Precalculus · G.SRT.D.10 · G.SRT.D.11

Law of sines

Sine, cosine and tangent need a right angle. What can you say about a triangle that has none?

You need

Sine, cosine and tangentGeometry

Unlocks

Area from two sides and the included anglePrecalculus The ambiguous casePrecalculus Law of cosinesPrecalculus

Where you are 6 in Laws of sines and cosines

Trigonometry7

Laws of sines and cosines6

Law of sines

Area from two sides and the included angle The ambiguous case Law of cosines Heron's formula Applications of oblique triangles

When you reach for it

The sine, cosine and tangent ratios solve right triangles. Most triangles have no right angle: two lookout towers and a smoke plume, three stakes in a field, a triangle in a bridge truss. A triangle with no right angle is called an oblique triangle, and solving a triangle means finding all three of its angles and all three of its sides.

The law of sines solves an oblique triangle whenever you know two angles and one side. That happens in two arrangements:

AAS (angle, angle, side): two angles and a side that is not between them.
ASA (angle, side, angle): two angles and the side between them, the included side.

Either way the triangle is fixed: AAS and ASA are congruence criteria, so only one triangle has those parts, and the law of sines computes the rest of it.

Throughout, the triangle is ABC, and each side is named by the lowercase letter of the angle across from it: side a is opposite A, side b is opposite B, side c is opposite C.

The procedure

The law, from an altitude

Figure 1The altitude from C is a side of two right triangles at once, so it can be written two ways, h = b \sin A and h = a \sin B.

Start with a triangle whose angles are all acute. Drop the altitude from C straight down to side AB, meeting it at D, and call its length h. The altitude splits the triangle into two right triangles, ADC and BDC, and h is a leg of both.

In right triangle ADC the hypotenuse is b and the side opposite A is h, so by the sine ratio \sin A = \dfrac{h}{b}, which gives h = b \sin A. In right triangle BDC the hypotenuse is a and the side opposite B is h, so \sin B = \dfrac{h}{a} and h = a \sin B.

Both expressions are the same length h, so they are equal:

b \sin A = a \sin B.

Divide both sides by \sin A \sin B, which is not zero because A and B are angles of a triangle, strictly between 0^\circ and 180^\circ:

\frac{a}{\sin A} = \frac{b}{\sin B}.

Nothing about C was special. The altitude from A to side BC gives b \sin C = c \sin B in the same way, and so \dfrac{b}{\sin B} = \dfrac{c}{\sin C}.

When an angle is obtuse

Figure 2When B is obtuse the altitude from C lands outside the triangle, and the right triangle CDB holds the angle 180^\circ - B instead of B.

Suppose B is obtuse. The altitude from C now misses side AB and lands on the line AB extended past B, at a point D outside the triangle. Right triangle ADC still has hypotenuse b and the angle A, so h = b \sin A exactly as before. The other right triangle, BDC, has hypotenuse a, but its angle at B is not B: it is the angle between side BC and the extension, and the two angles at B together make a straight line, so it is 180^\circ - B. The sine ratio gives

h = a \sin(180^\circ - B).

Now the unit circle finishes the job. The terminal points of the angles x and 180^\circ - x are mirror images across the y-axis, so they share a y-coordinate, and the y-coordinate is the sine. That is the identity from Unit circle symmetry and even-odd identities:

\sin(180^\circ - x) = \sin x.

So h = a \sin B after all, and b \sin A = a \sin B holds just as in the acute case. The other altitude needs no fix: the altitude from the obtuse vertex B falls inside the triangle, and it gives \dfrac{a}{\sin A} = \dfrac{c}{\sin C} directly. If B = 90^\circ the altitude from C is side a itself, and h = a = a \sin 90^\circ again. Every triangle obeys the same law.

The law of sines

In any triangle ABC with sides a, b, c opposite the angles A, B, C,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.

Each side divided by the sine of the angle across from it gives the same number. The reciprocals are equal too, \dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}, and either form may be used. In words: sides are proportional to the sines of their opposite angles.

