The law that finishes the job the Law of Sines cannot start: Side-Angle-Side and Side-Side-Side. Proves it by dropping a triangle into standard position and applying the distance formula, shows how it generalises the Pythagorean Theorem, and explains why the sign of the cosine makes it immune to the ambiguous case. Closes with Heron's Formula for the area of a triangle from its three sides alone.
Subject: Trigonometry · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.3 The Law of Cosines, pp. 910-916
Objectives
Five outcomes, and the fourth is the one that keeps answers accurate.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 910-916 — the pages these objectives are drawn from
Warm-up
The previous lesson ended by naming two configurations the Law of Sines cannot start from. Here is one of them.
Discussion prompt
A triangle has sides of 7 and 2 with an angle of 50 degrees between them. Write down the Law of Sines for this triangle and see how far you get.
Hint: Which side is opposite the 50 degree angle?
Answer:
The side opposite the 50 degree angle is the third one, which was not given. So every ratio in the law contains at least one unknown, and no equation can be solved.
But the triangle is clearly determined — you could build exactly one triangle from a 7, a 2 and a 50 degree angle between them, using nothing but a ruler and a protractor. The information is sufficient; the tool is wrong.
What is needed is a relation among three sides and one angle rather than among matched pairs. That is the Law of Cosines, and it comes from the distance formula.
Concept
Place the triangle with the angle at the origin and one adjacent side along the x-axis. The third vertex then has coordinates given by the circular functions, and the remaining side is just the distance between two known points.
Law of Cosines — For a triangle with angle-side opposite pairs, the square of any side equals the sum of the squares of the other two, minus twice their product times the cosine of the angle between them.
\[ a^2 = b^2 + c^2 - 2bc\cos(\alpha) \]
Solving for the cosine instead gives the form used when three sides are known: cosine alpha equals b squared plus c squared minus a squared, all over 2bc.
Figure (svg): A triangle placed in standard position with the angle alpha at the origin and side b along the positive x-axis, so that the third vertex has coordinates given by the circular functions
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 910-911
Section
Section 1
Concept
With the triangle in standard position, two vertices are at known coordinates and the third is given by the circular functions. The remaining side is the distance between two of them, and expanding that distance gives the law.
The setup holds for any alpha strictly between 0 and 180 degrees, obtuse ones included, because the coordinates of B come from the circular functions rather than from a right triangle.
Figure (svg): A triangle placed in standard position with the angle alpha at the origin and side b along the positive x-axis, so that the third vertex has coordinates given by the circular functions
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 910-911
Picture it
The correction term measures the departure from a right angle.
Figure (svg): The Law of Cosines shown reducing to the Pythagorean Theorem when the angle is a right angle, because the cosine of ninety degrees is zero
That is not a coincidence. The proof runs through the distance formula, which is itself the Pythagorean Theorem, so the two results are the same fact in different clothes.
Worked example
The whole proof, from the coordinates to the law.
\[ \text{Derive } a^2 = b^2 + c^2 - 2bc\cos(\alpha). \]
Write the distance from B to C
Why: B is at c cosine alpha comma c sine alpha, and C is at b comma 0.
\[ a ^{2} = (c \cos a - b) ^{2} + (c \sin a) ^{2} \]
Expand the first square
Why: The middle term is the one that survives.
\[ c ^{2} \cos ^{2} - 2 b c \cos + b ^{2} + c ^{2} \sin ^{2} \]
Group the two squared cosine and sine terms
Why: They share a factor of c squared.
\[ c ^{2}(\cos ^{2} + \sin ^{2}) + b ^{2} - 2 b c \cos \]
Apply the Pythagorean Identity
Why: The bracket is 1.
\[ a ^{2} = b ^{2} + c ^{2} - 2 b c \cos \alpha \]
Figure (svg): The solution to Worked example the derivation shown as a ladder of expressions, one row per legal move
\[ a^2 = b^2 + c^2 - 2bc\cos(\alpha) \]
Verify: test on an equilateral triangle
Why: With all sides 1 and all angles 60 degrees, the right side is 1 plus 1 minus 2 times one half, which is 1. The left side is also 1. The law is consistent.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 911-911
Sorting
The deciding question is whether a matched angle-side pair is available.
