SOH-CAH-TOA only works in right triangles. For any other triangle you need the law of sines and the law of cosines. Between them, they solve every triangle where you know three parts, as long as at least one is a side.
The labels used here: angles A, B, C, with side a opposite angle A, side b opposite B, and side c opposite C.
The law of sines
a/sin A = b/sin B = c/sin C
Each side divided by the sine of its opposite angle gives the same number. Use it when you know a side and its opposite angle, plus one more piece.
Example A = 40°, B = 65°, a = 10. Find b.
- Set up: 10/sin 40° = b/sin 65°.
- b = 10 sin 65°/sin 40° = 10(0.9063)/0.6428 ≈ 14.1.
Example A = 50°, C = 70°, c = 12. Find a and b.
- B = 180° − 50° − 70° = 60°.
- a = 12 sin 50°/sin 70° ≈ 12(0.7660)/0.9397 ≈ 9.78.
- b = 12 sin 60°/sin 70° ≈ 12(0.8660)/0.9397 ≈ 11.06.
The law of cosines
c² = a² + b² − 2ab cos C
It works the same way for any side: a² = b² + c² − 2bc cos A, and so on. When C = 90°, cos C = 0 and it turns into the Pythagorean theorem.
Use it when you know two sides and the angle between them (SAS), or all three sides (SSS).
Example a = 7, b = 9, C = 52°. Find c.
- c² = 49 + 81 − 2(7)(9) cos 52°.
- c² = 130 − 126(0.6157) ≈ 130 − 77.58 = 52.42.
- c ≈ 7.24.
Example Sides 5, 8 and 10. Find the largest angle.
- The largest angle is opposite the longest side, 10. Call it C.
- 100 = 25 + 64 − 80 cos C.
- 11 = −80 cos C, so cos C = −0.1375.
- C = cos⁻¹(−0.1375) ≈ 97.9°.
A negative cosine means the angle is obtuse. That is a good reason to find the largest angle first with the law of cosines: the law of sines can’t tell an obtuse angle from its acute partner.
Which law to use
| What you know | Use |
|---|---|
| Two angles and any side (AAS or ASA) | Law of sines |
| Two sides and the angle between them (SAS) | Law of cosines |
| All three sides (SSS) | Law of cosines |
| Two sides and an angle not between them (SSA) | Law of sines, but check the ambiguous case |
The ambiguous case (SSA)
With two sides and a non-included angle, there may be zero, one or two triangles. The reason is that sin θ = sin(180° − θ), so the law of sines gives two possible angles.
Example A = 30°, a = 6, b = 10. Find B.
- sin B = 10 sin 30°/6 = 5/6 ≈ 0.8333.
- B ≈ 56.4° or B ≈ 180° − 56.4° = 123.6°.
- Check each with A: 30° + 56.4° and 30° + 123.6° are both under 180°.
- Two triangles are possible.
If sin B comes out above 1, there’s no triangle. If the second angle plus A is 180° or more, there’s only one.
Calculator in radian mode
sin 30 in radian mode is about −0.988, not 0.5. If your answers look strange, check the mode first.
Area with two sides and an angle
Area = ½ab sin C
This comes from ½ × base × height, where the height is b sin C. It saves you from finding the height separately.
Example Find the area of a triangle with sides 8 and 11 and a 40° angle between them.
Area = ½(8)(11) sin 40° ≈ 44(0.6428) ≈ 28.3 square units.
A real-world setup
Example Two hikers leave camp. One walks 3 km on a bearing of 040°, the other 5 km on a bearing of 115°. How far apart are they?
- The angle between their paths is 115° − 40° = 75°.
- d² = 9 + 25 − 2(3)(5) cos 75° = 34 − 30(0.2588) ≈ 26.24.
- d ≈ 5.12 km.
Check yourself
A = 35°, B = 80°, a = 7. Find b.
Show answer
About 12.0. b = 7 sin 80°/sin 35°.
a = 6, b = 8, C = 60°. Find c.
Show answer
About 7.21. c² = 36 + 64 − 48 = 52.
Sides 4, 6 and 7. Find the angle opposite the side of length 7.
Show answer
About 86.4°. cos C = (16 + 36 − 49)/48 = 0.0625.
A = 40°, a = 3, b = 8. How many triangles are possible?
Show answer
None. sin B = 8 sin 40°/3 ≈ 1.71, which is more than 1.
Find the area of a triangle with sides 10 and 14 and a 30° angle between them.
Show answer
35. ½(10)(14)(0.5).