A complete study guide for every oblique triangle case — AAS, ASA, SSA (ambiguous case), SAS, and SSS — with all formulas, decision rules, 8 fully worked examples, and real-world applications in navigation, surveying, and engineering.
An oblique triangle is any triangle that is not a right triangle. Because there is no 90-degree angle, SOH-CAH-TOA does not apply directly. Instead, the Law of Sines and Law of Cosines generalize trigonometry to all triangles.
Standard notation: sides a, b, c are opposite angles A, B, C respectively. The sum A + B + C = 180° always. Every problem gives you some combination of sides and angles; the first task is to identify which case you have.
| Case | Given | Law to Use | Special Note |
|---|---|---|---|
| AAS | 2 angles + side not between them | Law of Sines | Find third angle first (sum = 180°) |
| ASA | 2 angles + side between them | Law of Sines | Find third angle first |
| SSA | 2 sides + non-included angle | Law of Sines | Ambiguous — may yield 0, 1, or 2 triangles |
| SAS | 2 sides + included angle | Law of Cosines | Use side form to find missing side |
| SSS | All 3 sides | Law of Cosines | Use angle form; find largest angle first |
Key Insight: The Matched Pair Test
If your given information includes at least one angle and its opposite side (e.g., you know angle B and side b), you have a matched pair and Law of Sines will work. If you have no matched pair (SAS or SSS), you need Law of Cosines.
Identify what information is given, then follow the decision tree below.
Law of Sines
Find the third angle first (A + B + C = 180°), then use a/sin A to build your ratio.
Law of Sines
Ambiguous case — compute h = b sin A and compare to a before proceeding.
Law of Cosines
Use c squared = a squared + b squared minus 2ab cos C to find the third side, then switch to Law of Sines.
Law of Cosines
Rearrange to cos C = (a squared + b squared minus c squared) / (2ab) to find angles. Find the largest angle first.
Quick Memory Rule
SSA and AAS/ASA all have angle-side pairs available — use Sines. SAS and SSS do not have a usable matched pair at the start — use Cosines. Once you find one new side or angle in an SAS/SSS problem, you can freely switch to Sines for the remaining unknowns.
Equivalently (reciprocal form): sin A / a = sin B / b = sin C / c
Drop a perpendicular of height h from vertex C to side c. In the left sub-triangle, h = b sin A. In the right sub-triangle, h = a sin B. Setting these equal: b sin A = a sin B, which rearranges to a/sin A = b/sin B. The third ratio follows by the same argument with a different altitude.
Given: A = 40°, B = 75°, a = 12
Find: sides b, c and angle C
Step 1 — Find the third angle
Step 2 — Identify the known matched pair
We know angle A = 40° and its opposite side a = 12. Build the constant ratio:
Step 3 — Solve for b
b / sin B = 18.67
b = 18.67 × sin 75° = 18.67 × 0.9659 ≈ 18.04
Step 4 — Solve for c
c / sin C = 18.67
c = 18.67 × sin 65° = 18.67 × 0.9063 ≈ 16.92
Step 5 — Verify (sanity check)
Given: A = 30°, C = 110°, b = 20
Find: sides a, c and angle B. Note: side b is between angles A and C (ASA).
Step 1 — Find B
Step 2 — Build the ratio using (b, B)
Step 3 — Solve for a and c
a = 31.12 × sin 30° = 31.12 × 0.5 ≈ 15.56
c = 31.12 × sin 110° = 31.12 × 0.9397 ≈ 29.24
SSA is called the ambiguous case because two sides and a non-included angle do not uniquely determine a triangle. You must compute the altitude h = b sin A and compare it to side a before you can know how many triangles exist.
| Condition | Result | Geometric Meaning |
|---|---|---|
| a < h (a < b sin A) | 0 triangles | Side a is too short to reach the base line |
| a = h (a = b sin A) | 1 triangle (right) | Side a just touches the base — forms a right angle |
| h < a < b | 2 triangles | Side a swings to two different positions |
| a ≥ b | 1 triangle | Side a is long enough that only one position works |
| Condition | Result |
|---|---|
| a ≤ b | 0 triangles — an obtuse angle cannot be opposite a shorter or equal side |
| a > b | 1 triangle — the longest side must be opposite the largest angle |
Example 3a (SSA — No Triangle): A = 50°, a = 4, b = 7
Compute the altitude: h = b sin A = 7 × sin 50° = 7 × 0.7660 ≈ 5.36
Compare: a = 4 < h ≈ 5.36
Conclusion: No triangle exists. Side a cannot reach the base line.
