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Law of Sines & Law of Cosines

You’ll be able to

Which law, and why

The choice is decided entirely by what you are given. The Law of Sines pairs a side with its opposite angle, so it needs at least one complete pair: use it for AAS, ASA, or SSA. The Law of Cosines relates all three sides to one angle, so it needs no pair: use it for SAS and SSS. Identify the configuration first and the law selects itself.

The two laws
a/sin A = b/sin B = c/sin C · c² = a² + b² − 2ab·cos C
In the Law of Cosines, the angle C must be the one *opposite* the side c. With C = 90° the cosine term vanishes and it reduces to the Pythagorean theorem.
Worked example

In triangle ABC, a = 12, b = 9 and C = 47°. Find c and then angle A.

  1. 1.Two sides and the included angle: SAS, so start with the Law of Cosines.
  2. 2.c² = 144 + 81 − 2(12)(9)cos 47° = 225 − 216(0.68200) ≈ 225 − 147.31 = 77.69.
  3. 3.c ≈ 8.814.
  4. 4.Now use the Law of Sines for angle A: sin A/12 = sin 47°/8.814.
  5. 5.sin A = 12(0.73135)/8.814 ≈ 0.99565, so A ≈ 84.7°.
  6. 6.Check the angle sum: B = 180 − 47 − 84.7 = 48.3°, and B should be the second largest since b = 9 is the second longest side. ✓
Answer: c ≈ 8.81 and A ≈ 84.7°. Solving for the largest remaining angle via the Law of Sines carries a risk — sin A near 1 means A could be either 84.7° or 95.3°. Here the angle sum with C = 47° permits both, so the safer route is to find the smaller angle B first, which is never ambiguous.

The ambiguous case

Given SSA — two sides and a non-included angle — the triangle may not be unique. The reason is that sin θ = sin(180° − θ), so the Law of Sines cannot distinguish an acute angle from its obtuse supplement. Depending on the numbers there may be two triangles, exactly one, or none. Every SSA problem therefore requires you to test the supplement explicitly and check whether the angles still sum to less than 180°.

Resolving SSA
find the acute angle α from the Law of Sines, then test 180° − α · the supplement is valid only if the three angles still sum to 180°
If the side opposite the given angle is shorter than the height a·sin A, no triangle exists — the arc cannot reach the base.
Checkpoint

Which configuration can produce two different triangles?

Worked example

In triangle ABC, a = 15, b = 20 and A = 40°. Determine how many triangles are possible and solve each.

  1. 1.SSA — check for ambiguity. Law of Sines: sin B/20 = sin 40°/15.
  2. 2.sin B = 20(0.64279)/15 ≈ 0.85705.
  3. 3.The acute solution is B ≈ 58.99°. The obtuse alternative is 180 − 58.99 = 121.01°.
  4. 4.Test the acute case: A + B = 40 + 58.99 = 98.99 < 180, so C ≈ 81.01°. Valid.
  5. 5.Test the obtuse case: A + B = 40 + 121.01 = 161.01 < 180, so C ≈ 18.99°. Also valid.
  6. 6.Find c in each: c = 15 sin C/sin 40°. For C ≈ 81.01°: c ≈ 15(0.98787)/0.64279 ≈ 23.05. For C ≈ 18.99°: c ≈ 15(0.32548)/0.64279 ≈ 7.59.
Answer: Two triangles exist. Triangle 1: B ≈ 59.0°, C ≈ 81.0°, c ≈ 23.1. Triangle 2: B ≈ 121.0°, C ≈ 19.0°, c ≈ 7.6. Both satisfy every given condition, so reporting only one is an incomplete answer.
Area from two sides and the included angle
Area = ½ab·sin C
C must be the angle *between* the sides a and b. With C = 90°, sin C = 1 and this reduces to ½(base)(height).
Checkpoint

Find the area of a triangle with sides 8 and 11 enclosing an angle of 62°.

Answer the 2 checkpoints as you read.

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