The Central Idea

In any triangle, match each side with the angle directly opposite it:

\[ a\leftrightarrow A, \qquad b\leftrightarrow B, \qquad c\leftrightarrow C. \]

The Law of Sines compares side lengths with the sines of their opposite angles:

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

The Law of Cosines connects all three sides with one angle:

\[ \boxed{ a^2=b^2+c^2-2bc\cos A }. \]

Cyclic versions give corresponding formulas for \(b\) and \(c\).

Triangle Notation

Standard triangle notation pairs each capital-letter angle with the lowercase side opposite it:

\[ \boxed{ A\leftrightarrow a,\qquad B\leftrightarrow b,\qquad C\leftrightarrow c }. \]

This pairing is essential. A frequent error is to match a side with an adjacent angle instead of its opposite angle.

The Angle Sum Still Applies

Every Euclidean triangle satisfies

\[ \boxed{A+B+C=180^\circ}. \]

Therefore, if two angles are known, the third is found immediately:

\[ C=180^\circ-A-B. \]

The Law of Sines

The Law of Sines may be written as

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

An equivalent reciprocal form is

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

Either form is valid. Choose the arrangement that makes the unknown easy to isolate.

What Makes the Law of Sines Useful

The Law of Sines is especially effective when at least one complete opposite side-angle pair is known.

Typical cases include:

ASA: Angle–Side–Angle

In an ASA configuration, two angles and the side between them are known.

Because two angles are known, the third angle follows from

\[ A+B+C=180^\circ. \]

Then the Law of Sines determines the remaining sides.

AAS: Angle–Angle–Side

In an AAS configuration, two angles and a side not between them are known.

Again, find the third angle first, then use the known side-angle pair in the Law of Sines.

Example: Law of Sines with Two Angles

Suppose

\[ A=40^\circ, \qquad B=65^\circ, \qquad a=10. \]

Find \(b\).

Use

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

Substitute:

\[ \frac{10}{\sin40^\circ} = \frac{b}{\sin65^\circ}. \]

Solve:

\[ b = \frac{10\sin65^\circ}{\sin40^\circ} \approx \boxed{14.10}. \]

Completing the Same Triangle

The third angle is

\[ C=180^\circ-40^\circ-65^\circ = \boxed{75^\circ}. \]

The remaining side could then be found from

\[ \frac{10}{\sin40^\circ} = \frac{c}{\sin75^\circ}. \]

Solving for an Angle with the Law of Sines

If a side-angle pair and another side are known, we may rearrange:

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

Then

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

Applying inverse sine gives a principal candidate:

\[ \boxed{ B=\arcsin\left(\frac{b\sin A}{a}\right) }. \]

In an SSA case, however, this principal value may not be the only possible angle.

The Ambiguous SSA Case

The configuration SSA means that two sides and a non-included angle are known.

This case is called ambiguous because the information may determine:

The ambiguity occurs because

\[ \sin B=\sin(180^\circ-B). \]

Why Two Angles Can Have the Same Sine

For an acute angle \(B\),

\[ \sin B = \sin(180^\circ-B). \]

Thus, if inverse sine gives an acute candidate \(B_1\), a second candidate may be

\[ \boxed{B_2=180^\circ-B_1}. \]

The second candidate is valid only if the triangle angle sum remains less than \(180^\circ\).

SSA Example with Two Triangles

Suppose

\[ A=30^\circ, \qquad a=7, \qquad b=10. \]

Use the Law of Sines:

\[ \frac{7}{\sin30^\circ} = \frac{10}{\sin B}. \]

Therefore,

\[ \sin B = \frac{10\sin30^\circ}{7} = \frac57. \]

First SSA Solution

The principal inverse-sine value is

\[ B_1 = \arcsin\left(\frac57\right) \approx 45.58^\circ. \]

Then

\[ C_1 = 180^\circ-30^\circ-45.58^\circ \approx 104.42^\circ. \]

This is a valid triangle.

Second SSA Solution

The supplementary angle is

\[ B_2 = 180^\circ-45.58^\circ \approx 134.42^\circ. \]

Then

\[ C_2 = 180^\circ-30^\circ-134.42^\circ \approx 15.58^\circ. \]

This angle is also positive, so a second triangle exists.