Sort into buckets
Sort each set of given data.
Worked example
One substitution turns the general law into the familiar one.
\[ \text{Show that the law reduces to } c^2 = a^2 + b^2 \text{ when } \gamma = 90^\circ. \]
Write the version with gamma
Why: The angle between sides a and b.
\[ c ^{2} = a ^{2} + b ^{2} - 2 a b \cos \gamma \]
Substitute the right angle
Why: The cosine of 90 degrees.
\[ \cos 90 = 0 \]
The correction term vanishes
Why: Anything times zero is zero.
\[ -2 a b(0) = 0 \]
Read off the result
Why: What remains is the Pythagorean Theorem.
\[ c ^{2} = a ^{2} + b ^{2} \]
Figure (svg): The solution to Worked example recovering Pythagoras shown as a ladder of expressions, one row per legal move
\[ \cos(90^\circ) = 0 \;\Longrightarrow\; c^2 = a^2 + b^2 \]
Verify: check the sign of the correction elsewhere
Why: For an acute gamma the cosine is positive, so c squared comes out less than a squared plus b squared, and the side opposite is shorter than the right-angled case. For an obtuse gamma the cosine is negative and c is longer. Both match the geometry.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 911-911
Trap
\[ a = 5, \; b = 8, \; \alpha = 40^\circ \;\Longrightarrow\; c^2 = 5^2 + 8^2 - 2(5)(8)\cos(40^\circ) \]
Substitute the two sides and the given angle
Why: The formula has slots for two sides and an angle, and all three are available.
But the version used is the one for gamma, the angle between a and b. Here the given angle is alpha, which is opposite side a rather than between the two given ones.
\[ a^2 = b^2 + c^2 - 2bc\cos(\alpha) \;\Longrightarrow\; 25 = 64 + c^2 - 16c\cos(40^\circ) \]
Use the version whose angle is the one you were given
Why: That version has the given angle's opposite side alone on the left.
Here the equation is quadratic in c, which is the algebraic signature of the ambiguous case appearing again: two positive roots would mean two triangles. The angle in the cosine and the side alone on the left must be an opposite pair.
Prediction
In a triangle, the angle gamma is obtuse.
Predict first
How does c squared compare with a squared plus b squared?
Correct: c squared is larger.
Why: The law says c squared equals a squared plus b squared minus 2ab cosine gamma. For an obtuse gamma the cosine is negative, so the term being subtracted is negative and the whole correction adds. Hence c squared exceeds a squared plus b squared, which is geometrically right: opening the angle past 90 degrees pushes the two far endpoints further apart. This gives a quick test for whether a triangle with three known sides contains an obtuse angle.
Faded example
The expanded distance formula has been grouped.
Fill in the blanks
a^2 = c^2\left(\cos^2\alpha + \sin^2\alpha\right) + b^2 - 2bc\cos\alpha = c^2 \cdot 1 + b^2 - 2bc\cos\alpha
Why: The bracket is the Pythagorean Identity, which equals 1 for every angle. That single substitution is what collapses the expanded distance formula into the law, and it is the only trigonometric fact the proof uses.
Socratic
The proof begins by moving the triangle rather than by reasoning about it where it stands.
Discussion prompt
What does placing alpha at the origin with b on the axis buy, and why is that legitimate?
Hint: What do you know about a point on a circle of radius c?
Answer:
It buys coordinates. With alpha at the origin, the vertex B sits on a circle of radius c at angle alpha, so its coordinates are immediately c cosine alpha and c sine alpha — the definition of the circular functions, needing no right triangle.
It is legitimate because congruence is preserved by translation and rotation. Any triangle can be moved into this position without changing any of its side lengths or angles, so a relation proved here holds everywhere.
This is a general technique worth naming: choose coordinates that make the unknowns readable. The same move proved the distance formula, the equations of the conics, and will prove the polar identities two lessons from now.
Section
Section 2
Concept
The angle is included, meaning it is adjacent to both given sides. The law gives the third side directly, and from there the remaining angles follow.