Example 3b (SSA — Two Triangles): A = 35°, a = 9, b = 12
Step 1: Compute altitude h = b sin A = 12 × sin 35° = 12 × 0.5736 ≈ 6.88
Since h (6.88) < a (9) < b (12) → two triangles exist.
Step 2: Use Law of Sines to find B.
sin B / b = sin A / a → sin B = 12 × sin 35° / 9 = 12 × 0.5736 / 9 ≈ 0.7648
Step 3: Find both values of B.
B₁ = arcsin(0.7648) ≈ 49.9° (acute)
B₂ = 180° − 49.9° = 130.1° (obtuse)
Step 4: Check both are geometrically valid (A + B < 180°).
Triangle 1: 35° + 49.9° = 84.9° < 180° ✓ → C₁ = 95.1°
Triangle 2: 35° + 130.1° = 165.1° < 180° ✓ → C₂ = 14.9°
Step 5: Solve for c in each triangle.
c₁ = 9 × sin 95.1° / sin 35° = 9 × 0.9961 / 0.5736 ≈ 15.63
c₂ = 9 × sin 14.9° / sin 35° = 9 × 0.2571 / 0.5736 ≈ 4.04
Example 3c (SSA — One Triangle, a ≥ b): A = 35°, a = 15, b = 12
Since a (15) ≥ b (12), exactly one triangle exists.
sin B = b sin A / a = 12 × sin 35° / 15 ≈ 0.4589
B = arcsin(0.4589) ≈ 27.3°
C = 180° − 35° − 27.3° = 117.7°
c = 15 × sin 117.7° / sin 35° ≈ 23.25
a² = b² + c² − 2bc cos A
b² = a² + c² − 2ac cos B
c² = a² + b² − 2ab cos C
The side on the left is the one being found. The angle on the right is the angle opposite that side.
cos A = (b² + c² − a²) / (2bc)
cos B = (a² + c² − b²) / (2ac)
cos C = (a² + b² − c²) / (2ab)
These are algebraic rearrangements of the side form. The angle on the left is isolated using arccos.
Connection to the Pythagorean Theorem
When C = 90°, cos C = 0, so the term 2ab cos C vanishes and we get c² = a² + b² — the Pythagorean theorem. The Law of Cosines is the generalization of the Pythagorean theorem to all triangles.
Given: b = 8, c = 11, A = 60°
Angle A is between sides b and c (the included angle). Find: side a, angles B and C.
Step 1 — Apply Law of Cosines to find a
a² = b² + c² − 2bc cos A
a² = 8² + 11² − 2(8)(11) cos 60°
a² = 64 + 121 − 176 × 0.5
a² = 185 − 88 = 97
a = √97 ≈ 9.85
Step 2 — Switch to Law of Sines to find B (find the smaller angle first)
Side b = 8 is smaller than c = 11, so B is smaller and cannot be obtuse — no ambiguity in arcsin.
sin B / b = sin A / a
sin B = 8 × sin 60° / 9.85 = 8 × 0.8660 / 9.85 ≈ 0.7034
B = arcsin(0.7034) ≈ 44.7°
Step 3 — Find C by subtraction
Given: a = 7, b = 10, c = 13
Find all three angles A, B, C.
Step 1 — Find the largest angle first (C, opposite longest side c = 13)
Largest angle first is the SSS rule. If the largest angle is obtuse, arccos returns the correct obtuse value automatically — no ambiguity issue. Once you confirm C is obtuse, A and B must be acute, so arcsin or arccos is unambiguous for them.
cos C = (a² + b² − c²) / (2ab)
cos C = (49 + 100 − 169) / (2 × 7 × 10)
cos C = −20 / 140 ≈ −0.1429
C = arccos(−0.1429) ≈ 98.2° (obtuse ✓)
Step 2 — Find angle B
cos B = (a² + c² − b²) / (2ac)
cos B = (49 + 169 − 100) / (2 × 7 × 13)
cos B = 118 / 182 ≈ 0.6484
B = arccos(0.6484) ≈ 49.6°
Step 3 — Find A by subtraction
The formula K = (1/2)bh (base times height over 2) requires knowing the perpendicular height. For oblique triangles you usually don't have h directly, but if you know two sides and the included angle, you can compute h implicitly.
