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Lesson 15 — Law of sines and law of cosines

Solving general (non-right) triangles. Applications in surveying, navigation, and physics.

Used in: 1.º ano do EM (15 anos) · Math II japonês (cap. 図形と計量) · Trigonometry — US precalc

asinA=bsinB=csinC=2R,c2=a2+b22abcosC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R, \quad c^2 = a^2 + b^2 - 2ab\cos C
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Rigorous notation, full derivation, hypotheses

Proofs and use

Law of sines

what this means · Holds for any triangle (acute, obtuse, right). R is the radius of the circumscribed circle.

Proof (for an acute triangle): construct the altitude hh from vertex CC to side AB\overline{AB}. Then h=bsinA=asinBh = b \sin A = a \sin B. Therefore a/sinA=b/sinBa/\sin A = b/\sin B. Same argument for cc. ∎

Special case (right angle at CC): sinC=1\sin C = 1, so c=2Rc = 2R — the hypotenuse is the diameter of the circumscribed circle. Thales' theorem (geometric).

Law of cosines

what this means · Generalizes Pythagoras. When C = 90°, cos C = 0 and we recover c² = a² + b².

Proof: by the dot product of vectors CBCA=AB\vec{CB} - \vec{CA} = \vec{AB}: AB2=CB2+CA22CBCA|\vec{AB}|^2 = |\vec{CB}|^2 + |\vec{CA}|^2 - 2 \vec{CB} \cdot \vec{CA}

Since CBCA=CBCAcosC=abcosC\vec{CB} \cdot \vec{CA} = |\vec{CB}||\vec{CA}|\cos C = ab \cos C, we obtain c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab \cos C. ∎

When to use each law

You haveYou wantUse
2 angles + 1 side (AAS, ASA)the other sidesLaw of sines
2 sides + angle opposite to one (SSA)remaining (ambiguous!)Law of sines
2 sides + angle between them (SAS)third sideLaw of cosines
3 sides (SSS)any angleLaw of cosines inverted

Ambiguous case (SSA)

Given aa, bb, and AA (angle opposite to aa): there may be 0, 1, or 2 triangles. Decision:

  • If aba \geq b: 1 triangle.
  • If a<bsinAa < b \sin A: 0 triangles (geometrically impossible).
  • If bsinA<a<bb \sin A < a < b: 2 triangles.

Exercise list

35 exercises · 8 with worked solution (25%)