Once the third side is known, the remaining angles could be found with either law, but the Law of Cosines has an advantage explained in the next section.
Figure (svg): Two columns separating the cases handled by the Law of Sines from those handled by the Law of Cosines
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 911-912
Picture it
Between them, the two laws cover every triangle that can be determined at all.
Figure (svg): Two columns separating the cases handled by the Law of Sines from those handled by the Law of Cosines
The Law of Cosines is the one that needs no matched pair, which is exactly why it can start where the other cannot.
Worked example
Example 11.3.1, part 1. Beta is 50 degrees, a is 7, and c is 2.
\[ \text{Solve the triangle with } \beta = 50^\circ, \; a = 7, \; c = 2. \]
Use the version with beta
Why: Beta is between sides a and c, so b stands alone.
\[ b ^{2} = 49 + 4 - 28 \cos 50 \]
Take the positive root
Why: Exactly the square root of 53 minus 28 cosine 50 degrees.
\[ b = 5.92 \]
Find the largest unknown angle first
Why: The longest side is a, so alpha is the largest angle.
\[ \cos \alpha = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c} \]
Then find gamma from the original data
Why: Using the cosine version again rather than the angle sum.
\[ \alpha = 114.99, \gamma = 15.01 \]
Figure (svg): The solution to Worked example an SAS triangle shown as a ladder of expressions, one row per legal move
\[ b = \sqrt{53 - 28\cos(50^\circ)} \approx 5.92, \quad \alpha \approx 114.99^\circ, \quad \gamma \approx 15.01^\circ \]
Verify: check the angle sum and the ordering
Why: The three angles are 114.99, 50 and 15.01, summing to 180 exactly. The sides opposite them are 7, 5.92 and 2, decreasing as the angles decrease. Both checks pass.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 911-912
Faded example
A triangle has b equal to 9, c equal to 5, and alpha equal to 110 degrees between them.
Fill in the blanks
a^2 = 81 + 25 - 2(9)(5)\cos(110^\circ) = 106 + 30.78 \;\Longrightarrow\; a \approx 11.69
Why: The cosine of 110 degrees is about negative 0.342, so minus 90 times it is plus 30.78. Note the sign: an obtuse included angle makes the correction term positive, so a comes out longer than it would with a right angle. The square root of 136.78 is about 11.69.
Worked example
Example 11.3.2. The pond cannot be crossed, so the width is computed rather than measured.
\[ \text{From } P, \text{ two shores are } 950 \text{ and } 1000 \text{ feet away with } 60^\circ \text{ between the sight lines. Find the width.} \]
Recognise the case
Why: Two distances and the angle between them: SAS.
Write the law for the missing side
Why: The width is opposite the 60 degree angle.
\[ w ^{2} = 950 ^{2} + 1000 ^{2} - 2(950) (1000) \cos 60 \]
Evaluate the cosine and multiply
Why: The cosine of 60 degrees is one half.
\[ w ^{2} = 1902500 - 950000 \]
Take the root
Why: The square root of 952500.
\[ w = 976\text{ feet} \]
Figure (svg): The solution to Worked example a length across an obstacle shown as a ladder of expressions, one row per legal move
\[ w = \sqrt{952500} \approx 976 \text{ feet} \]
Verify: bound the answer
Why: With an angle of 60 degrees the width must lie between the difference of the two distances, 50 feet, and their sum, 1950 feet. It should also be near the shorter side, since 60 degrees is a fairly closed angle. Nine hundred seventy-six feet sits sensibly between 950 and 1000, as expected for an angle just under a right angle.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 913-913
Error analysis
A student computes the pond width.
Annotate
On: \( w^2 = 950^2 + 1000^2 - 2(950)(1000)\cos(60^\circ) \;\Longrightarrow\; w = 950 + 1000 - 976 \)
This error survives because in this particular problem the wrong method lands near the right answer. Change the angle to 120 degrees and the two diverge by hundreds of feet. Always evaluate the whole right side before rooting.
Prediction
Two sides of a triangle are 6 and 10, and the angle between them varies.
Predict first
What is the range of possible lengths for the third side?
Correct: Between 4 and 16, exclusive.