K = ½ ab sin C
K = ½ ac sin B
K = ½ bc sin A
Use whichever form matches the two sides and included angle you have. All three give the same area.
The standard base-height formula is K = (1/2) × base × height. If you take side a as the base, the altitude h from vertex B satisfies h = c sin A (from the definition of sine in the sub-triangle). Substituting: K = (1/2) × a × c sin A, which is one of the three equivalent forms above.
Example: Find the area of the triangle with b = 8, c = 11, A = 60°
K = ½ × b × c × sin A
K = ½ × 8 × 11 × sin 60°
K = ½ × 88 × (√3 / 2)
K = 44 × 0.8660 ≈ 38.1 square units
| Given Information | Area Formula |
|---|---|
| Base and height | K = (1/2) bh |
| Two sides and included angle | K = (1/2) ab sin C |
| All three sides (no angles) | Heron's Formula |
| Three angles and one side | Find sides via Law of Sines, then use K = (1/2) ab sin C |
Heron's formula gives the area of a triangle from its three side lengths alone — no angles required. It was known to Heron of Alexandria around 60 CE and remains one of the most elegant results in elementary geometry.
s = (a + b + c) / 2
K = √[s(s − a)(s − b)(s − c)]
s is the semi-perimeter — half the total perimeter of the triangle.
Example: Triangle with a = 7, b = 10, c = 13
Step 1: s = (7 + 10 + 13) / 2 = 30 / 2 = 15
Step 2: s − a = 15 − 7 = 8
s − b = 15 − 10 = 5
s − c = 15 − 13 = 2
Step 3: K = √[15 × 8 × 5 × 2]
K = √[1200] = √[400 × 3]
K = 20√3 ≈ 34.64 square units
Example: A triangular plot with sides 120 ft, 80 ft, 95 ft
s = (120 + 80 + 95) / 2 = 295 / 2 = 147.5
s − 120 = 27.5, s − 80 = 67.5, s − 95 = 52.5
K = √[147.5 × 27.5 × 67.5 × 52.5]
K = √[14,394,140.6] ≈ 3794 sq ft
Pro Tip
Heron's formula is most useful in SSS problems when you need the area immediately without first computing angles. If you already found an angle (via Law of Cosines), use K = (1/2) ab sin C instead — it has fewer computation steps.
A classic surveying problem: a river, cliff, or other obstacle makes it impossible to measure a distance directly. By measuring accessible angles and a known baseline, the inaccessible distance is computed using the Law of Sines or Cosines.
Example 8 — Width of a River (ASA)
Surveyors set two stakes A and B on one bank, 200 m apart. A flagpole P on the far bank is sighted from A at 70° and from B at 55° (interior angles of triangle ABP). Find the distance AP and the width of the river.
Step 1: Find the third angle at P.
P = 180° − 70° − 55° = 55°
Step 2: ASA with baseline AB = 200 m. Use Law of Sines.
AP / sin B = AB / sin P
AP = 200 × sin 55° / sin 55° = 200 m
(Because angle B = angle P = 55°, triangle ABP is isosceles — AP = AB = 200 m)
Step 3: Width is the perpendicular from P to AB.
Width = AP × sin A = 200 × sin 70° = 200 × 0.9397 ≈ 187.9 m
| Application | Typical Case | Law Used |
|---|---|---|
| Width of river / canyon | ASA | Law of Sines |
| Height of a cliff or building | AAS with right angle | Law of Sines / right trig |
| Distance between two ships | SAS | Law of Cosines |
| Property boundary calculations | SSS | Law of Cosines for angles |
| GPS triangulation | ASA | Law of Sines |
| Structural truss analysis | SSS | Law of Cosines |
| Radar position fixing | SSA | Law of Sines (check ambiguous) |
Two forces acting on an object create a triangle of forces. The resultant (net force) is the third side of a triangle whose other two sides are the individual forces and whose included angle determines the direction. This is a direct SAS application of the Law of Cosines.
