Law of Sines: Solving Triangles, the Ambiguous Case and Surveying

What does the law of sines say?

The law of sines says that in every triangle the ratio of a side to the sine of the angle opposite it is the same for all three sides: (a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C). The letters follow one convention: A, B, C are the angles, and a, b, c the sides opposite them, so a is the side across from A. Take the 3-4-5 right triangle: the angle opposite the side 3 is 36.87^\circ, whose sine is 0.6, and (3)/(0.6)=5; opposite 4 is 53.13^\circ with sine 0.8 and (4)/(0.8)=5; and opposite 5 is the right angle, with sine 1 and 5/1=5. All three ratios are 5.

It is usually written in one of three equivalent ways. As a chain of ratios (above). As proportions, a:b:c=\sin A:\sin B:\sin C, which says that the sides are in the same proportion as the sines of their opposite angles (not as the angles themselves). Or upside down, (\sin A)/(a)=(\sin B)/(b)=(\sin C)/(c), which is the form to use when the unknown is an angle. It holds for obtuse triangles too, because the sine of an obtuse angle is positive and equals the sine of its supplement, \sin(180^\circ-x)=\sin x.

What it needs is a pair: a side together with the angle opposite it. Once you have one pair, any other side or angle gives you a second, and the equation can be solved for the missing one. This article shows why the law is true, what the common value means, how to use it in each case, and what goes wrong in one case, the ambiguous one. For solving a triangle in general, with the law of cosines as well, see the article on triangles.

Why is the law of sines true? A proof with one altitude

The law of sines follows from writing the same altitude of the triangle in two ways. Draw the altitude h from C down to the line AB. It splits the triangle into two right triangles that share the side h. In the one with the angle A and hypotenuse b, the definition of the sine (see the right triangle) gives \sin A=h/b, so h=b\sin A. In the one with the angle B and hypotenuse a, \sin B=h/a, so h=a\sin B.

The two expressions are the same length, so b\sin A=a\sin B. Divide both sides by \sin A\sin B and you get (a)/(\sin A)=(b)/(\sin B). Draw an altitude from another vertex and the same argument ties b and c, or a and c, so all three ratios are equal.

If the angle at A or at B is obtuse, the foot of the altitude falls outside the side AB, on its extension, and the right triangle that contains h has the angle 180^\circ-A instead of A. Nothing changes, because \sin(180^\circ-A)=\sin A. That is also why the law holds for every triangle, and it is the seed of the ambiguous case later: an angle and its supplement have the same sine.

What is the common value of the three ratios? The circle through the vertices

The common value (a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C) is the diameter 2R of the circle that passes through the three vertices (the circumscribed circle), where R is its radius. To see why, draw that circle with centre O and a diameter BB' from the vertex B. Two facts about circles do the work. An angle that sits on a diameter is a right angle (Thales), so in the triangle BCB' the angle at C is 90^\circ. And two angles that look at the same chord BC from the same arc are equal, so the angle at B' equals the angle A. In the right triangle BCB' the side BC=a is opposite the angle B'=A and the hypotenuse is BB'=2R, hence \sin A=(a)/(2R),\qquadthat is (a)/(\sin A)=2R. (If A is obtuse, B' is on the other arc and its angle is 180^\circ-A, with the same sine.) The same argument works from any vertex, so all three ratios are 2R.

This gives a formula for the circumradius, R=(a)/(2\sin A), and a few quick consequences. In a right triangle with C=90^\circ, \sin C=1, so c=2R: the hypotenuse is a diameter, and the centre is the midpoint of the hypotenuse (the 3-4-5 triangle has R=2.5). For the triangle with sides 13, 14, 15, the angles are 53.13^\circ, 59.49^\circ and 67.38^\circ and (13)/(\sin53.13^\circ)=16.25, so R=8.125. The centre of this circle, the circumcenter, is one of the triangle's four classical centres (see the centres of a triangle).