Why: The law gives the third side squared as 136 minus 120 cosine of the included angle. The cosine ranges over negative one to one exclusively for an angle strictly between 0 and 180 degrees, so the expression ranges over 16 to 256, and the side over 4 to 16. Those endpoints are the difference and the sum of the two sides — the degenerate cases where the triangle collapses to a line. This is the triangle inequality, recovered from the Law of Cosines.
Two truths and a lie
Three of these are true of the Side-Angle-Side case and one is false.
Eliminate the wrong options
One of these statements about the Side-Angle-Side case is wrong.
Survives elimination: C
Why: The false statement is C. The whole reason this lesson exists is that SAS gives no matched pair, so the Law of Sines has nothing to work with. It becomes usable only after the Law of Cosines has supplied the missing side.
Edge cases
The included angle in an SAS problem can be anything strictly between 0 and 180 degrees.
Discussion prompt
What happens to the triangle as the angle approaches 0, and as it approaches 180?
Hint: What does the law give for the third side at each end?
Answer:
As the angle approaches 0, the cosine approaches 1 and the third side squared approaches a squared minus 2ab plus b squared, which is a minus b, all squared. So the third side approaches the difference of the two given sides and the triangle flattens with the two sides folded on top of one another.
As the angle approaches 180, the cosine approaches negative 1 and the third side approaches the sum of the two given sides, with the triangle flattening the other way into a straight segment.
Neither limit is a triangle, which is why the angle must be strictly between. But the two limits are exactly the bounds of the triangle inequality, so that geometric fact falls out of this law as a special case rather than being an extra rule to remember.
Section
Section 3
Concept
With three sides given, every angle is found from the cosine form. That is fortunate, because the cosine distinguishes acute angles from obtuse ones and the sine does not.
\[ \cos(\alpha) = \frac{b^2 + c^2 - a^2}{2bc} \]
Compare the arcsine, whose range stops at 90 degrees and which therefore can never report an obtuse angle. That limitation caused the entire ambiguous case in the previous lesson, and it does not arise here.
Figure (svg): The cosine taking positive values for acute angles and negative values for obtuse angles, contrasted with the sine which is positive for both
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 911-913
Picture it
There is at most one obtuse angle in a triangle, and its position is known in advance.
Figure (svg): A triangle with its three sides ordered by length, showing that the largest angle sits opposite the longest side and is the only one that could be obtuse
Identifying it first means it is found by a method that can see it, and the remaining two are then guaranteed acute.
Worked example
Example 11.3.1, part 2. Sides 4, 7 and 5, no angles given.
\[ \text{Solve the triangle with } a = 4, \; b = 7, \; c = 5. \]
Find the largest angle first
Why: The longest side is b, so beta is the largest.
\[ \cos \beta = \frac{a ^{2} + c ^{2} - b ^{2}}{2 a c} \]
Substitute
Why: Sixteen plus 25 minus 49, over 40.
\[ \cos \beta = -\frac{8}{40} = -\frac{1}{5} \]
Read the sign
Why: The cosine is negative, so beta is obtuse.
\[ \beta = 101.54 ^\circ \]
Find the other two from the ORIGINAL sides
Why: Both come out acute, as they must.
\[ \gamma = 44.42, \alpha = 34.05 \]
Figure (svg): The solution to Worked example an SSS triangle shown as a ladder of expressions, one row per legal move
\[ \beta = \arccos\left(-\tfrac{1}{5}\right) \approx 101.54^\circ, \quad \gamma = \arccos\left(\tfrac{5}{7}\right) \approx 44.42^\circ, \quad \alpha = \arccos\left(\tfrac{29}{35}\right) \approx 34.05^\circ \]
Verify: check the ordering, and notice the rounding
Why: The largest angle is opposite the longest side and the smallest opposite the shortest, as required. The three approximations sum to 180.01 degrees, which is geometrically impossible — the excess is pure rounding, and the book flags it deliberately as a reminder that decimal answers carry error.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 912-913
Sorting
Compare the longest side squared against the sum of the other two squared.
Sort into buckets
Sort each triangle by its largest angle.
