DWG NO. 6.1 — Lesson 1 of 6

Law of Sines

Unit 6: Additional Topics in Trigonometry · ~25 min

Objective Use the Law of Sines to find missing sides and angles in oblique triangles, including recognizing the ambiguous SSA case.

Unit 4.3 handled right triangles: one 90° angle made the six trig ratios do all the work. Most triangles that show up in surveying, navigation, or engineering aren't right triangles at all — they're oblique. Solving an oblique triangle means finding every missing side and angle from the ones you're given, and the first general tool for that job is the Law of Sines.

The Law of Sines

Label a triangle the usual way: side a opposite angle A, side b opposite angle B, side c opposite angle C. The Law of Sines says the ratio of each side to the sine of its opposite angle is the same all the way around:

\(\frac{a}{sinA}\) = \(\frac{b}{sinB}\) = \(\frac{c}{sinC}\)

This works whenever you know an angle and its opposite side, plus one more piece of information — that covers three cases: AAS (two angles and a non-included side), ASA (two angles and the included side — find the third angle first, then it's really AAS), and SSA (two sides and a non-included angle), which needs extra care.

Dropping altitude h from vertex A shows h = c·sinB = a·sinC, which rearranges into the Law of Sines

The ambiguous case (SSA)

Given two sides and a non-included angle, the picture isn't always unique — depending on the lengths involved, there may be zero, one, or two triangles that fit. Suppose you're given side a, side b, and angle A (opposite side a). Compute the height of the triangle if it were a right triangle: h = b·sinA, then compare:

a < h — no triangle exists
a = h — exactly one triangle (a right triangle)
h < a < b — two triangles exist
a ≥ b — exactly one triangle

Worked Example 1 · AAS case

ProblemIn triangle ABC, A = 42°, B = 75°, a = 12. Find C, b, and c.

1Angle sum: C = 180° − 42° − 75° = 63°.

2b = \(\frac{a\cdotsinB}{sinA}\) = \(\frac{12\cdotsin75^\circ}{sin42^\circ}\) = 12\(\frac{0.9659}{0.6691}\) ≈ 17.32.

3c = \(\frac{a\cdotsinC}{sinA}\) = \(\frac{12\cdotsin63^\circ}{sin42^\circ}\) = 12\(\frac{0.8910}{0.6691}\) ≈ 15.98.

C = 63°, b ≈ 17.3, c ≈ 16.0

Worked Example 2 · Ambiguous SSA case

ProblemGiven a = 15, b = 20, A = 40°, determine how many triangles exist and solve each.

1Check the height: h = b·sinA = 20·sin40° ≈ 20(0.6428) = 12.86.

2Since h < a < b (12.86 < 15 < 20), there are two triangles.

3sinB = \(\frac{b\cdotsinA}{a}\) = 20\(\frac{0.6428}{15}\) = 0.8570, so B ≈ 59.0° or its supplement B′ ≈ 121.0°.

4Case 1: C = 180° − 40° − 59.0° = 81.0°, c = \(\frac{15\cdotsin81.0^\circ}{sin40^\circ}\) ≈ 23.05.

5Case 2: C′ = 180° − 40° − 121.0° = 19.0°, c′ = \(\frac{15\cdotsin19.0^\circ}{sin40^\circ}\) ≈ 7.60.

Triangle 1: B ≈ 59.0°, C ≈ 81.0°, c ≈ 23.1 | Triangle 2: B ≈ 121.0°, C ≈ 19.0°, c ≈ 7.6

Guided practice

1.AAS: A = 50°, B = 60°, a = 10. Find b.
Show answer

b = \(\frac{10\cdotsin60^\circ}{sin50^\circ}\) = 10\(\frac{0.8660}{0.7660}\) ≈ 11.31.
2.ASA: A = 35°, C = 95°, c = 18. Find a.
Show answer

B = 180° − 35° − 95° = 50° (not needed directly). a = \(\frac{18\cdotsin35^\circ}{sin95^\circ}\) = 18\(\frac{0.5736}{0.9962}\) ≈ 10.36.
3.SSA: a = 9, b = 14, A = 30°. How many triangles exist?
Show answer

h = 14·sin30° = 7. Since h < a < b (7 < 9 < 14), there are two triangles.
4.SSA: a = 5, b = 12, A = 50°. How many triangles exist?
Show answer

h = 12·sin50° ≈ 9.19. Since a = 5 < h, no triangle exists.
5.Explain why the Law of Sines can't be applied directly to an SAS or SSS triangle.
Show answer

The Law of Sines needs a known angle paired with its opposite side. In SAS and SSS, none of the given angles (if any are given at all) is paired with a known opposite side, so none of the three ratios \(\frac{a}{sinA,}\) \(\frac{b}{sinB,}\) \(\frac{c}{sinC}\) can be computed to start the chain.

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