Set two angles and a side below and read the three ratios, the circle and the area.

How do you solve a triangle when you know two angles and a side?

When two angles and any side are known (ASA or AAS), the triangle is completely determined: find the third angle from the angle sum, then use the law of sines for the other two sides. The third angle is free, because the three angles of a triangle add up to 180^\circ (see the angle sum). After that there is always a side with its opposite angle, which is the pair the law needs.

1. Third angle. C=180^\circ-A-B. If this is not positive, there is no triangle.
2. The common ratio. Compute k=(a)/(\sin A) from the side and angle you were given.
3. The other sides. b=k\sin B and c=k\sin C.
4. Check. The longest side must be opposite the largest angle, and the three ratios must agree.

Example. A=40^\circ, B=65^\circ and a=12. Then C=180^\circ-40^\circ-65^\circ=75^\circ and k=(12)/(\sin40^\circ)=18.6687, so b=18.6687\sin65^\circ=16.92 and c=18.6687\sin75^\circ=18.03. The largest angle, 75^\circ, is opposite the longest side, 18.03, as it should be. The same triangle is described by ASA: given A=40^\circ, the side c=18.03 between the angles A and B, and B=65^\circ, the same steps give the same sides.

Two practical points. Keep the full value of k in your calculator, and use the given data, not a rounded result of an earlier step, so that errors do not pile up. And if you know only angles, the triangle is not determined: every triangle similar to it has the same angles, and one side sets the size.

Why do two sides and an angle sometimes give two triangles, or none?

Two sides and an angle opposite one of them (SSA) can give no triangle, one or two, because the sine of the unknown angle has two answers between 0^\circ and 180^\circ, B and 180^\circ-B, and each one is a triangle only if it leaves room for the third angle. Say a, b and A are given. The law of sines gives \sin B=(b\sin A)/(a). Three things can happen.

• If (b\sin A)/(a)>1, no angle has that sine: no triangle. The side a is too short to reach.
• If the value is exactly 1, B=90^\circ: one right triangle.
• Otherwise there are two angles, B_1=\arcsin\left((b\sin A)/(a)\right) (the acute one, the only one that a calculator reports) and B_2=180^\circ-B_1 (obtuse). The second one is a triangle only if A+B_2<180^\circ.

The picture shows it. Put A at the origin with a ray going right, and C at distance b from A at the angle A. The unknown vertex B is on the ray at distance a from C: that is where the circle of radius a around C meets the ray. A circle and a ray can meet in 0, 1 or 2 points, and the height of C above the ray, h=b\sin A, decides which.

Example. a=3, b=5, A=30^\circ. The height is h=5\sin30^\circ=2.5, and 2.5<3<5: two triangles. \sin B=(5\sin30^\circ)/(3)=\frac56=0.8333, so B_1=56.44^\circ and B_2=123.56^\circ. For B_1: C=180^\circ-30^\circ-56.44^\circ=93.56^\circ and c=(3\sin93.56^\circ)/(\sin30^\circ)=5.99. For B_2: C=26.44^\circ and c=2.67. Both fit the data. With a=9, b=5 and A=120^\circ (obtuse) only one triangle exists: B=28.76^\circ, C=31.24^\circ, c=5.39, since a second angle B of 151.24^\circ would leave no room next to 120^\circ. And a=2, b=5, A=30^\circ is shorter than h=2.5: no triangle.

The ambiguity is exactly the equation \sin B=c of the article on trigonometric equations: an inverse sine returns one angle, and the other is yours to find. Slide a below and watch the circle.

How does the law of sines connect to the area of a triangle?

The area of a triangle is half the product of two sides and the sine of the angle between them, K=\frac12ab\sin C, and putting it together with the law of sines gives K=(abc)/(4R). The first formula is the altitude formula with the height written as h=b\sin C instead of measured: area is \frac12× base a × height h, and the height from A is b\sin C. It holds for any pair, K=\frac12bc\sin A=\frac12ac\sin B=\frac12ab\sin C, which is a good check on a solved triangle.