Worked example
Sometimes only the shape of the answer is wanted.
\[ \text{Does the triangle with sides } 5, 6, 10 \text{ contain an obtuse angle?} \]
Identify the candidate
Why: Only the angle opposite the longest side can be obtuse.
\[ \text{the angle opposite } 10 \]
Write its cosine
Why: Twenty-five plus 36 minus 100, over 60.
\[ \cos = -\frac{39}{60} \]
Read the sign
Why: Negative, so the angle is obtuse.
Note the shortcut
Why: It was enough to compare 10 squared against 5 squared plus 6 squared.
\[ 100 > 61 \]
Figure (svg): The solution to Worked example detecting an obtuse angle without solving shown as a ladder of expressions, one row per legal move
\[ \cos = -\tfrac{39}{60} < 0 \;\Longrightarrow\; \text{obtuse, about } 130.5^\circ \]
Verify: confirm by the shortcut
Why: The numerator of the cosine is b squared plus c squared minus a squared, so its sign is decided by comparing the longest side squared with the sum of the other two squared. One hundred exceeds 61, so the numerator is negative and the angle is obtuse — no arccosine needed.
Trap
\[ \frac{\sin\beta}{7} = \frac{\sin(34.05^\circ)}{4} \;\Longrightarrow\; \beta = \arcsin(0.9798) \approx 78.46^\circ \]
Find one angle by the cosine form, then use the Law of Sines for the rest
Why: It is fewer keystrokes, and the ratio is easy to set up.
But beta is really 101.54 degrees, and 78.46 is its supplement. The arcsine cannot return an obtuse value, so it silently gave the wrong one of the two candidates.
\[ \cos(\beta) = \frac{a^2 + c^2 - b^2}{2ac} = -\tfrac{1}{5} \;\Longrightarrow\; \beta = \arccos\left(-\tfrac{1}{5}\right) \approx 101.54^\circ \]
Use the cosine form for any angle that might be obtuse
Why: The negative cosine identifies it unambiguously.
If the Law of Sines is used at all, use it only for the smallest unknown angle, which is guaranteed acute because a triangle has at most one obtuse angle. The book states exactly this rule, and it is the safe way to mix the two laws.
Prediction
You compute an angle and your calculator's arccosine returns 118 degrees.
Predict first
Should you also consider its supplement, 62 degrees?
Correct: No; the arccosine already gives the only valid angle.
Why: The arccosine's range is 0 to 180 degrees, which is exactly the range of possible angles in a triangle, and the cosine is one-to-one on that interval. So each cosine value corresponds to exactly one triangle angle and there is no second candidate to consider. This is precisely what the arcsine cannot promise, and it is why the Law of Cosines has no ambiguous case.
Fill the middle
Find the largest angle in the triangle with sides 5, 12 and 14.
Fill in the blanks
\cos\theta = \frac-27obtuse = \frac___}___ \;\Longrightarrow\; \theta \text___ ___
Why: The numerator is negative 27, so the cosine is about negative 0.225 and the angle is about 103 degrees. The sign alone answered the question before the arccosine was ever taken, which is the practical value of the cosine form.
Explain it to yourself
Both laws can finish a triangle once one angle is known.
Discussion prompt
Explain why the Law of Cosines is preferred, and state the one rule that makes the Law of Sines safe to use anyway.
Hint: What does each inverse function refuse to tell you?
Answer:
The cosine of an acute angle is positive and the cosine of an obtuse angle is negative, so the sign itself identifies the angle type. The sine is positive for both, so a sine value alone cannot distinguish them, and the arcsine will always report the acute one.
Since the arccosine's range is 0 to 180 degrees — exactly the range of triangle angles — every cosine value maps to exactly one possible angle. There is nothing to disambiguate.
The rule that makes the Law of Sines safe: use it only for the smallest unknown angle. A triangle contains at most one obtuse angle, so the smallest is certainly acute and the arcsine will report it correctly. The book states this explicitly, and following it lets you use the quicker law without risk.
Section
Section 4
Concept
In a Side-Angle-Side problem the third side must be computed before any angle can be found, so some propagation is unavoidable. Good practice minimises the rest.
