How to Use the Law of Sines, Step by Step

You have a triangle to solve and no right angle in sight. The law of sines gives you a direct path to the missing sides and angles, as long as you know two angles and a side (AAS or ASA) or two sides and a non-included angle (SSA). Here is the clear procedure to work through your homework triangle.

To know how to use law of sines correctly, start by labeling every vertex and side. Use A, B, and C for the vertices and write the side across from each in the matching lowercase letter. Side a is across from angle A, side b across from angle B, and side c across from angle C. This labeling is not optional, the law of sines only works when you respect the opposite-side pairing.

Write the standard form: a / sin A = b / sin B = c / sin C. The alternative form, sin A / a = sin B / b = sin C / c, is also valid. Pick whichever you prefer and keep it consistent through the calculation.

Draw and Mark Every Corner

Draw your triangle and put a letter at each corner. Label the side across from each corner with the same letter in lowercase. If your triangle is not drawn to scale, that is fine, the law of sines does not care about the shape as long as the angle-side pairs are correct.

Check that you have marked the known angles and the known side length. If you are given an ASA case (two angles and the included side), the known side sits between the two given angles. If you are given AAS, the known side is not between them. This distinction changes which ratio you set up first.

For an oblique triangle, any triangle with no right angle, the law of sines is your primary tool. Avoid using the law of cosines unless you have SAS or SSS, because the law of sines is faster for AAS, ASA, and SSA.

Given AAS, Two Angles and a Non-Included Side

You know angle A = 40°, angle B = 60°, and side a = 10 cm. This is AAS because the known side a is across from angle A, which is one of the known angles. The missing side b is across from the other known angle B.

Find angle C first: 180° − 40° − 60° = 80°. The triangle angle sum guarantees this step. Now set up the law of sines proportion: a / sin A = b / sin B. Rearrange to isolate b: b = (a × sin B) / sin A.

Using a calculator in degree mode, sin 40° ≈ 0.6428 and sin 60° ≈ 0.8660. Then b = (10 × 0.8660) / 0.6428 ≈ 13.47 cm. Round to the same number of decimal places as your given side. If the problem expects whole numbers, 13.5 cm is fine.

You can find side c the same way: c / sin C = a / sin A, so c = (a × sin C) / sin A = (10 × sin 80°) / sin 40°. Sin 80° ≈ 0.9848, giving c ≈ 15.3 cm.

Verify your work by checking that the sum of the three sides makes sense with the angles, the largest side should be across from the largest angle.

Given ASA, Two Angles and the Included Side

You know angle A = 40°, angle B = 60°, and side c = 10 cm. Side c is the included side, it lies between the two given angles. The procedure is nearly identical to AAS: find the third angle first, then apply the law of sines.

Angle C = 180° − 40° − 60° = 80°. Now find side a: a / sin A = c / sin C, so a = (c × sin A) / sin C = (10 × sin 40°) / sin 80°. Sin 40° ≈ 0.6428, sin 80° ≈ 0.9848, giving a ≈ 6.53 cm.

Find side b: b = (c × sin B) / sin C = (10 × sin 60°) / sin 80° = (10 × 0.8660) / 0.9848 ≈ 8.79 cm.

The only difference between ASA and AAS is which side you know. In ASA you start with the included side c; in AAS you start with a non-included side. The law of sines works the same way in both cases because you always have two angles and one side.

Given SSA, Two Sides and a Non-Included Angle

You know side a = 8 cm, side b = 12 cm, and angle A = 30°. This is SSA because the known angle A is not between sides a and b, side b does not touch angle A. This is where the ambiguous case lives. You must check for zero, one, or two possible triangles.

Set up sin B / b = sin A / a. Rearrange: sin B = (b × sin A) / a = (12 × sin 30°) / 8 = (12 × 0.5) / 8 = 6 / 8 = 0.75. Take the inverse sine: B₁ = sin⁻¹(0.75) ≈ 48.6°. This is the first possible angle.

Because sin B = 0.75 has a second solution, B₂ = 180° − 48.6° = 131.4°. Both B₁ and B₂ produce a valid sine of 0.75. Check if each forms a valid triangle by summing with angle A. For B₁ = 48.6°, C = 180° − 30° − 48.6° = 101.4°, a valid triangle. For B₂ = 131.4°, C = 180° − 30° − 131.4° = 18.6°, also valid. That gives two possible triangles.

The condition for two solutions in SSA is: angle A is acute, and b × sin A < a < b. Here 12 × sin 30° = 6, and 6 < 8 < 12, so two triangles exist. If a were less than or equal to 6, no triangle would exist. If a were greater than or equal to 12, only one triangle would exist.

For each valid triangle, use the law of sines again to find side c. For B₁ = 48.6°, C = 101.9799) / 0.5 ≈ 15.7 cm. For B₂ = 131.4°, C = 18.6°, sin 18.6° ≈ 0.3190, giving c ≈ 5.1 cm.

If your calculator gives an error for sin⁻¹(0.75), or if the sine value exceeds 1, no triangle exists for the given input. That is the zero-solution case.

These mistakes cause most homework errors. Avoid them.

• Calculator mode wrong. A degree-mode calculator used on a radian problem gives a completely different sine value. Set your calculator to degrees unless the problem specifies radians. The law of sines holds for radians, but you must be in the matching mode.
• Forgetting the ambiguous case. SSA is not a single-answer situation. Always compute the second possible angle (180° − first angle) and check whether it forms a valid triangle. Many calculators silently return only one solution.
• Rounding intermediate steps. Truncating sin 40° to 0.64 instead of 0.6428 can shift your final side by several tenths. Keep at least four decimal places until the last step, then round to the problem's requested precision.
• Mismatching side-angle pairs. The side in the numerator must be across from the angle inside the sine. A side labeled a goes with sin A, not sin B. Double-check your labeling.
• Using law of cosines instead of law of sines. If you have two angles and a side, law of sines is faster. Save law of cosines for SAS and SSS.

When you need only a missing side and you already have two angles and one side, the procedure is the same regardless of whether the given side is included or not. Find the third angle first, then set up the proportion that connects the known side to the unknown side.

If you know side a, angle A, and angle B, then side b = (a × sin B) / sin A. If you know side c, angle A, and angle B, then side a = (c × sin A) / sin C and side b = (c × sin B) / sin C. The formula never changes, only the numbers you plug in.

For SSA, finding a missing side requires the extra step of checking both possible angle B values first. Only after you confirm which triangles exist do you compute the missing side.