Replace \sin C by (c)/(2R) (the law of sines) and it becomes K=\frac12ab·(c)/(2R)=(abc)/(4R), so the radius of the circle through the vertices is R=(abc)/(4K). For the 13, 14, 15 triangle the area is 84 (see the area of a triangle), so R=(13·14·15)/(4·84)=(2730)/(336)=8.125, the same value as before. And \frac12·13·14·\sin67.38^\circ=84 again.

When two angles and a side are known (AAS) you do not even need the other sides: K=(a^2\sin B\sin C)/(2\sin A). For A=40^\circ, B=65^\circ, a=12 this is (144·0.9063·0.9659)/(2·0.6428)=98.06, the same as \frac12·12·16.92·\sin75^\circ. This is how a plot of land is measured from a baseline and angles.

How do you measure the width of a river with the law of sines?

To measure a distance you cannot walk, measure a baseline you can walk and two angles, and the law of sines does the rest; this is triangulation. A tree T stands on the far bank. On your bank you mark two points P and Q a known distance L apart, and from each you read the angle between the baseline and the line of sight to the tree, α at P and β at Q. The three angles of the triangle PQT add up to 180^\circ, so the angle at the tree is \gamma=180^\circ-α-β, and the law of sines pairs the side L with the angle \gamma: (PT)/(\sinβ)=(L)/(\sin\gamma) ⇒ PT=(L\sinβ)/(\sin(α+β)), QT=(L\sinα)/(\sin(α+β)). (because \sin\gamma=\sin(180^\circ-α-β)=\sin(α+β)). The width of the river is the perpendicular distance from T to the baseline, h=PT\sinα.

Example. L=100 m, α=60^\circ and β=70^\circ. Then \gamma=50^\circ, PT=(100\sin70^\circ)/(\sin50^\circ)=122.67 m, QT=(100\sin60^\circ)/(\sin50^\circ)=113.05 m and the river is h=122.67\sin60^\circ=106.23 m wide. No one crossed the water.

How good is the answer? The angles are read with an instrument, and every reading is a little off. If each of the two angles may be wrong by 1^\circ, the width above could be anywhere from 101.2 m to 111.3 m, about 5\% either way. The error depends on the shape of the triangle: with α=β=85^\circ the lines of sight are nearly parallel, the angle at the tree is only 10^\circ and the same 1^\circ error can change the answer by 25\%; with α=β=45^\circ, a right angle at the tree, it is only 3.6\%. A surveyor therefore chooses the baseline so that the two lines of sight cross at a healthy angle, near 90^\circ. The method is older than any instrument: Thales' shadow measurement used similar triangles, as in the article on triangles, and triangulation uses angles instead of shadows. Try your own baseline and angles below.

When do you use the law of sines and when the law of cosines?

Use the law of sines when you know a side and its opposite angle (a pair), and the law of cosines when you do not. Look at what the three given measurements are:

The law of cosines, c^2=a^2+b^2-2ab\cos C, is the Pythagorean theorem with a correction. It is the safer tool for finding an angle, since the arccosine returns a single angle between 0^\circ and 180^\circ and so tells acute from obtuse, while the arcsine cannot. A common plan for SAS: use the law of cosines for the third side, then find the smaller of the two other angles with the law of sines, because the smaller angle is always acute and the arcsine is then right.

How do you use the law of sines in a program?

A program solves AAS directly and has to treat SSA as a list: compute \sin B, return nothing if it is above 1, and otherwise try both B and 180^\circ-B. Three details. Convert degrees to radians before calling sin. Do not call asin with a value that is 1.0000000000000002 by rounding: allow a tiny margin and clamp. And keep the second angle only if the third, 180^\circ-A-B, is positive. A program that returns the single acute B silently loses the second triangle.