Solving AAS and ASA triangles

The law links each side to its opposite angle, so to use it you need one complete pair: a side and the angle across from it. Two angles always give you the third, because the angles of a triangle add to 180^\circ (Triangle angle sum). With all three angles known, the one given side is now paired with its opposite angle, whether it started as the included side (ASA) or not (AAS). That pair fixes the common ratio k = \dfrac{a}{\sin A}, and every other side is k times the sine of its own opposite angle.

Keep the full calculator value of each result and round only at the end; rounding early drifts the last digit. Answers here are rounded to two decimal places.

Worked examples

Example 1

The law on a triangle you already know

A 30-60-90 triangle has sides 1, \displaystyle \sqrt{3} and 2. Check that the law of sines holds for it exactly.

Solution

The side 1 is opposite 30^\circ, the side \displaystyle \sqrt{3} is opposite 60^\circ, and the hypotenuse 2 is opposite 90^\circ (the longest side is across from the largest angle). Form each ratio with exact sine values:

\begin{aligned} \frac{1}{\sin 30^\circ} &= \frac{1}{1/2} = 2, \\ \frac{\sqrt{3}}{\sin 60^\circ} &= \frac{\sqrt{3}}{\sqrt{3}/2} = 2, \\ \frac{2}{\sin 90^\circ} &= \frac{2}{1} = 2. \end{aligned}

All three ratios equal 2. For a right triangle the common ratio is the hypotenuse, because \sin 90^\circ = 1.

Example 2

Two angles and a side across from one of them

Figure 3Example 2. Two angles and the side across from one of them.

In triangle ABC, A = 38^\circ, B = 71^\circ and a = 14.5. Solve the triangle.

Solution

The side given, a, is not between A and B (that would be c), so this is AAS.

The third angle comes from the angle sum: C = 180^\circ - 38^\circ - 71^\circ = 71^\circ.

The complete pair is a = 14.5 with A = 38^\circ. Set up the law with b:

\frac{b}{\sin 71^\circ} = \frac{14.5}{\sin 38^\circ},

b = \frac{14.5 \sin 71^\circ}{\sin 38^\circ} \approx 22.27.

For c the same steps give c = \dfrac{14.5 \sin 71^\circ}{\sin 38^\circ}, the same expression, because C = B = 71^\circ. So c \approx 22.27 as well. That is right: two equal angles make the triangle isosceles, and the law of sines says so on its own, since equal angles have equal sines.

So C = 71^\circ and b = c \approx 22.27. Check: the shortest side, a, is across from the smallest angle, 38^\circ.

Example 3

Two angles and the side between them

In triangle ABC, A = 44^\circ, C = 57^\circ and b = 25. Solve the triangle.

Solution

Side b joins the vertices A and C, so it is the included side: ASA. The law cannot start yet, because the angle across from b is unknown. The angle sum supplies it: B = 180^\circ - 44^\circ - 57^\circ = 79^\circ.

Now b = 25 is paired with B = 79^\circ, and

\begin{aligned} a &= \frac{25 \sin 44^\circ}{\sin 79^\circ} \approx 17.69, \\ c &= \frac{25 \sin 57^\circ}{\sin 79^\circ} \approx 21.36. \end{aligned}

So B = 79^\circ, a \approx 17.69, c \approx 21.36. The order of the sides, a < c < b, matches the order of their angles, 44^\circ < 57^\circ < 79^\circ.

Example 4

An obtuse angle

In triangle ABC, A = 24^\circ, B = 128^\circ and b = 40. Solve the triangle.

Solution

Side b is across from B, not between A and B, so this is AAS, and b already has its partner angle. First C = 180^\circ - 24^\circ - 128^\circ = 28^\circ.

