Law of Sines Calculator

a ÷ sin A = b ÷ sin B = c ÷ sin C sin B = b · sin A ÷ a C = 180° − A − B c = a · sin C ÷ sin A

a and A

The side you know together with the angle across from it. These two are the pair that sets the ratio for the whole triangle, and one of them must be present to use this law at all

b

The second side you know. It is the one that decides how many triangles the pieces allow, and the whole ambiguous case turns on how its length compares with side a

B

The angle across from side b. Taking the inverse sine of the ratio always returns the acute angle between zero and ninety degrees, and when the obtuse partner also fits inside a straight angle there is a second triangle using it

C and c

The third angle and the third side, both obtained after B is settled: the angle by subtracting the other two from a straight angle, the side from the same shared ratio

Use this page when you know two sides and an angle that is not between them — two bearings and a distance across a bay, two legs of a survey traverse where the angle is at the far end — and you want the rest of the triangle. It is the only angle-and-side shape the law of cosines cannot take comfortably, and it is also the only one where the answer can be a choice rather than a measurement, which is why this page leads with the count. If the pieces admit two triangles, look at the angles as well as the sides before you decide which one you are standing in: a surveyor sighting two landmarks usually knows whether the corner in front of them is sharp or blunt, and that single fact picks the triangle. If you have all three sides, or two sides and the angle between them, the law of cosines page is the right tool and it has no ambiguity at all. And if all you want is one side of a right triangle, the right triangle page is shorter than either.

1. Two sides and a wide-open angle, 6, 30 degrees and 8

   1. The shared ratio is side a over the sine of angle A, which is 6 divided by 0.5, or 12
   2. The sine of angle B is side b over that ratio, which is 8 divided by 12, or 0.6667
   3. The angle whose sine is 0.6667 is 41.81 degrees — and its supplement of 138.19 degrees has the same sine, so both are candidates
   4. The angle C is whatever is left of a straight angle: 180 minus 30 minus 41.81 is 108.19
   5. Side c comes from the same ratio, 12 times the sine of 108.19 degrees, which is 11.4

   The page's own starting values, and the textbook picture of the ambiguous case: the second side is longer than the first, so it can swing to either side of the perpendicular and still close the figure. The four readings describe the triangle built on the acute angle B — the one where the second side leans back towards the base. The other triangle uses the obtuse partner of 41.81 degrees, which is 138.19, and it is in the last column of the table below; its third angle is a narrow 11.81 degrees rather than the 108.19 printed here. Both are correct answers to the same three pieces of information.

2. The same angle with a shorter second side, 8, 30 degrees and 6

   1. The shared ratio is 8 divided by the sine of 30 degrees, or 16
   2. The sine of angle B is 6 divided by 16, or 0.375
   3. The angle whose sine is 0.375 is 22.02 degrees; its supplement of 157.98 leaves no room for angle A, since 30 plus 157.98 already passes a straight angle, so only the acute option survives
   4. Angle C is 180 minus 30 minus 22.02, which is 127.98
   5. Side c is 16 times the sine of 127.98 degrees, which is 12.61

   Swap the two sides and the ambiguity disappears: 6 is shorter than 8, so the second position cannot close the figure any more and only one triangle is left. The count drops from two to one without any of the other numbers changing their nature, which is the point of printing the count first — the same three kinds of input produce a different answer depending only on how the two lengths compare. Note that the surviving triangle is the obtuse one at C, at 127.98 degrees, because with a shorter second side the figure has to be completed by reaching further along.

3. The boundary case, 5, 30 degrees and 10

   1. The shared ratio is 5 divided by the sine of 30 degrees, or 10
   2. The sine of angle B is 10 divided by 10, which is exactly 1
   3. A sine of 1 fixes the angle at 90 degrees with no second option, because there is no other angle between zero and 180 with the same sine
   4. Angle C is 180 minus 30 minus 90, which is 60, and side c is 10 times the sine of 60 degrees, which is 8.66

   The second side is long enough to reach exactly once: it comes down square onto the third side and the triangle is a right triangle with B at exactly 90 degrees. This is the dividing line between no triangle and two, and it is worth watching what happens as you cross it from below — the two solutions appear together the moment the side is a hair longer than this, one of them just past ninety degrees and the other just short of it, then separate as the side grows. At the exact value there is one, and its sine comes out at exactly 1 with nothing rounded.

