When a triangle gives you one side together with the angle directly across from it, the law of sines is usually the fastest way in. It connects every side to the angle facing it, so a single known pair unlocks the rest.

The formula and its symbols

In any triangle with sides aa, bb, cc opposite angles AA, BB, CC,

asin(A)=bsin(B)=csin(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}

The symbols travel in opposite pairs: side aa with angle AA, side bb with angle BB, side cc with angle CC. Matching those pairs correctly is the whole game. If you pair a side with the wrong angle, the setup is broken even when the algebra that follows is flawless.

You may also see the relationship flipped:

asin(A)=bsin(B)orsin(A)a=sin(B)b\frac{a}{\sin(A)} = \frac{b}{\sin(B)} \quad \text{or} \quad \frac{\sin(A)}{a} = \frac{\sin(B)}{b}

Both say the same thing. Pick whichever version isolates your unknown most cleanly, and keep the opposite-pair rule unchanged.

Why the ratio holds

The formula says every side-to-opposite-angle pair shares one common ratio. That is why a wider angle always faces a longer side, and a narrower angle faces a shorter side.

This gives you a built-in sanity check. If one angle opens wider, the side across from it should come out longer. When an answer breaks that pattern, you have almost certainly matched the wrong side and angle.

Worked example

Suppose A=42A = 42^\circ, B=71B = 71^\circ, and a=8a = 8. Find side bb.

Match opposite pairs:

asin(A)=bsin(B)\frac{a}{\sin(A)} = \frac{b}{\sin(B)}

Substitute the known values:

8sin(42)=bsin(71)\frac{8}{\sin(42^\circ)} = \frac{b}{\sin(71^\circ)}

Solve for bb:

b=8sin(71)sin(42)b = 8 \cdot \frac{\sin(71^\circ)}{\sin(42^\circ)}

Using decimal approximations,

b80.94550.669111.30b \approx 8 \cdot \frac{0.9455}{0.6691} \approx 11.30

So

b11.3b \approx 11.3

This passes the check. Since BB is larger than AA, side bb should exceed side aa, and 11.3>811.3 > 8.

Practice it yourself

Set up your own triangle with A=35A = 35^\circ, C=95C = 95^\circ, and a=12a = 12. First find angle BB from the fact that the three angles sum to 180180^\circ, then apply the law of sines to find side cc. Before you compute, predict whether cc should be longer or shorter than aa — that quick prediction catches setup slips before they cost you.

When you finish, confirm that the longest side faces the largest angle. If C=95C = 95^\circ is the widest angle, side cc should be the longest side in the triangle.

Where it applies and what trips students up

The law of sines works for any triangle, but it shines on non-right triangles where at least one opposite side-angle pair is already known. The common entry points are:

  • AAS or ASA: two angles and one side
  • SSA: two sides and a non-included angle, where the known angle is opposite one of the known sides

If instead you know two sides and the included angle, start with the law of cosines, not the law of sines.

The pitfalls cluster around three slips. The first is pairing a side with the wrong angle — the law uses opposite pairs, never adjacent ones. The second is reaching for it too early; if no opposite pair is known, it is rarely the right first equation. The third is the SSA ambiguous case: if you get sin(B)=k\sin(B) = k with 0<k<10 < k < 1, two angles may satisfy it, BB and 180B180^\circ - B. That does not guarantee two triangles. You still must check that each choice keeps the angle sum below 180180^\circ and stays consistent with the given side data.

Frequently Asked Questions

When should you use the law of sines instead of the law of cosines?
Use the law of sines when you already know at least one opposite side-angle pair, which happens in AAS, ASA, and SSA setups. If you know two sides and the included angle instead, no opposite pair is available, so the law of cosines is the better first equation. The law of sines works for any triangle but shines with non-right triangles.
What does the law of sines actually say?
It says every side-to-opposite-angle pair in a triangle follows the same ratio: each side divided by the sine of its opposite angle gives the same value. That is why a larger angle always faces a longer side and a smaller angle faces a shorter side, which also gives you a quick way to sanity-check answers.
Does the law of sines work for right triangles?
Yes. The law of sines holds for any triangle, including right triangles. In practice, though, it is most useful for non-right triangles, because right triangles are usually solved more directly with basic trigonometric ratios. The key requirement is knowing at least one matching opposite side-angle pair before you set up the ratio.
What is the most common mistake when using the law of sines?
Pairing a side with the wrong angle. The law of sines uses opposite pairs: side a goes with angle A, side b with angle B, and side c with angle C, not adjacent ones. If you mix up the pairs, the setup is wrong even when the algebra is correct. Check that the larger angle faces the longer side.
Why does a larger angle face a longer side in a triangle?
Because every side-to-opposite-angle pair shares the same ratio, a side grows as the sine of its opposite angle grows. If one angle opens wider, the side across from it must be longer to keep the ratio equal. This pattern is the fastest intuition check: an answer that breaks it usually means a mismatched side-angle pair.

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