Law of Sines vs Law of Cosines: Which One to Use (and the Ambiguous SSA Case)

Example 1 - AAS

In triangle ABC, A = 35°, B = 65° and a = 12. Solve the triangle.

1. Third angle by subtraction.
   $$C = 180^{\circ} - 35^{\circ} - 65^{\circ} = 80^{\circ}$$
2. The pair (a, A) is complete, so the law of sines gives both remaining sides.
   $$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} \quad\Longrightarrow\quad b = \dfrac{a \sin B}{\sin A}$$
3. Substitute for b.
   $$b = \dfrac{12 \sin 65^{\circ}}{\sin 35^{\circ}} = \dfrac{12 \times 0.9063}{0.5736} \approx 18.96$$
4. Substitute for c on the same ratio.
   $$c = \dfrac{12 \sin 80^{\circ}}{\sin 35^{\circ}} = \dfrac{12 \times 0.9848}{0.5736} \approx 20.60$$
5. Check the ordering: the largest angle C = 80° faces the largest side c ≈ 20.60, and the smallest angle A = 35° faces the smallest side a = 12.

   In any triangle the side ordering matches the angle ordering. If yours does not, you have made a slip.

Answer. C = 80°, b ≈ 18.96, c ≈ 20.60.

Example 2 - SAS, and why the closing law matters

In triangle ABC, b = 7, c = 10 and the included angle A = 40°. Solve the triangle.

1. Apply the law of cosines to the side opposite the known angle.
   $$a^{2} = b^{2} + c^{2} - 2bc\cos A = 7^{2} + 10^{2} - 2(7)(10)\cos 40^{\circ}$$
2. Evaluate, with cos 40° ≈ 0.7660.
   $$a^{2} = 49 + 100 - 140(0.7660) = 149 - 107.25 = 41.75 \quad\Longrightarrow\quad a \approx 6.462$$
3. Now find angle B, using the rearranged law of cosines rather than the law of sines.
   $$\cos B = \dfrac{a^{2} + c^{2} - b^{2}}{2ac} = \dfrac{41.75 + 100 - 49}{2(6.462)(10)} = \dfrac{92.75}{129.24} \approx 0.7177$$
4. Take the arccosine.
   $$B = \arccos(0.7177) \approx 44.1^{\circ}$$
5. Finish by subtraction.
   $$C = 180^{\circ} - 40^{\circ} - 44.1^{\circ} \approx 95.9^{\circ}$$
6. The trap, concretely. Solving for C with the law of sines gives sin C ≈ 0.9948, and arcsin(0.9948) ≈ 84.1° - the supplement of the true answer.
   $$\sin C = \dfrac{c \sin A}{a} = \dfrac{10 \times 0.6428}{6.462} \approx 0.9948$$

   The check that catches it: c = 10 is the longest side, so C must be the largest angle, and 84.1° is smaller than 95.9°.

Answer. a ≈ 6.46, B ≈ 44.1°, C ≈ 95.9°. The triangle is obtuse at C.

Example 3 - SSS

In triangle ABC, a = 8, b = 11 and c = 15. Solve the triangle.

1. Check the triangle inequality first: 8 + 11 = 19 > 15, so a triangle exists.

   If the two shorter sides do not exceed the longest, the law of cosines returns a cosine outside [−1, 1].

2. Solve for the largest angle first - opposite c = 15 - because it is the only one that can be obtuse.
   $$\cos C = \dfrac{a^{2} + b^{2} - c^{2}}{2ab} = \dfrac{64 + 121 - 225}{2(8)(11)} = \dfrac{-40}{176} \approx -0.2273$$
3. The cosine is negative, so C is obtuse. Arccosine reports this correctly.
   $$C = \arccos(-0.2273) \approx 103.1^{\circ}$$
4. The other two angles must be acute, so the law of sines is now safe.
   $$\sin B = \dfrac{b \sin C}{c} = \dfrac{11 \sin 103.1^{\circ}}{15} \approx \dfrac{11 \times 0.9738}{15} \approx 0.7141$$
5. Take the arcsine and subtract for the last angle.
   $$B \approx 45.6^{\circ}, \qquad A = 180^{\circ} - 103.1^{\circ} - 45.6^{\circ} \approx 31.3^{\circ}$$

Answer. A ≈ 31.3°, B ≈ 45.6°, C ≈ 103.1°. Largest side opposite largest angle, as expected.

