Lesson 8: The Law of Sines and the Law of Cosines

Trigonometry so far required a right angle. Real triangles — fields, flight paths, mountain profiles — rarely cooperate. The Law of Sines and the Law of Cosines generalise everything to *any* triangle. Choosing between them is a data question: what do you know, and what must you find?

The two laws

For a triangle with sides a, b, c opposite angles A, B, C respectively:

Law of Sines: a/sin A = b/sin B = c/sin C. Use it when you know an angle-side pair plus one more piece (ASA, AAS, or SSA). Each ratio equals the diameter of the triangle's circumcircle, though you rarely need that fact.

Law of Cosines: c² = a² + b² − 2ab · cos C. Use it when you know two sides and the included angle (SAS, to find the third side) or all three sides (SSS, to find an angle: cos C = (a² + b² − c²)/(2ab)). Notice it is Pythagoras with a correction term: when C = 90°, cos C = 0 and the formula collapses to c² = a² + b².

The ambiguous case (SSA). Given an angle, an adjacent side and the opposite side, there may be two valid triangles, one, or none — because sin θ = sin(180° − θ). Always check whether the supplementary solution also fits (the two candidate angles plus the given angle must stay under 180°).

A practical decision rule: if you have an angle-side opposite pair, reach for the Law of Sines; if not, reach for the Law of Cosines first, then switch to Sines once you own a pair.

Worked example 1: Law of Sines (AAS)

Triangle with A = 40°, B = 75° and a = 10. First, C = 180° − 40° − 75° = 65°. The pair (a, A) = (10, 40°) sets the common ratio: 10/sin 40° ≈ 10/0.6428 ≈ 15.56. Then b = 15.56 · sin 75° ≈ 15.56 · 0.9659 ≈ 15.03, and c = 15.56 · sin 65° ≈ 15.56 · 0.9063 ≈ 14.10. Sanity check: the largest side (b ≈ 15.03) sits opposite the largest angle (75°). ✓

Worked example 2: Law of Cosines (SAS)

Two sides measure 8 and 11 with an included angle of 60°. The third side: c² = 8² + 11² − 2 · 8 · 11 · cos 60° = 64 + 121 − 176 · 0.5 = 185 − 88 = 97. So c = √97 ≈ 9.85. Notice c is shorter than the 11 side — reasonable, since the 60° angle "pulls" the ends closer than a right angle would (Pythagoras would have given √185 ≈ 13.6).

Worked example 3: Law of Cosines for an angle (SSS)

A triangle has sides 5, 7 and 10. Find the angle opposite the side of length 10. cos C = (5² + 7² − 10²)/(2 · 5 · 7) = (25 + 49 − 100)/70 = −26/70 ≈ −0.3714. Then C = cos⁻¹(−0.3714) ≈ 111.80°. A negative cosine signals an obtuse angle — the sign check is built into the method, which is one reason the Law of Cosines is preferred over Sines for finding angles: the Law of Sines alone cannot distinguish an angle from its supplement.

Worked example 4: The ambiguous case, decided

Given A = 30°, b = 10 and a = 8. Solve for angle B: sin B = (b · sin A)/a = (10 · 0.5)/8 = 0.625. So B ≈ 38.68° *or* B ≈ 180° − 38.68° = 141.32°. Check the second candidate: 141.32° + 30° = 171.32°, still under 180°, so both triangles exist: one with B ≈ 38.68°, C ≈ 111.32°, and a skinny one with B ≈ 141.32°, C ≈ 8.68°. Had a been 4 instead of 8, sin B would be 1.25 — impossible, so no triangle. Had a been exactly 5, sin B = 1 exactly: one right triangle. Always run this check on SSA data.

Geometry at work

Surveyors measure land they cannot walk by triangulation: measure one accessible baseline and the angles from its ends to a distant landmark, then the Law of Sines produces the inaccessible distances. Air traffic and shipping navigation solve the "wind triangle" with the Law of Cosines — ground speed results from air speed, wind speed and the angle between them. Astronomers historically used the same laws on the Earth–Moon–Sun triangle to estimate cosmic distances.

Common mistakes

• Reaching for the Law of Sines with no angle-side pair. If no side is matched with its opposite angle, start with Cosines.
• Forgetting the ambiguous case. SSA data may hide a second triangle; check the supplementary angle every time.
• Using the Law of Sines to find the largest angle. sin⁻¹ returns the acute candidate; use the Law of Cosines when an obtuse angle is possible.
• Dropping the −2ab·cos C term's sign. For an obtuse included angle, cos C is negative and the correction *adds* length.

Exam Experience

1. In △ABC, a/sin A = b/sin B is an example of which theorem?

• A) Pythagorean Theorem
• B) Law of Cosines
• C) Law of Sines
• D) Triangle Angle-Sum Theorem

Answer: C.