The ratio uses \sin 128^\circ. A calculator gives about 0.7880, the same as \sin 52^\circ, which is the identity \sin(180^\circ - x) = \sin x at work: this is the obtuse case of the derivation. Then

\begin{aligned} a &= \frac{40 \sin 24^\circ}{\sin 128^\circ} \approx 20.65, \\ c &= \frac{40 \sin 28^\circ}{\sin 128^\circ} \approx 23.83. \end{aligned}

So C = 28^\circ, a \approx 20.65, c \approx 23.83. The side across from the obtuse angle, 40, is the longest, as it must be.

Example 5

Two lookout towers and a smoke plume

Figure 4Example 5. Each tower measures one angle; the baseline between them is known.

Thorvaldur and Ketill staff fire lookout towers T and K, 9.6 miles apart. Both see the same smoke plume S. Thorvaldur measures the angle between the line to Ketill’s tower and his line of sight to the smoke as 52^\circ; Ketill measures the matching angle at his tower as 61^\circ. How far is the smoke from each tower, and how far is it from the straight line joining the towers?

Solution

The triangle TKS has two known angles and the included side TK = 9.6: ASA. The angle at the smoke is S = 180^\circ - 52^\circ - 61^\circ = 67^\circ, and that is the angle across from the baseline, so the common ratio is \dfrac{9.6}{\sin 67^\circ}.

The distance TS is across from Ketill’s angle, 61^\circ, and KS is across from Thorvaldur’s angle, 52^\circ:

\begin{aligned} TS &= \frac{9.6 \sin 61^\circ}{\sin 67^\circ} \approx 9.12, \\ KS &= \frac{9.6 \sin 52^\circ}{\sin 67^\circ} \approx 8.22. \end{aligned}

The distance from the smoke to the line TK is the altitude d from S, which is the derivation’s own move. In the right triangle with hypotenuse TS and angle 52^\circ at T, d = TS \sin 52^\circ \approx 7.19.

The smoke is about 9.12 miles from Thorvaldur’s tower, 8.22 miles from Ketill’s, and 7.19 miles from the line between them. Check: from Ketill’s side, KS \sin 61^\circ also gives about 7.19.

Example 6

Angles and a perimeter

A triangle has angles 40^\circ, 60^\circ and 80^\circ, and its perimeter is 50. Find its three sides.

Solution

No single side is known, but the law of sines says every side is the same number k times the sine of its opposite angle:

a = k \sin 40^\circ, \quad b = k \sin 60^\circ, \quad c = k \sin 80^\circ.

The perimeter adds them: k(\sin 40^\circ + \sin 60^\circ + \sin 80^\circ) = 50. The sum of the sines is about 2.4936, so k \approx 20.05. Then

a \approx 12.89, \qquad b \approx 17.36, \qquad c \approx 19.75.

Check: 12.89 + 17.36 + 19.75 = 50.00. The angles fixed the shape; the perimeter fixed the size.

Boundaries

The law needs one complete pair. Every ratio in it is a side over the sine of the angle across from it. If no side is known together with its opposite angle, and no two angles are known (so the angle sum cannot make a pair), the law has no starting point. Three sides (SSS) or two sides and the angle between them (SAS) are exactly those cases; they belong to the law of cosines.

Two sides and an angle not between them (SSA) is a different problem. Then the law is used to find an angle from its sine, and since \sin x = \sin(180^\circ - x), one sine can belong to an acute angle and an obtuse one. Zero, one or two triangles may fit; The ambiguous case sorts them out. AAS and ASA never meet this, because their angles come from the angle sum, never from a sine.

The two given angles must add to less than 180^\circ. Otherwise no third angle is left and no triangle exists.

Angles strictly between 0^\circ and 180^\circ. The division by \sin A in the derivation is allowed only because \sin A > 0 for such angles.

Decimal answers are approximations. Exact values appear only when the sines are exact, as in a 30-60-90 or 45-45-90 triangle.

Where this goes next

The ambiguous case takes the law into SSA, where the altitude from this lesson decides how many triangles exist. The law of cosines covers the SSS and SAS triangles this law cannot start. The altitude h = b \sin A also gives the area of a triangle from two sides and the angle between them. And the lookout problem is the seed of Applications of oblique triangles: locating a point from the two ends of a known baseline, which is how surveyors and navigators fix positions.