The programs below solve A=40^\circ, B=65^\circ, a=12 (giving b=16.9196 and c=18.0326), then a=3, b=5, A=30^\circ (two triangles, B=56.44^\circ and B=123.56^\circ), count the triangles for a=2 and for the obtuse case a=9, b=5, A=120^\circ, and finally find the river width 106.23 m.

import math


def solve_aas(A, B, a):
    """Two angles (degrees) and the side a opposite A: the sides and angles (a, b, c, A, B, C), or None."""
    C = 180 - A - B
    if min(A, B, C, a) <= 0:
        return None                                       # the angles leave nothing for the third
    k = a / math.sin(math.radians(A))                     # the common ratio, 2R
    return a, k * math.sin(math.radians(B)), k * math.sin(math.radians(C)), A, B, C


def solve_ssa(a, b, A):
    """Two sides and the angle A opposite a: a list with 0, 1 or 2 triangles (a, b, c, A, B, C)."""
    s = b * math.sin(math.radians(A)) / a                 # sin B
    if s > 1 + 1e-12:
        return []                                         # no angle has a sine above 1
    B1 = math.degrees(math.asin(min(1.0, s)))             # asin gives only the acute angle
    found = []
    for B in sorted({B1, 180 - B1}):                      # the other angle with the same sine
        C = 180 - A - B
        if C > 1e-9:                                      # it is a triangle only if something is left for C
            found.append((a, b, a * math.sin(math.radians(C)) / math.sin(math.radians(A)), A, B, C))
    return found


def river_width(L, alpha, beta):
    """Baseline L and the angles alpha, beta at its ends: the distance from the target to the baseline."""
    gamma = 180 - alpha - beta
    pt = L * math.sin(math.radians(beta)) / math.sin(math.radians(gamma))
    return pt * math.sin(math.radians(alpha))


print([round(x, 4) for x in solve_aas(40, 65, 12)])      # [12, 16.9196, 18.0326, 40, 65, 75]
for t in solve_ssa(3, 5, 30):
    print(round(t[4], 2), round(t[5], 2), round(t[2], 2))   # 56.44 93.56 5.99, then 123.56 26.44 2.67
print(len(solve_ssa(2, 5, 30)), len(solve_ssa(9, 5, 120)))  # 0 1
print(round(river_width(100, 60, 70), 2))                    # 106.23

const rad = (d) => (d * Math.PI) / 180
const deg = (r) => (r * 180) / Math.PI

function solveAas(A, B, a) {
  // two angles (degrees) and the side a opposite A: [a, b, c, A, B, C], or null
  const C = 180 - A - B
  if (Math.min(A, B, C, a) <= 0) return null                    // the angles leave nothing for the third
  const k = a / Math.sin(rad(A))                                // the common ratio, 2R
  return [a, k * Math.sin(rad(B)), k * Math.sin(rad(C)), A, B, C]
}

function solveSsa(a, b, A) {
  // two sides and the angle A opposite a: a list with 0, 1 or 2 triangles
  const s = (b * Math.sin(rad(A))) / a                          // sin B
  if (s > 1 + 1e-12) return []                                  // no angle has a sine above 1
  const B1 = deg(Math.asin(Math.min(1, s)))                     // asin gives only the acute angle
  const found = []
  for (const B of [...new Set([B1, 180 - B1])].sort((p, q) => p - q)) {   // the other angle with the same sine
    const C = 180 - A - B
    if (C > 1e-9) found.push([a, b, (a * Math.sin(rad(C))) / Math.sin(rad(A)), A, B, C])   // a triangle only if something is left for C
  }
  return found
}

function riverWidth(L, alpha, beta) {
  // baseline L and the angles alpha, beta at its ends: the distance from the target to the baseline
  const gamma = 180 - alpha - beta
  const pt = (L * Math.sin(rad(beta))) / Math.sin(rad(gamma))
  return pt * Math.sin(rad(alpha))
}