Side Lengths in the Two SSA Triangles

For the first triangle,

\[ \frac{7}{\sin30^\circ} = \frac{c_1}{\sin104.42^\circ}, \]

giving

\[ c_1\approx\boxed{13.56}. \]

For the second,

\[ \frac{7}{\sin30^\circ} = \frac{c_2}{\sin15.58^\circ}, \]

giving

\[ c_2\approx\boxed{3.76}. \]

When the SSA Case Gives No Triangle

If the Law of Sines leads to

\[ \sin B>1, \]

then no real angle \(B\) exists.

Therefore the given measurements cannot form a triangle.

When the SSA Case Gives One Triangle

A single triangle may occur if:

A Geometric Height Test for Acute SSA

Suppose \(A\) is acute and \(a\) is opposite \(A\), while \(b\) is another known side.

Define the altitude

\[ \boxed{h=b\sin A}. \]

Then:

The Law of Cosines

The Law of Cosines is

\[ \boxed{ a^2=b^2+c^2-2bc\cos A }. \]

The corresponding cyclic forms are

\[ \boxed{ b^2=a^2+c^2-2ac\cos B } \]

and

\[ \boxed{ c^2=a^2+b^2-2ab\cos C }. \]

Law of Cosines as a Generalized Pythagorean Theorem

If

\[ A=90^\circ, \]

then

\[ \cos90^\circ=0. \]

Therefore

\[ a^2=b^2+c^2-2bc(0) \]

becomes

\[ \boxed{a^2=b^2+c^2}. \]

The Pythagorean relationship is therefore a special case of the Law of Cosines.

When to Use the Law of Cosines

The Law of Cosines is especially useful for:

It can also be useful as a verification tool after solving with the Law of Sines.

SAS: Side–Angle–Side

If two sides and the angle between them are known, the Law of Cosines directly finds the opposite side.

For example, if \(b\), \(c\), and \(A\) are known,

\[ \boxed{ a=\sqrt{b^2+c^2-2bc\cos A} }. \]

SAS Example

Suppose

\[ b=8, \qquad c=11, \qquad A=60^\circ. \]

Then

\[ a^2 = 8^2+11^2-2(8)(11)\cos60^\circ. \]

Since

\[ \cos60^\circ=\frac12, \]

we get

\[ a^2=64+121-88=97. \]

Therefore,

\[ \boxed{a=\sqrt{97}\approx9.85}. \]

After an SAS Calculation

Once the third side is known, the triangle can be completed with either law.

Often the Law of Sines becomes convenient because a complete side-angle pair is now available.

Alternatively, another Law of Cosines formula can determine a remaining angle.

SSS: Side–Side–Side

If all three sides are known, begin by solving for an angle using the Law of Cosines.

Rearranging

\[ c^2=a^2+b^2-2ab\cos C \]

gives

\[ \boxed{ \cos C = \frac{a^2+b^2-c^2}{2ab} }. \]

Then apply inverse cosine.

SSS Example

Suppose

\[ a=7, \qquad b=9, \qquad c=12. \]

Find \(C\).

\[ \cos C = \frac{7^2+9^2-12^2}{2(7)(9)}. \]

Simplify:

\[ \cos C = \frac{49+81-144}{126} = -\frac19. \]

Therefore,

\[ C=\arccos\left(-\frac19\right) \approx \boxed{96.38^\circ}. \]

Largest Side Opposes Largest Angle

In any triangle, longer sides lie opposite larger angles.

In the SSS example,

\[ c=12 \]

is the largest side.

Therefore \(C\) should be the largest angle.

The computed value

\[ C\approx96.38^\circ \]

is consistent with that expectation.

Acute, Right, and Obtuse Triangles from the Law of Cosines

Suppose \(c\) is the longest side.

Compare

\[ c^2 \]

with

\[ a^2+b^2. \]

Then:

Why the Classification Works

From

\[ c^2=a^2+b^2-2ab\cos C, \]

we see that:

Choosing between the Two Laws

A useful first question is:

Do I already know a complete opposite side-angle pair?