4. An angle of 90 degrees given instead of found

   1. The shared ratio is 10 divided by the sine of 90 degrees, which is 10 itself
   2. The sine of angle B is 6 divided by 10, or 0.6, giving 36.87 degrees
   3. Angle C is 180 minus 90 minus 36.87, which is 53.13
   4. Side c is 10 times the sine of 53.13 degrees, which is 8

   Here the right angle is one of the pieces you supplied rather than something the arithmetic discovers, and the shared ratio becomes the hypotenuse over one — which is the hypotenuse itself. From there the law of sines gives the two acute angles directly and the third side comes out at exactly 8, a clean number because this is nothing other than the three-four-five triangle scaled by two. It is a useful check on the whole page: whenever the side facing the right angle is the longest of the three, everything should agree with the Pythagorean theorem, and here it does.

Three limits worth stating plainly. The panel prints one triangle, not two. When the count is 2 the four readings below it belong to the triangle built on the acute angle B, and the obtuse partner is named in the last column of the reference table rather than shown in the panel — the panel has one place for an answer and no way to present two side by side, so the choice was to print the smaller angle and point at the other. Second, the pieces have to make sense before anything can be solved: the given angle must be strictly between zero and one hundred and eighty degrees, and if the second side is too short to reach the third one there is no triangle and this page reports that instead of inventing one. Angles of exactly zero or exactly a straight angle are refused rather than approximated, because a triangle with a zero angle has no shape to solve for. Third, the readings are given to two decimal places, which stops meaning much on a very large triangle: a side of several hundred million centimetres cannot carry hundredths in a double, so treat the decimals of the printed side length as indicative once the figure is that big. The angles stay accurate, because a hundredth of a degree is still a hundredth of a degree whatever the size of the triangle.

What is the ambiguous case of the law of sines?

It is the situation where two sides and a non-included angle allow two different triangles rather than one. The given angle fixes the direction of one side but not which way the other one swings, so the second side can lean back and close the figure early or lean forward and close it late — one triangle is small with an obtuse angle where the other is large and acute. Both use the same three pieces of information. This page counts the possibilities first and then works out the acute one, because whether the second position exists at all depends on how the two lengths compare.

How do I know whether there is one triangle or two?

Compare the second side with the first. Work out the second side multiplied by the sine of the given angle and divide by the first side: if that number is above one the pieces make no triangle at all, because no angle has a sine greater than one. If it is below one you have the acute angle, and there is a second triangle whenever its obtuse partner still leaves room for the given angle — that is, whenever the two angles together stay under 180 degrees. The page does that comparison for you and prints the count as its headline answer.

Which of the two triangles does the calculator show me?

The one built on the smaller angle B, which is the triangle whose third angle is larger. The panel prints the count and then the four readings for that triangle, and the obtuse partner of B appears in the last column of the reference table beside it, so both are on the page. The reason for showing one rather than both is that a result panel has one place for an answer; printing two sets of four numbers with no room to label them would be worse than printing one and naming the other.

Why does the page refuse an angle of 0 or 180 degrees?

Because a triangle with a zero angle is not a triangle — it is a line segment with a point on it, and the side facing the zero angle would have to be zero too. Those two values sit at the boundary where the shape stops existing rather than at the edge of the arithmetic, and the honest answer is to say so rather than to return an angle of zero and a side of zero that look like a solution. Angles in between are fine, including angles very close to either end, where the triangle becomes very long and very thin but stays a real triangle.

Do I use the law of sines or the law of cosines?

Use the law of sines when you have a side together with the angle across from it and one other piece, and the law of cosines when you have all three sides or two sides and the angle between them. The case that decides it is two sides and an angle that is not between them: that is this page's shape, and it is the one the law of cosines takes awkwardly because it has to solve a quadratic. It is also the shape that carries the ambiguity, which is why this page spends its headline on the count.

Why is the printed side length only accurate to about two decimals?

Two decimals is what the page rounds to, and for an ordinary triangle that is well inside what a double can represent. On a very large triangle it is not: a side of several hundred million has exhausted the precision of the number before the decimals get anywhere, so the last digits of such a reading are along for the ride rather than measured. The angles do not have this problem, because a hundredth of a degree remains a hundredth of a degree however big the triangle is.

• Law of Sines — the shared ratio of a side to the sine of the opposite angle, and the conditions under which the relation admits two triangles — Wolfram MathWorld (United States)
• Law of Cosines — the companion relation for three sides, and the shape of triangle this page cannot take — Wolfram MathWorld (United States)
• Triangle — the triangle inequality, the rule that the largest side faces the largest angle, and the cases in which a triangle is not determined by its parts — Wolfram MathWorld (United States)
• The Law of Sines — the theorem worked through from first principles, including the ambiguous case and how the number of solutions follows from the lengths — Math is Fun (United Kingdom)