Exercises

Two angles and a side

Solve each triangle: find the missing angle and the two missing sides, to two decimal places unless an exact value is asked for.

A = 50^\circ, B = 70^\circ, a = 12.
B = 35^\circ, C = 100^\circ, b = 20.
A = 42^\circ, C = 73^\circ, b = 18.
In triangle PQR, P = 28^\circ, Q = 117^\circ, and the side PQ = 9. Find R, QR and PR.
A = 105^\circ, B = 30^\circ, c = 16. Give b exactly as well.
A = 45^\circ, B = 60^\circ, a = 10. Find b exactly.
A = 30^\circ, C = 120^\circ, a = 7. Find B, b, and c exactly.

Heights and the obtuse case

In an acute triangle ABC, a = 10, B = 40^\circ and A = 58^\circ. Find the altitude h from C to side AB, then use h = b \sin A to find b.
In triangle ABC, B = 130^\circ and a = 8. Find the length h of the altitude from C to the line AB.
The triangle in Exercise 9 also has A = 20^\circ. Use h to find b, and confirm it with the law of sines.

Lookouts and other uses

Ketill’s tower K and Hallveig’s tower H are 12.4 miles apart. A plume makes an angle of 48^\circ with the baseline at K and 73^\circ at H. How far is the plume from each tower?
In Exercise 11, how far is the plume from the straight line joining the two towers?
Hallveig drives east on a straight road. At point P the line of sight to a plume north of the road makes an angle of 32^\circ with the road ahead. After 3.5 more miles, at point Q, the angle with the road ahead is 51^\circ. How far is the plume from Q, and from P?
A triangle has angles 35^\circ, 65^\circ and 80^\circ, and its perimeter is 60. Find its longest side.

Reasoning

Use the unit circle to explain why \sin(180^\circ - x) = \sin x, and say why the obtuse case of the derivation needs it.
Show that for a right triangle with C = 90^\circ, the law of sines reduces to the sine ratio \sin A = \dfrac{a}{c}.
Explain why the law of sines cannot start on a triangle given by three sides, or by two sides and the angle between them.
Use the law of sines to prove that in any triangle, if A < B then a < b: the larger side is across from the larger angle.
Solving an AAS or ASA triangle never needs an inverse sine. Explain why, and explain why finding an angle from its sine would cause trouble.