If yes, the Law of Sines is often natural.

If no, but the data are SAS or SSS, the Law of Cosines is usually the correct starting tool.

A Quick Decision Table

\[ \begin{array}{c|c} \text{Given information} & \text{Typical first method}\\ \hline \text{ASA} & \text{Law of Sines}\\ \text{AAS} & \text{Law of Sines}\\ \text{SSA} & \text{Law of Sines, check ambiguity}\\ \text{SAS} & \text{Law of Cosines}\\ \text{SSS} & \text{Law of Cosines} \end{array} \]

Why AAA Does Not Determine a Unique Triangle

Three angles determine the shape of a triangle but not its size.

Infinitely many similar triangles can share the same three angle measures.

Therefore AAA alone cannot determine unique side lengths.

Why Three Sides Must Satisfy the Triangle Inequality

Before solving an SSS triangle, check that the sum of any two side lengths exceeds the third:

\[ a+b>c, \] \[ a+c>b, \]

and

\[ b+c>a. \]

If this fails, no nondegenerate triangle exists.

Area from Two Sides and the Included Angle

Trigonometry also gives a useful area formula.

If sides \(b\) and \(c\) include angle \(A\), then

\[ \boxed{ K=\frac12bc\sin A }. \]

Similarly,

\[ \boxed{ K=\frac12ac\sin B } \]

and

\[ \boxed{ K=\frac12ab\sin C }. \]

Deriving the Trigonometric Area Formula

Begin with

\[ K=\frac12(\text{base})(\text{height}). \]

If \(c\) is the base and \(b\) forms angle \(A\) with it, the altitude is

\[ h=b\sin A. \]

Therefore,

\[ K = \frac12c(b\sin A) = \boxed{\frac12bc\sin A}. \]

Area Example

Suppose

\[ b=8, \qquad c=11, \qquad A=60^\circ. \]

Then

\[ K = \frac12(8)(11)\sin60^\circ. \]

Therefore,

\[ \boxed{ K=22\sqrt3\approx38.11 }. \]

Law of Sines and the Circumradius

A deeper form of the Law of Sines is

\[ \boxed{ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R }, \]

where \(R\) is the radius of the triangle's circumcircle.

This connects general triangle trigonometry directly with circle geometry.

Finding the Circumradius

If a side and its opposite angle are known, then

\[ \frac{a}{\sin A}=2R. \]

Therefore,

\[ \boxed{ R=\frac{a}{2\sin A} }. \]

Verification with the Angle Sum

Whenever all three angles have been found, verify

\[ \boxed{A+B+C=180^\circ}. \]

A large discrepancy signals an arithmetic, inverse-function, or rounding error.

Verification with Side–Angle Order

Compare the side lengths and angle measures.

If

\[ a>b, \]

then we should also have

\[ A>B. \]

A mismatch is a strong warning that the triangle has been solved incorrectly.

Verification with the Other Law

Once a triangle is solved, substitute the values into the other law when a high-confidence check is useful.

A triangle found by the Law of Sines should also satisfy the Law of Cosines, and vice versa.

Preserve Precision during Multi-Step Solutions

Triangle solving often requires several dependent calculations.

If an angle is rounded too early, later side lengths may accumulate avoidable error.

Keep full calculator precision through intermediate steps and round final answers according to the problem's requirements.

Degree Mode and Radian Mode

Triangle problems often state angles in degrees, but the formulas work in either degrees or radians.

The calculator must interpret the angle unit consistently.

If the input is

\[ 60^\circ, \]

use degree mode unless the angle has first been converted to radians.

Applied Example: Surveying across an Obstacle

Suppose two accessible points and a distant target form a triangle.

If the baseline between the accessible points is known and the two angles to the target are measured, the situation is an ASA or AAS problem.

The Law of Sines can determine the inaccessible distances without directly crossing the obstacle.

Applied Example: Navigation

Two travel segments with a known included angle form an SAS triangle.

The Law of Cosines can determine the direct displacement between the starting and ending positions.

A remaining direction angle can then be found using the Law of Sines or Cosines.

Applied Example: Structural Bracing

Suppose two structural members of known lengths meet at a known angle.