In the SSS case no propagation is needed at all: all three angles can be found from the three given sides, which is the book's reason for calling it a rare opportunity.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 912-913
Picture it
Every arrow in a solution is a chance for error to grow.
Figure (svg): A flow showing given data feeding directly into each computed answer rather than each answer feeding into the next
The cost of the good habit is nothing; the cost of the bad one only shows up in the problems where it matters most.
Worked example
Example 11.3.1, part 1 revisited. Gamma from the angle sum, and gamma from the law.
\[ \text{With } \beta = 50^\circ, \; a = 7, \; c = 2, \text{ find } \gamma \text{ two ways.} \]
The quick way: the angle sum
Why: One hundred eighty minus the rounded alpha minus beta.
\[ \gamma = 180 - 114.99 - 50 \]
Note what it depends on
Why: Every rounding error in alpha lands in gamma.
The careful way: the law again
Why: Using a, b and c with the cosine form for gamma.
\[ \cos \gamma = \frac{a ^{2} + b ^{2} - c ^{2}}{2 a b} \]
Compare
Why: Both give about 15.01 degrees here.
\[ \gamma = 15.01 \]
Figure (svg): The solution to Worked example the same angle, two ways shown as a ladder of expressions, one row per legal move
\[ \gamma = \arccos\left(\frac{7 - 2\cos(50^\circ)}{\sqrt{53 - 28\cos(50^\circ)}}\right) \approx 15.01^\circ \]
Verify: ask when the two would differ
Why: They agree to two places here because alpha was accurate. Had alpha been rounded to a whole degree, the angle-sum route would give 15 degrees and the law would still give 15.01. The gap widens with every additional chained step.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 912-912
Prediction
You solve an SSS triangle and your three rounded angles sum to 180.01 degrees.
Predict first
What does that indicate?
Correct: Normal rounding error.
Why: Each angle was rounded independently to two decimal places, so each can be off by up to half a hundredth, and three such errors can accumulate to a hundredth in the sum. The book raises exactly this example and points out that the sum is geometrically impossible while the working is entirely correct. It is a sign that decimals are approximations, not that anything went wrong.
Worked example
The same shape as the pond problem, with an obtuse angle.
\[ \text{From camp it is } 4 \text{ miles to one rim and } 2 \text{ to the other, with } 117^\circ \text{ between. Find the diameter.} \]
Recognise the case
Why: Two distances and the angle between them: SAS again.
Write the law
Why: The diameter is opposite the 117 degree angle.
\[ d ^{2} = 16 + 4 - 16 \cos 117 \]
Evaluate the cosine
Why: About negative 0.454, so the term adds.
\[ d ^{2} = 20 + 7.26 \]
Take the root
Why: The square root of 27.26.
\[ d = 5.22\text{ miles} \]
Figure (svg): The solution to Worked example the diameter of a crater shown as a ladder of expressions, one row per legal move
\[ d = \sqrt{20 - 16\cos(117^\circ)} \approx 5.22 \text{ miles} \]
Verify: check against the bounds
Why: The diameter must lie between the difference of the distances, 2 miles, and their sum, 6 miles. It should be near the top of that range because the angle is obtuse and the two rims are nearly opposite. Five point two two miles fits.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 916-916
Error analysis
A student solving the SAS triangle rounds the third side before using it.
Annotate
On: \( b \approx 5.9 \;\Longrightarrow\; \cos\alpha = \frac{5.9^2 + 4 - 49}{2(5.9)(2)} \approx -0.4157 \)
Round once, at the end, to the precision the question asks for. Every intermediate rounding is error deliberately introduced, and squaring an intermediate value makes it worse.
Sorting
In each situation, decide whether to use given or computed data.
Sort into buckets
Sort each choice.
Faded example
A clock's hour hand is 4 inches and its minute hand 5.5 inches. At four o'clock, find the distance between their tips.
Fill in the blanks
\text120 deg = 4 \times 30^\circ = 8.24 \;\Longrightarrow\; d = \sqrt___})} \approx ___
Why: Each hour mark is 30 degrees around the face, so four hours is 120 degrees. The cosine of 120 degrees is negative one half, so minus 44 times it is plus 22, giving 68.25 under the root and about 8.24 inches. Note the obtuse angle again makes the distance longer than the two hands are individually.
