)/sin(35°).
20: F. sin²θ = 1 − cos²θ = 1 − 1/4 = 3/4, so sinθ = √3/2. Sine is positive between 90° and 180° (the second quadrant). Choice G reports the negative root, overlooking that sine is positive throughout the second quadrant.