The distance between their free endpoints is opposite the included angle.

This is an SAS configuration, so the Law of Cosines determines the brace or span length directly.

Common Error: Pairing the Wrong Side and Angle

The Law of Sines requires opposite pairs:

\[ \frac{a}{\sin A}, \qquad \frac{b}{\sin B}, \qquad \frac{c}{\sin C}. \]

A side must never be paired with an adjacent angle simply because the labels appear near each other in a diagram.

Common Error: Using the Law of Sines for SAS Too Early

In an SAS configuration, no complete opposite side-angle pair is initially known.

The Law of Cosines is therefore the natural starting method.

Common Error: Forgetting the Ambiguous SSA Case

If inverse sine gives

\[ B_1, \]

always consider whether

\[ B_2=180^\circ-B_1 \]

could also form a valid triangle.

Stopping at the principal inverse-sine value can miss a second solution.

Common Error: Using the Supplement When the Triangle Sum Fails

A supplementary SSA candidate is valid only if all three triangle angles remain positive.

If

\[ A+B_2\geq180^\circ, \]

then the second candidate cannot form a triangle.

Common Error: Forgetting the Square Root

After using the Law of Cosines,

\[ a^2=b^2+c^2-2bc\cos A, \]

the equation gives \(a^2\), not \(a\).

The final side length is

\[ \boxed{ a=\sqrt{b^2+c^2-2bc\cos A} }. \]

Common Error: Using the Wrong Included Angle

In

\[ a^2=b^2+c^2-2bc\cos A, \]

angle \(A\) must be the angle between sides \(b\) and \(c\), and side \(a\) must lie opposite \(A\).

Common Error: Rounding Angles Too Early

If an angle found by inverse sine or inverse cosine is immediately rounded to a whole degree, subsequent side calculations may become noticeably less accurate.

Keep extra digits until the final answers.

Common Error: Ignoring Triangle Inequality

Three positive numbers do not automatically form a triangle.

For example,

\[ 2,\quad3,\quad6 \]

fail because

\[ 2+3<6. \]

No nondegenerate SSS triangle exists.

A Reliable Procedure for ASA and AAS

  1. Label opposite side-angle pairs carefully.
  2. Find the third angle using \(A+B+C=180^\circ\).
  3. Identify a known side-angle pair.
  4. Use the Law of Sines to find a missing side.
  5. Repeat if another side is needed.
  6. Check side-angle order and the angle sum.

A Reliable Procedure for SSA

  1. Identify the known opposite pair.
  2. Use the Law of Sines to calculate the sine of a second angle.
  3. If the sine value exceeds \(1\), conclude that no triangle exists.
  4. Use inverse sine to find the principal candidate.
  5. Compute the supplementary candidate \(180^\circ-B_1\).
  6. Test whether each candidate keeps the triangle angle sum below \(180^\circ\).
  7. Complete every valid triangle separately.
  8. Preserve precision until the end.

A Reliable Procedure for SAS

  1. Identify the two known sides and their included angle.
  2. Use the Law of Cosines to find the opposite side.
  3. Use the Law of Sines or Cosines to find a remaining angle.
  4. Find the final angle from the triangle angle sum.
  5. Check side-angle order and reasonableness.

A Reliable Procedure for SSS

  1. Check the triangle inequality.
  2. Use the Law of Cosines to find one angle, preferably opposite the largest side.
  3. Use the Law of Sines or another Law of Cosines equation for a second angle.
  4. Find the third angle from the angle sum.
  5. Check that the largest side opposes the largest angle.

Practice 1: Law of Sines

Suppose

\[ A=35^\circ, \qquad B=70^\circ, \qquad a=9. \]

Find \(b\).

Practice 2: Find the Third Angle

A triangle has

\[ A=48^\circ \]

and

\[ B=67^\circ. \]

Find \(C\).

Practice 3: SAS with the Law of Cosines

Suppose

\[ b=6, \qquad c=10, \qquad A=60^\circ. \]

Find \(a\).

Practice 4: SSS Angle

Suppose

\[ a=5, \qquad b=7, \qquad c=8. \]

Find angle \(C\) to the nearest tenth of a degree.