Parallelogram Method
Resultant Formula (SAS)
R² = F1² + F2² − 2(F1)(F2)cos(180° − θ)
Since cos(180° − θ) = −cos θ:
R² = F1² + F2² + 2(F1)(F2)cos θ
Example — Two Forces at an Angle
Two forces of 80 N and 50 N act at an angle of 40° to each other. Find the magnitude of the resultant force.
Using the resultant formula with theta = 40°:
R² = 80² + 50² + 2(80)(50) cos 40°
R² = 6400 + 2500 + 8000 × 0.7660
R² = 8900 + 6128 = 15028
R = √15028 ≈ 122.6 N
Direction (angle from F1, using Law of Sines):
sin alpha / 50 = sin(180° − 40°) / 122.6 = sin 140° / 122.6
sin alpha = 50 × 0.6428 / 122.6 ≈ 0.2621
alpha ≈ 15.2° from the 80 N force
Component Method vs Triangle Method
Precalculus courses teach both approaches. The triangle method (Law of Cosines) is faster for two forces. The component method (resolving into x/y components using right triangle trig, then adding) is more systematic for three or more forces. Both yield identical results.
Law of Sines
a / sin A = b / sin B = c / sin C
Use for: AAS, ASA, SSA
Law of Cosines
c² = a² + b² − 2ab cos C
cos C = (a² + b² − c²) / (2ab)
Use for: SAS, SSS
Area (Sine)
K = ½ ab sin C
Use when: two sides + included angle known
Heron's Formula
s = (a + b + c) / 2
K = √[s(s−a)(s−b)(s−c)]
Use when: all three sides known
Master Decision Table
| Case | First Step | Then | Watch For |
|---|---|---|---|
| AAS | Find 3rd angle (sum = 180°) | Law of Sines ratio | None |
| ASA | Find 3rd angle (sum = 180°) | Law of Sines ratio | None |
| SSA | Compute h = b sin A; compare to a | Law of Sines (if triangle exists) | Ambiguous: 0, 1, or 2 triangles |
| SAS | Law of Cosines (side form) | Law of Sines for angles | Use smaller remaining angle in arcsin |
| SSS | Law of Cosines (largest angle first) | Law of Sines or subtract | Negative cosine = obtuse angle (valid) |
Skipping the SSA ambiguous case analysis
Always compute h = b sin A first. Compare a to h and b. Never just jump in with Law of Sines. If you find two valid values of sin B, you must report both triangles or explain why one is rejected.
Using Law of Sines for SAS or SSS
Law of Sines requires a known angle-side pair. In SAS and SSS, you have no matched pair to start with. Use Law of Cosines first; once you find one angle, you can switch to Law of Sines for the rest.
Wrong arrangement of the Law of Cosines
The side squared on the left is opposite the angle on the right. Write it as: (target side)² = (other side)² + (other side)² − 2(other)(other) cos(angle opposite target side).
Calculator in Radians mode instead of Degrees
All precalculus triangle problems use degrees unless explicitly stated otherwise. sin 35° in radians gives the wrong number. Check your mode every session.
Rounding intermediate values too early
Keep at least 4 decimal places throughout each calculation. Rounding 0.7648 to 0.76 early can shift B by more than 1°, cascading into large final errors.
Not verifying angles sum to 180°
Add A + B + C at the end. If the sum is not 180° (within 0.5° rounding tolerance), you made an error. This sanity check catches most mistakes before you turn in the exam.
Using arcsin in SSS without checking for obtuse angles
In SSS problems, find the largest angle first with arccos. Once the largest angle is found, the other two must be acute, so arcsin is safe for them. Never use arcsin to find the largest angle — you might get the acute supplement instead.
Forgetting units in applied problems
Navigation distances in nautical miles, surveying in feet or meters, forces in newtons — always carry and state units. The method works identically regardless of units, but a unitless answer is incomplete.
