Trigonometric Identities and Triangles
Solving Triangles: the Sine and Cosine Rules
Finding every side and angle of a triangle from partial information — right triangles by ratio, the rest by the sine and cosine rules.
What this page covers
- Solve a right triangle from one side and one acute angle
- Apply the sine rule and know when it is usable
- Apply the cosine rule and know when it is required
- Recognise the ambiguous case
What solving a triangle means
A triangle has six elements: three sides and three angles. Solving means finding all six from a sufficient subset. The source phrases the task as given some information, find the rest.
Three pieces are generally needed, and at least one must be a side — three angles fix the shape but not the size.
| Given | Determined? | Method |
|---|---|---|
| Right angle plus one side and one angle | Yes | The basic ratios |
| Two angles and any side (AAS, ASA) | Yes | Sine rule |
| Two sides and the angle between them (SAS) | Yes | Cosine rule |
| Three sides (SSS) | Yes | Cosine rule |
| Two sides and a non-included angle (SSA) | Not always | Sine rule, ambiguous case |
| Three angles (AAA) | No | Shape only, not size |
The source records the summary directly: we can solve all triangles where we have a RHS or SAS; we may not be able to with SSA, with the letters expanded in the margin as right angle, hypotenuse, side, side, angle.
Right triangles
With a right angle present the basic ratios suffice and no rule is needed.
Worked example — the source's first
Given A = 26° and hypotenuse c = 3 m, find the other sides and angles.
Pythagoras: 1.3152 + 2.6962 = 9.000 = 32 ✓. And independently, B = arctanba = arctan(2.050) = 63.999° ✓. Using a relation that was not used in the solving is what makes a check meaningful.
Worked example — the source's second
Given A = 34° and the opposite side a = 4.
√42 + 5.932 = √51.16 = 7.153 ✓.
The sine rule
Each side is paired with the angle opposite it. The rule is usable whenever one complete pair is known plus one further element.
The source observes that it contains the right-triangle case: with C = 90°, sin C = 1 and the rule gives sin Aa = 1c, that is sin A = ac — opposite over hypotenuse.
Worked example — the source's case
Given B = 64°, b = 4 and c = 3.
The third angle follows from the angle sum:
And the third side from the sine rule again:
It verifies with the cosine rule, which was not used in the solution: a2 = b2 + c2 - 2bccos A = 16 + 9 - 24cos 73.616° = 25 - 6.77 = 18.23, giving a = 4.27 ✓.
The cosine rule
The pattern is uniform: the squared side on the left, the other two squared on the right, and a correction term involving the angle between them.
With A = 90°, cos A = 0 and the first form becomes a2 = b2 + c2. Pythagoras is the special case, and the term -2bccos A measures the departure from a right angle.
| Known | Rule | Reason |
|---|---|---|
| A side and its opposite angle, plus one more | Sine | A complete pair is available |
| Three sides (SSS) | Cosine | No angle is known, so no pair exists |
| Two sides and the included angle (SAS) | Cosine | The known angle is not opposite a known side |
| Two sides and a non-included angle (SSA) | Sine, with care | The ambiguous case |
For SSS the rule is rearranged to give the angle: cos A = b2 + c2 - a22bc. The sign of the numerator then tells you at once whether A is acute or obtuse.
The ambiguous case
Two sides and a non-included angle may describe two different triangles, one triangle, or none. The source flags it by saying SSA may not be solvable.
The reason is that sin θ = sin(180° - θ). An inverse sine returns the acute value, but the obtuse one may be the intended angle, and both can produce a valid triangle.
| Situation | Triangles | Detection |
|---|---|---|
| sin C > 1 from the rule | None | The value is impossible; no such triangle |
| sin C = 1 | One | C = 90° exactly |
| sin C < 1 and both angles fit | Two | Check whether 180° - C also gives a valid angle sum |
| sin C < 1 but the obtuse option overflows | One | The angle sum would exceed 180° |
Compute the acute angle from the inverse sine, then test whether its supplement also leaves a positive third angle. In the worked example above, C = 42.384°; the supplement 137.616° plus B = 64° exceeds 180°, so only one triangle exists.
Checking a solved triangle
- Angle sum. The three angles must total 180°.
- Ordering. The longest side must face the largest angle, and the shortest the smallest.
- An unused relation. Verify with a rule that was not used in the solving — the source checks a sine-rule solution with the cosine rule.
- Plausibility. No side may exceed the sum of the other two.
The second and fourth cost nothing and catch gross errors instantly. The third is the real test, because it uses information the solution did not.
Common mistakes
| Mistake | Correct | Check |
|---|---|---|
| Pairing a side with the wrong angle | Each side pairs with the angle opposite it | Label the diagram before starting |
| Using the sine rule with no complete pair | Use the cosine rule | A pair means a side and its opposite angle |
| Taking only the acute inverse sine | Test the supplement too | The ambiguous case |
| Wrong angle in the cosine rule | The angle is between the two named sides | a2 pairs with cos A |
| Calculator in the wrong mode | Match degrees or radians | sin 34 = 0.559 in degrees, 0.529 in radians |
| Checking with the rule already used | Use an unused relation | Otherwise the check is circular |
Frequently asked questions
Which rule should I use?
The sine rule when a side and its opposite angle are both known. The cosine rule when they are not — that is, for three sides, or two sides and the angle between them.
Why is the sine rule sometimes ambiguous?
Because sin θ = sin(180° - θ), so an inverse sine returns an acute angle when an obtuse one may be intended. The source notes this by saying two sides and a non-included angle may not determine the triangle.
Does the cosine rule reduce to Pythagoras?
Yes. With C = 90°, cos C = 0 and c2 = a2 + b2 - 2abcos C becomes c2 = a2 + b2.
How do I check a solved triangle?
Three ways: the angles must sum to 180°; the longest side must face the largest angle; and any relation not used in the solving can be verified independently. The source does all three.
Source. Handwritten teaching notes, Week 9, pages 6-9.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.