Application 18Understanding 2Modeling 12Proof 3
  1. Ex. 15.1Application
    Triangle with a=8a = 8, A=30°A = 30°, B=45°B = 45°. Calculate bb.
  2. Ex. 15.2Application
    Triangle with a=12a = 12, A=50°A = 50°, B=70°B = 70°. Calculate bb and cc.
  3. Ex. 15.3Application
    Triangle with a=5a = 5, b=8b = 8, A=30°A = 30°. How many triangles are possible?
  4. Ex. 15.4ApplicationAnswer key
    Triangle with b=10b = 10, B=45°B = 45°, A=60°A = 60°. Calculate aa.
  5. Ex. 15.5ApplicationAnswer key
    In a triangle ABCABC, A=40°A = 40°, B=80°B = 80°, a=7a = 7. Calculate CC and cc.
  6. Ex. 15.6Application
    Triangle with a=6a = 6, A=35°A = 35°, B=50°B = 50°. Calculate area.
  7. Ex. 15.7ApplicationAnswer key
    Law of sines: a/sin30°=c/sin90°a/\sin 30° = c/\sin 90°. For a=4a = 4, calculate cc.
  8. Ex. 15.8Application
    In a triangle, a=10a = 10, b=7b = 7, A=90°A = 90°. Confirm with the law of sines.
  9. Ex. 15.9Application
    Triangle: A=50°A = 50°, a=12a = 12. Determine the radius of the circumscribed circle RR.
    Solve onlineref: OpenStax A&T §10.1
  10. Ex. 15.10Understanding
    Show that in an equilateral triangle (A=B=C=60°A = B = C = 60°), a=b=ca = b = c.
  11. Ex. 15.11Application
    Triangle with a=5a = 5, b=7b = 7, C=60°C = 60°. Calculate cc.
  12. Ex. 15.12Application
    Triangle with a=8a = 8, b=6b = 6, C=90°C = 90°. Calculate cc. (Recover Pythagoras.)
  13. Ex. 15.13Application
    Triangle with a=4a = 4, b=3b = 3, C=120°C = 120°. Calculate cc.
  14. Ex. 15.14Application
    Triangle with a=5a = 5, b=6b = 6, c=7c = 7. Calculate CC.
  15. Ex. 15.15Application
    Triangle with a=10a = 10, b=12b = 12, c=15c = 15. Determine the 3 angles.
  16. Ex. 15.16ApplicationAnswer key
    In a triangle, a=12a = 12, b=8b = 8, A=80°A = 80°. Use the law of sines for BB and then calculate cc.
  17. Ex. 15.17Application
    Triangle ABCABC: a=4a = 4, b=5b = 5, c=6c = 6. Calculate area using Heron's formula.
  18. Ex. 15.18Application
    In an equilateral triangle of side \ell, show via the law of cosines that each angle is 60°60°.
  19. Ex. 15.19Application
    Triangle with sides 7,24,257, 24, 25. Verify it is a right triangle via the law of cosines.
  20. Ex. 15.20Understanding
    When C0C \to 0, what does the law of cosines c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab\cos C tend to? Interpret geometrically.
  21. Ex. 15.21Modeling
    You walk 5 km east, then turn 60°60° north and walk 3 more km. Distance from the origin?
  22. Ex. 15.22Modeling
    A ship leaves port, sails 12 km northwest, then 8 km northeast. Distance from origin?
  23. Ex. 15.23Modeling
    A drone observes two points AA and BB on the ground at angles of 50°50° and 70°70°. Drone at 200 m altitude. Calculate distance ABAB.
  24. Ex. 15.24ModelingAnswer key
    Two sides of a triangular plot measure 80 m and 100 m, forming a 75°75° angle. Length of the third side?
  25. Ex. 15.25ModelingAnswer key
    In a soccer field, an attacker shoots from the position seeing the 6-meter goal at a 20°20° angle from position AA (dist. to goal = 30 m). Distance from goal to attacker from AA? (Goal geometry and angle.)
  26. Ex. 15.26Modeling
    Surveying: you need to measure the distance between two points AA and BB separated by a river. You're at CC, with ACB^=60°\hat{ACB} = 60°, AC=50AC = 50 m, BC=70BC = 70 m. Distance ABAB?
  27. Ex. 15.27ModelingAnswer key
    Astronomy: stellar parallax of a star measures an angle of π/(3606060)\pi/(360 \cdot 60 \cdot 60) rad (1 arc-second) from one side to the other of Earth's orbit. What is the distance to the star in AU? (Ans: 206,265 AU = 1 parsec.)
  28. Ex. 15.28Modeling
    An irrigation triangle has sides 100m, 120m, 80m. Area?
  29. Ex. 15.29Modeling
    Inverse kinematics: a robotic arm with 2 segments 1=30\ell_1 = 30 cm, 2=25\ell_2 = 25 cm needs to reach a point at distance r=40r = 40 cm. Angle between segments?
  30. Ex. 15.30ModelingAnswer key
    Resultant velocity of a boat at 55 km/h in a river with 33 km/h perpendicular current: magnitude and angle?
  31. Ex. 15.31Modeling
    A plane travels at 500 km/h on heading 60°60° NE. Wind blows at 100 km/h from the east. Resultant velocity?
  32. Ex. 15.32Modeling
    In 2D GPS, two satellites at (0,100)(0, 100) and (50,80)(50, 80) km see you at angles 30°30° and 45°45° — describe (do not calculate) the triangulation.
  33. Ex. 15.33Proof
    Prove the law of sines for an acute triangle, using the altitude from vertex CC.
  34. Ex. 15.34Proof
    Prove the law of cosines for any triangle, using the dot product.
  35. Ex. 15.35Proof
    Prove Heron's formula using the law of cosines + area = (1/2)ab sin C.

Sources for this lesson

Updated on 2026-04-29 · Author(s): Clube da Matemática

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