solveAas(40, 65, 12).map((x) => +x.toFixed(4))                  // [12, 16.9196, 18.0326, 40, 65, 75]
solveSsa(3, 5, 30).map((t) => [+t[4].toFixed(2), +t[5].toFixed(2), +t[2].toFixed(2)])   // [[56.44, 93.56, 5.99], [123.56, 26.44, 2.67]]
;[solveSsa(2, 5, 30).length, solveSsa(9, 5, 120).length]        // [0, 1]
+riverWidth(100, 60, 70).toFixed(2)                              // 106.23

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

const double PI = 3.14159265358979323846;
double rad(double d) { return d * PI / 180.0; }
double deg(double r) { return r * 180.0 / PI; }

double Bs[2], Cs[2], cs[2];                         // the triangles of the last solveSsa call

// Two sides and the angle A opposite a: fills Bs, Cs, cs and returns how many triangles (0, 1 or 2).
int solveSsa(double a, double b, double A) {
  double s = b * sin(rad(A)) / a;                   // sin B
  if (s > 1 + 1e-12) return 0;                      // no angle has a sine above 1
  if (s > 1) s = 1;
  double B1 = deg(asin(s));                         // asin gives only the acute angle
  int count = 0;
  for (int i = 0; i < 2; i++) {
    double B = i == 0 ? B1 : 180 - B1;              // the other angle with the same sine
    if (i == 1 && fabs(B - B1) < 1e-9) continue;
    double C = 180 - A - B;
    if (C > 1e-9) {                                 // a triangle only if something is left for C
      Bs[count] = B;
      Cs[count] = C;
      cs[count] = a * sin(rad(C)) / sin(rad(A));
      count++;
    }
  }
  return count;
}

double riverWidth(double L, double alpha, double beta) {
  double gamma = 180 - alpha - beta;
  double pt = L * sin(rad(beta)) / sin(rad(gamma));
  return pt * sin(rad(alpha));
}

int main() {
  double A = 40, B = 65, a = 12;                    // two angles and a side
  double C = 180 - A - B;
  double k = a / sin(rad(A));                       // the common ratio, 2R
  printf("%.4f %.4f %.0f\n", k * sin(rad(B)), k * sin(rad(C)), C);
  int n = solveSsa(3, 5, 30);
  for (int i = 0; i < n; i++) printf("%.2f %.2f %.2f\n", Bs[i], Cs[i], cs[i]);
  int none = solveSsa(2, 5, 30);
  int obtuse = solveSsa(9, 5, 120);
  printf("%d %d\n", none, obtuse);
  printf("%.2f\n", riverWidth(100, 60, 70));
  return 0;
}

/* output:
16.9196 18.0326 75
56.44 93.56 5.99
123.56 26.44 2.67
0 1
106.23
*/

Where does the law of sines come from?

The law of sines grew out of astronomy, was stated in its plane form by mathematicians writing in Arabic, and became the working tool of the surveyors who measured whole countries.

• 10th to 13th century. Mathematicians writing in Arabic, among them Abu al-Wafa, al-Jayyani and al-Tusi, state and prove the law of sines, first for triangles on a sphere (needed for astronomy and for the direction of Mecca) and then for plane triangles. Al-Tusi's treatise on the quadrilateral is the best-known source for the plane form.
• 1464. Regiomontanus writes De triangulis omnimodis (printed in 1533), the first European book to treat triangles, with the law of sines, as a subject of its own and not as a chapter of astronomy.
• 1533. Gemma Frisius describes triangulation, fixing the position of a point from a measured baseline and two angles.
• 1617. Willebrord Snellius measures a degree of the meridian by a chain of triangles, the first measurement of the Earth by triangulation, in Eratosthenes Batavus.
• From 1802. The Great Trigonometrical Survey of India, begun by William Lambton, carries a chain of triangles across the subcontinent from the south to the Himalaya, with angles read by instruments and sides computed by the law of sines. In 1856 it announced the height of Peak XV, later called Everest, as about 8.8 km.