Stewart Precalculus (7th/8th editions) covers oblique triangles in Chapter 6. Use this map to find the right section for each topic.
| Section | Topic | Key Concepts |
|---|---|---|
| 6.1 | Law of Sines | AAS, ASA, SSA — the ambiguous case decision table |
| 6.2 | Law of Cosines | SAS, SSS — side form and angle form |
| 6.3 | Area of a Triangle | K = (1/2) ab sin C and Heron's formula |
| 6.4 | Vectors | Magnitude, direction, component form, resultant |
| 6.5 | The Dot Product | u dot v = |u||v| cos theta, work, projections |
Work these problems without a calculator first to identify which law applies, then verify with a calculator.
Level 1
Level 2
Level 3
The Law of Sines states: a/sin A = b/sin B = c/sin C, where a, b, c are the side lengths opposite angles A, B, C respectively. You can also write it as sin A/a = sin B/b = sin C/c. Use this law when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA).
The ambiguous case (SSA) occurs when you know two sides and a non-included angle. Depending on the values, there may be 0, 1, or 2 valid triangles. Compute h = b sin A. If angle A is acute: no triangle when a < h; one right triangle when a = h; two triangles when h < a < b; one triangle when a is greater than or equal to b. If angle A is obtuse: no triangle when a is less than or equal to b; one triangle when a > b.
The Law of Cosines has three equivalent forms: a squared = b squared + c squared minus 2bc cos A; b squared = a squared + c squared minus 2ac cos B; c squared = a squared + b squared minus 2ab cos C. For finding an angle: cos A = (b squared + c squared minus a squared) divided by (2bc). Use this law when you know all three sides (SSS) or two sides and the included angle (SAS).
Use the Law of Sines for AAS (two angles plus any side), ASA (two angles plus included side), and SSA (two sides plus non-included angle — check the ambiguous case). Use the Law of Cosines for SAS (two sides plus included angle) and SSS (all three sides). The key test: if you have a matched angle-side pair, Law of Sines works; if you have SAS or SSS with no matched pair, use Law of Cosines.
The area of any triangle is K = (1/2) ab sin C, where a and b are two sides and C is the included angle between them. Three equivalent forms: K = (1/2) ab sin C = (1/2) ac sin B = (1/2) bc sin A. This formula works for any triangle — right or oblique — as long as you know two sides and the angle between them.
Heron's formula gives the area of a triangle when all three sides are known, without needing any angles. First compute the semi-perimeter s = (a + b + c)/2. Then K = square root of [s(s minus a)(s minus b)(s minus c)]. Use it in SSS problems when you need the area and do not want to first solve for an angle.
Convert compass bearings to interior triangle angles by subtracting or adding as needed. Draw a diagram labeling all known sides and angles. Identify which triangle case you have (usually ASA or SAS in two-leg navigation problems). Apply the appropriate law to find the missing side (distance) or angle (bearing to destination). Always verify the angles sum to 180 degrees.
Yes. When angle C equals 90 degrees, cos C equals 0, so the Law of Cosines c squared = a squared + b squared minus 2ab cos C becomes c squared = a squared + b squared, which is exactly the Pythagorean theorem. The Law of Cosines is the generalization of the Pythagorean theorem to non-right triangles.
Top mistakes: (1) Forgetting the ambiguous case in SSA — always compute h = b sin A and compare to a. (2) Using Law of Sines for SAS or SSS — you must use Law of Cosines there. (3) Wrong form of Law of Cosines — the side being found is isolated on the left. (4) Calculator in wrong mode (degrees vs radians). (5) Not checking that angles sum to 180 degrees as a sanity check. (6) Rounding intermediate answers before the final step.
SOH-CAH-TOA, special triangles 30-60-90 and 45-45-90, inverse trig
Pythagorean, double angle, half angle, sum and difference formulas
Component form, magnitude, direction, dot product, resultant forces
Conic sections and distance/angle applications in the coordinate plane
Unit circle, trig graphs, inverse functions, identities — full guide
General solutions, specific intervals, and equation strategies
Interactive problems for every case — AAS through SSS — with step-by-step solutions, spaced repetition, and private tutoring. Free to try.