2. You know two sides of a triangle and the included angle, and must find the third side. Which tool applies directly?

• A) Law of Sines
• B) Law of Cosines
• C) Tangent ratio only
• D) Exterior Angle Theorem

Answer: B. SAS with no angle-side pair calls for the Law of Cosines.

3. Given A = 35°, a = 7 and b = 10, about how many triangles satisfy the data?

• A) None
• B) Exactly one
• C) Two
• D) Infinitely many

Answer: C. sin B = 10 · sin 35°/7 ≈ 10 · 0.5736/7 ≈ 0.8194, so B ≈ 55.0° or B ≈ 125.0°; both fit with A = 35° (sums 90.0° and 160.0°), so two triangles.

Constructed response. A surveyor models a plot of land as a triangle with A = 52°, b = 14 m and c = 9 m. (a) Find side a. (b) Then find angle B. Show each law you use and justify the choice.

Scoring guide (4 points):

• 1 point: Chooses the Law of Cosines for part (a), noting there is no angle-side pair yet (SAS data).
• 1 point: Computes a² = 14² + 9² − 2 · 14 · 9 · cos 52° = 277 − 155.15 = 121.85, so a ≈ 11.04 m.
• 1 point: For part (b), uses the Law of Sines with the now-known pair (a, A): sin B = 14 · sin 52°/11.04 ≈ 0.9994, so B ≈ 88.0°.
• 1 point: Finds C ≈ 180° − 52° − 88.0° ≈ 40.0° and checks consistency (largest side b = 14 sits opposite the largest angle B ≈ 88°).

Going deeper: Completing a triangle after the Law of Cosines

Return to Worked Example 2: sides 8 and 11 with included angle 60°, third side c = √97 ≈ 9.85. The triangle is not finished until its angles are known. Now that a side-angle pair exists (c opposite the 60° angle), the Law of Sines takes over. For the angle A opposite the side of length 8: sin A = (8 · sin 60°)/9.85 ≈ (8 · 0.8660)/9.85 ≈ 0.7034, so A ≈ 44.7°. (The supplement 135.3° is impossible: 135.3° + 60° > 180°.) Then B = 180° − 60° − 44.7° ≈ 75.3°. Check the ordering: the longest side, 11, faces the largest angle, 75.3°; the shortest side, 8, faces the smallest angle, 44.7°. ✓ This two-law relay — Cosines to manufacture a pair, Sines to finish — is the standard solution pattern for SAS and SSS data, and the ordering check at the end catches most slips.

One more discipline point: find the *largest* angle first when you have a choice, preferably with the Law of Cosines (which never confuses an angle with its supplement). Once the largest angle is pinned down, the remaining angles are acute, and the Law of Sines' acute answers are safe.

The area formula (1/2) · a · b · sin C

The two laws have a sibling: a triangle's area equals half the product of two sides times the sine of the *included* angle. For the SAS triangle of Worked Example 2 (sides 8 and 11, included angle 60°), the area is (1/2) · 8 · 11 · sin 60° = 44 · (√3/2) = 22√3 ≈ 38.11 square units. Where does it come from? Take side 11 as the base; the height is the side of length 8 projected vertically, which is 8 · sin 60° — Lesson 7's opposite side. Then area = (1/2) · base · height = (1/2) · 11 · 8 sin 60°. The formula matters because it needs no right angle and no altitude construction: any SAS data yields an area in one line. Surveyors pricing an odd triangular plot, and game programmers computing collision areas for triangular meshes, both live on it. It also explains the ambiguous case's geometry: for fixed a, b and angle A, the two possible triangles have the same area formula inputs... but different shapes — area alone never distinguishes them.

Final drill: Choosing the law in three scenarios

Exam speed comes from instant classification. Scenario one: a triangle with two known angles (50° and 65°) and a known side between... a side anywhere known, since the third angle (65°) is free and any known side finds its opposite angle — that is ASA/AAS data: Law of Sines, no Cosines needed. Scenario two: three known sides (7, 9, 13) and the largest angle requested: no angle-side pair exists, so Law of Cosines first — and ask for the largest angle first, since cos⁻¹ reports obtuse angles honestly (here cos C = (49 + 81 − 169)/(2 · 7 · 9) = −39/126 ≈ −0.3095, so C ≈ 108.0°). Scenario three: angle 40°, side opposite it 9, another side 12 — SSA: Law of Sines plus the ambiguity check: sin B = 12 · sin 40°/9 ≈ 0.8571, so B ≈ 59.0° or 121.0°; both fit beside 40°, so *two* triangles, and an answer naming one is half an answer. Classify before you calculate: pair → Sines; no pair → Cosines; SSA → Sines with suspicion.