What are the common mistakes with the law of sines?

The most common mistakes with the law of sines are the nine below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: the law of sines at a glance

• The law: (a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C), with each side opposite its angle; true for every triangle, since \sin(180^\circ-x)=\sin x.
• Why: the altitude is h=b\sin A=a\sin B; the common value is the diameter 2R of the circle through the vertices.
• ASA and AAS: C=180^\circ-A-B, then k=(a)/(\sin A), b=k\sin B, c=k\sin C; one triangle.
• SSA: \sin B=(b\sin A)/(a) has two answers, B and 180^\circ-B; with h=b\sin A and an acute A: a<h none, a=h one right, h<a<b two, a\ge b one; with an obtuse A: a>b one, otherwise none.
• Area: K=\frac12ab\sin C=(abc)/(4R); with AAS, K=(a^2\sin B\sin C)/(2\sin A).
• Surveying: baseline L and angles α, β give PT=(L\sinβ)/(\sin(α+β)) and the width h=PT\sinα; lines of sight that cross near a right angle are steadiest.
• Which law: a side with its opposite angle, use sines; SAS and SSS, use cosines first.

Glossary: key terms

Law of sines

The rule that in every triangle each side divided by the sine of the opposite angle gives the same number, which is the diameter of the circle through the three vertices.

Opposite angle

The angle across the triangle from a given side, the one whose vertex is not an end of that side; side a is opposite angle A.

Circumscribed circle

The circle that passes through all three vertices of a triangle; its radius R satisfies a over sin A equal to 2R.

ASA and AAS

The cases in which two angles and a side of a triangle are known, which determine exactly one triangle.

Ambiguous case (SSA)

The case in which two sides and an angle opposite one of them are known, which can fit no triangle, one triangle or two triangles.

Supplementary angles

Two angles that add up to 180 degrees; they have the same sine, which is why the law of sines can give two answers.

Baseline

A measured distance between two observation points, from which the distance to a faraway point is found with two angles.

Triangulation

Finding the position of a point, or a distance, by measuring a baseline and the angles to the point and solving the triangle with the law of sines.

Frequently asked questions

References

1. James Stewart, Lothar Redlin, Saleem Watson (2016). Precalculus: Mathematics for Calculus (chapters on the law of sines and the law of cosines). Cengage Learning.
2. I. M. Gelfand, Mark Saul (2001). Trigonometry (2nd ed.). Birkhäuser.
3. Eli Maor (1998). Trigonometric Delights. Princeton University Press.
4. Carl B. Boyer, Uta C. Merzbach (2011). A History of Mathematics (3rd ed.). Wiley.
5. Johannes Regiomontanus (1533). De triangulis omnimodis. Johannes Petreius, Nuremberg.
6. Willebrord Snellius (1617). Eratosthenes Batavus. Elzevir, Leiden.
7. John Keay (2000). The Great Arc. HarperCollins.

Related articles

• Triangles: Types, Angles, Area, Centers and Congruence: Triangles explained: types by sides and angles, the 180° angle sum, the triangle inequality, area and Heron's formula, centers, congruence and the law of sines.
• Basic Trigonometry: Sine, Cosine and Tangent: Basic trigonometry explained: sine, cosine and tangent of a right triangle, exact values at 30°, 45° and 60°, the unit circle, identities, equations and graphs.
• Trigonometric Equations: How to Solve sin x = c and Harder Ones: How to solve trigonometric equations: solutions in an interval, multiple angles, quadratics, identities, a sin x + b cos x = c, false roots and numerical methods.
• The Pythagorean Theorem: Proofs, Triples and Applications: The Pythagorean theorem explained: the statement, four proofs, the converse and the 3-4-5 rule, Pythagorean triples, the distance formula, and generalizations.