Example 4 - SSA with two solutions

In triangle ABC, a = 8, b = 11 and A = 40°. Solve the triangle - all of it.

1. A is acute, so compute the altitude h and compare.
   $$h = b \sin A = 11 \sin 40^{\circ} = 11 \times 0.6428 \approx 7.07$$
2. Compare: h ≈ 7.07 < a = 8 < b = 11 - the two-triangle band, so expect two complete answers.

   Doing this check first is what stops you reporting half the solution.

3. Apply the law of sines for angle B.
   $$\sin B = \dfrac{b \sin A}{a} = \dfrac{11 \times 0.6428}{8} \approx 0.8838$$
4. Take both angles with this sine: the calculator's acute value and its supplement.
   $$B_{1} \approx 62.1^{\circ}, \qquad B_{2} = 180^{\circ} - 62.1^{\circ} \approx 117.9^{\circ}$$
5. Test each against the angle sum. Both survive: 40 + 62.1 = 102.1 and 40 + 117.9 = 157.9, both under 180°.

   This is the arithmetic that rejects the second triangle when a ≥ b - there the supplement pushes the sum past 180°.

6. Finish the first triangle.
   $$C_{1} = 180^{\circ} - 40^{\circ} - 62.1^{\circ} \approx 77.9^{\circ}, \qquad c_{1} = \dfrac{a \sin C_{1}}{\sin A} \approx \dfrac{8 \times 0.9778}{0.6428} \approx 12.17$$
7. Finish the second triangle.
   $$C_{2} = 180^{\circ} - 40^{\circ} - 117.9^{\circ} \approx 22.1^{\circ}, \qquad c_{2} = \dfrac{a \sin C_{2}}{\sin A} \approx \dfrac{8 \times 0.3764}{0.6428} \approx 4.68$$

Answer. Two triangles fit the data. Triangle 1: B ≈ 62.1°, C ≈ 77.9°, c ≈ 12.17. Triangle 2: B ≈ 117.9°, C ≈ 22.1°, c ≈ 4.68.

Can I always use the law of cosines instead of the law of sines?

Whenever the law of cosines can start, it is the safer choice, because arccosine distinguishes acute from obtuse and arcsine does not. But it cannot start on AAS or ASA data, where you have only one side, and it cannot start on SSA either - the law of cosines applied to SSA produces a quadratic in the unknown side, whose two roots are precisely the two ambiguous triangles.

Why is SSA ambiguous when SAS is not?

In SAS the known angle sits between the two known sides, so the triangle is rigid - there is nothing left to vary. In SSA the angle is at one end and the opposite side swings freely, so the arc it traces can meet the base line in two places. That is why SAS has one solution and SSA can have two.

How do I know if a set of three sides forms a triangle at all?

The sum of any two sides must exceed the third; checking the two shortest against the longest is enough. If it fails, the law of cosines will return a cosine outside [−1, 1] and the arccosine will be undefined - which is the algebra telling you the same thing.

What does the 2R in the law of sines mean?

R is the radius of the circle through all three vertices, the circumcircle. So a/sin A is not just equal to b/sin B by coincidence; the common ratio has a geometric meaning, and rearranging gives R = a / (2 sin A) - a one-line way to find the circumradius once a triangle is solved.

In an exam, should I report both SSA triangles?

Yes, unless the question rules one out - for example by stating that the triangle is acute, or that a particular angle is obtuse, or by giving a diagram. Where two triangles genuinely satisfy the data, a solution presenting only one is incomplete and is usually marked as such.