Answers

C = 60^\circ, b \approx 14.72, c \approx 13.57. C = 180^\circ - 50^\circ - 70^\circ; b = \dfrac{12 \sin 70^\circ}{\sin 50^\circ}, c = \dfrac{12 \sin 60^\circ}{\sin 50^\circ}.
A = 45^\circ, a \approx 24.66, c \approx 34.34. a = \dfrac{20 \sin 45^\circ}{\sin 35^\circ}, c = \dfrac{20 \sin 100^\circ}{\sin 35^\circ}.
B = 65^\circ, a \approx 13.29, c \approx 18.99. ASA: first B = 180^\circ - 42^\circ - 73^\circ, then a = \dfrac{18 \sin 42^\circ}{\sin 65^\circ}, c = \dfrac{18 \sin 73^\circ}{\sin 65^\circ}.
R = 35^\circ, QR \approx 7.37, PR \approx 13.98. PQ is across from R, so QR = \dfrac{9 \sin 28^\circ}{\sin 35^\circ} and PR = \dfrac{9 \sin 117^\circ}{\sin 35^\circ}.
C = 45^\circ, a \approx 21.86, \displaystyle b = 8\sqrt{2} \approx 11.31. b = \dfrac{16 \sin 30^\circ}{\sin 45^\circ}, which is \displaystyle \dfrac{8}{\sqrt{2}/2} = 8\sqrt{2}; a = \dfrac{16 \sin 105^\circ}{\sin 45^\circ}.
\displaystyle b = 5\sqrt{6} \approx 12.25. b = \dfrac{10 \sin 60^\circ}{\sin 45^\circ}, and the halves cancel: \displaystyle b = \dfrac{10\sqrt{3}}{\sqrt{2}}, which is \displaystyle 5\sqrt{6}.
B = 30^\circ, b = 7, \displaystyle c = 7\sqrt{3} \approx 12.12. A = B, so b = a = 7; c = \dfrac{7 \sin 120^\circ}{\sin 30^\circ}, and \displaystyle \sin 120^\circ = \dfrac{\sqrt{3}}{2}, so \displaystyle c = 7\sqrt{3}.
h \approx 6.43, b \approx 7.58. h = a \sin B = 10 \sin 40^\circ; then b = \dfrac{h}{\sin A} = \dfrac{10 \sin 40^\circ}{\sin 58^\circ}, the law of sines itself.
h \approx 6.13. The right triangle at B holds 180^\circ - 130^\circ = 50^\circ, so h = 8 \sin 50^\circ, which equals 8 \sin 130^\circ.
b \approx 17.92. h = b \sin A, so b = \dfrac{8 \sin 50^\circ}{\sin 20^\circ}. The law gives b = \dfrac{8 \sin 130^\circ}{\sin 20^\circ}, the same value since \sin 130^\circ = \sin 50^\circ.
About 13.83 miles from K and 10.75 miles from H. The angle at the plume is 59^\circ; distance from K = \dfrac{12.4 \sin 73^\circ}{\sin 59^\circ}, from H = \dfrac{12.4 \sin 48^\circ}{\sin 59^\circ}.
About 10.28 miles. The altitude from the plume is (distance from K) \times \sin 48^\circ; from H’s side, (distance from H) \times \sin 73^\circ agrees.
About 5.70 miles from Q and 8.35 miles from P. The angle at Q inside the triangle is 180^\circ - 51^\circ = 129^\circ, so the angle at the plume is 180^\circ - 32^\circ - 129^\circ = 19^\circ. Then QS = \dfrac{3.5 \sin 32^\circ}{\sin 19^\circ}, PS = \dfrac{3.5 \sin 129^\circ}{\sin 19^\circ}.
About 23.97. The sides are k \sin 35^\circ, k \sin 65^\circ, k \sin 80^\circ with k = \dfrac{60}{\sin 35^\circ + \sin 65^\circ + \sin 80^\circ}; the longest is across from 80^\circ.
The terminal point of x is (\cos x, \sin x). Reflecting it across the y-axis gives (-\cos x, \sin x), which is the terminal point of 180^\circ - x, since the reflection measures the same angle back from the negative x-axis. Same y-coordinate, same sine. In the obtuse case the right triangle at B contains 180^\circ - B, not B, so the sine ratio gives h = a \sin(180^\circ - B); the identity turns this into a \sin B.
\sin 90^\circ = 1, so \dfrac{c}{\sin C} = c, and the law says \dfrac{a}{\sin A} = c, that is \sin A = \dfrac{a}{c}: the side opposite A over the hypotenuse.
Each ratio needs a side and its opposite angle. With SSS no angle is known. With SAS the one known angle is between the known sides, so the side across from it is unknown, and each of the other two angles is unknown too. Every ratio has an unknown and the law gives no number; the law of cosines starts these.
Let k = \dfrac{a}{\sin A} > 0, so a = k \sin A and b = k \sin B; it is enough to show \sin A < \sin B. If B \le 90^\circ, sine increases from 0^\circ to 90^\circ (the terminal point climbs), so \sin A < \sin B. If B > 90^\circ, then A + B < 180^\circ gives A < 180^\circ - B < 90^\circ, so \sin A < \sin(180^\circ - B) = \sin B.
In AAS and ASA the third angle comes from the angle sum, and each side is k times the sine of a known angle: sines are only evaluated, never undone. Going back from a sine is ambiguous: \sin x = 0.5 holds for x = 30^\circ and x = 150^\circ, and both are possible triangle angles. That is the SSA situation of The ambiguous case.