Practice 5: Triangle Area

Find the area of a triangle with

\[ b=12, \qquad c=15, \qquad A=40^\circ. \]

Practice 6: Identify the Best Law

Which law should usually be used first for each case?

  1. ASA
  2. SSS
  3. SAS
  4. AAS

Practice 7: Triangle Classification

A triangle has side lengths

\[ 5,\quad7,\quad10. \]

Is the angle opposite the side of length \(10\) acute, right, or obtuse?

Practice 8: Triangle Inequality

Can side lengths

\[ 4,\quad5,\quad10 \]

form a nondegenerate triangle?

Practice 9: SSA Possibility Check

Suppose

\[ A=30^\circ, \qquad a=4, \qquad b=10. \]

Determine whether a triangle exists.

Practice 10: Circumradius

Suppose

\[ a=12 \]

and

\[ A=30^\circ. \]

Find the circumradius \(R\).

Practice 11: Compare Side and Angle Size

If

\[ a>b, \]

what must be true about \(A\) and \(B\)?

Practice 12: Identify the Error

A student has an SAS triangle and immediately writes a Law of Sines proportion even though no opposite side-angle pair is known.

Explain the problem and identify the better first method.

Practice Answers

Practice 1

\[ \frac9{\sin35^\circ} = \frac b{\sin70^\circ}. \] \[ b = \frac{9\sin70^\circ}{\sin35^\circ} \approx \boxed{14.74}. \]

Practice 2

\[ C=180^\circ-48^\circ-67^\circ. \] \[ \boxed{C=65^\circ}. \]

Practice 3

\[ a^2 = 6^2+10^2-2(6)(10)\cos60^\circ. \] \[ a^2=36+100-60=76. \] \[ \boxed{a=\sqrt{76}=2\sqrt{19}\approx8.72}. \]

Practice 4

\[ \cos C = \frac{5^2+7^2-8^2}{2(5)(7)} = \frac{10}{70} = \frac17. \] \[ C=\arccos\left(\frac17\right) \approx \boxed{81.8^\circ}. \]

Practice 5

\[ K = \frac12(12)(15)\sin40^\circ. \] \[ K = 90\sin40^\circ \approx \boxed{57.85}. \]

Practice 6

  1. ASA: \(\boxed{\text{Law of Sines}}\)
  2. SSS: \(\boxed{\text{Law of Cosines}}\)
  3. SAS: \(\boxed{\text{Law of Cosines}}\)
  4. AAS: \(\boxed{\text{Law of Sines}}\)

Practice 7

The longest side is \(10\). Compare:

\[ 10^2=100 \]

and

\[ 5^2+7^2=25+49=74. \]

Since

\[ 100>74, \]

the opposite angle is

\[ \boxed{\text{obtuse}}. \]

Practice 8

Check

\[ 4+5>10. \]

But

\[ 9>10 \]

is false.

Therefore,

\[ \boxed{\text{no nondegenerate triangle exists}}. \]

Practice 9

The Law of Sines would require

\[ \sin B = \frac{10\sin30^\circ}{4} = \frac54. \]

Since a real sine value cannot exceed \(1\),

\[ \boxed{\text{no triangle exists}}. \]

Practice 10

\[ R = \frac{a}{2\sin A} = \frac{12}{2\sin30^\circ}. \] \[ R = \frac{12}{1} = \boxed{12}. \]

Practice 11

The larger side lies opposite the larger angle.

Therefore,

\[ \boxed{A>B}. \]

Practice 12

The Law of Sines is most useful when a complete opposite side-angle pair is known. In an SAS problem, that pair is not initially available.

The better first method is

\[ \boxed{\text{the Law of Cosines}}. \]

What to Remember

The Law of Sines is

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

The Law of Cosines is

\[ \boxed{ a^2=b^2+c^2-2bc\cos A }, \]

together with its cyclic forms.

Where We Go Next

The Laws of Sines and Cosines extend trigonometry from right triangles to general triangles and strengthen the connection between angle, side length, and geometric measurement.

The next lesson, Periodic Models, Graphs, and Waves, returns to repeating behavior and uses trigonometric functions to model oscillations, cycles, signals, and wave-like phenomena.