Real world
A robotic arm has two segments, 40 cm and 30 cm, hinged at the elbow. A controller needs the distance from the shoulder to the gripper for any elbow angle.
Discussion prompt
Explain which law applies, why the answer is always unique, and what the extreme values of the reach are.
Hint: What is given, and what is the angle between?
Answer:
The two segment lengths are fixed and the elbow angle is the one between them, so this is Side-Angle-Side and the reach is the third side: the square root of 2500 minus 2400 cosine of the elbow angle.
The answer is always unique because SAS determines exactly one triangle. That matters for a controller: for each elbow angle there is one and only one reach, so the relation is a genuine function and can be tabulated or inverted safely.
The extremes are the triangle inequality bounds. Fully folded, at an elbow angle near 0, the reach approaches 10 cm; fully extended, near 180 degrees, it approaches 70 cm. Those are the difference and the sum of the segment lengths, and no elbow angle can take the gripper outside that annulus — which is exactly the working envelope a robotics engineer would specify.
Section
Section 5
Concept
Combining the Law of Cosines with the area formula from the previous lesson eliminates the angle entirely, leaving an expression in the three side lengths and their semiperimeter.
semiperimeter — Half the perimeter of the triangle, written s. Each factor in Heron's Formula is the semiperimeter minus one of the sides.
\[ A = \sqrt{s(s-a)(s-b)(s-c)}, \quad s = \tfrac{1}{2}(a+b+c) \]
A negative factor is a useful diagnostic: it means the three lengths violate the triangle inequality and no such triangle exists.
Figure (svg): Heron's Formula, showing the semiperimeter and the three differences that go into the product under the square root
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 914-915
Picture it
Four numbers multiplied, one square root taken.
Figure (svg): Heron's Formula, showing the semiperimeter and the three differences that go into the product under the square root
For the triangle with sides 4, 7 and 5 the semiperimeter is 8 and the product is 96, so the area is four root six, about 9.80 square units.
Worked example
Example 11.3.3, the area of the SSS triangle solved earlier.
\[ \text{Find the area of the triangle with } a = 4, \; b = 7, \; c = 5. \]
Compute the semiperimeter
Why: Half of 4 plus 7 plus 5.
\[ s = 8 \]
Compute the three differences
Why: Eight minus each side in turn.
\[ 4, 1, 3 \]
Multiply all four
Why: Eight times 4 times 1 times 3.
\[ 96 \]
Take the root
Why: The square root of 96 simplifies.
\[ A = 4 \sqrt{6} = 9.80 \]
Figure (svg): The solution to Worked example applying Heron's Formula shown as a ladder of expressions, one row per legal move
\[ A = \sqrt{(8)(4)(1)(3)} = \sqrt{96} = 4\sqrt{6} \approx 9.80 \]
Verify: cross-check with the sine formula
Why: From the earlier solution gamma is about 44.42 degrees, so half of 4 times 7 times the sine of 44.42 is about 9.80 as well. The two independent formulas agree, which checks both the area and the earlier angle.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 915-915
Faded example
A triangle has sides 6, 8 and 10.
Fill in the blanks
s = \tfrac1224(6+8+10) = ___ \;\Longrightarrow\; A = \sqrt___ = \sqrt___ = ___
Why: The semiperimeter is 12 and the differences are 6, 4 and 2, giving 576 under the root and an area of 24. This triangle is right-angled, since 36 plus 64 equals 100, so the area can be checked as half of 6 times 8, which is also 24.
Worked example
How the angle disappears.
\[ \text{Sketch the derivation of Heron's Formula from } A = \tfrac{1}{2}ab\sin(\gamma). \]
Square the area formula
Why: Squaring is what makes the sine tractable.
\[ A ^{2} = (\frac{1}{4}) a ^{2} b ^{2} \sin ^{2} \gamma \]
Replace the squared sine
Why: One minus the squared cosine, by the Pythagorean Identity.
\[ A ^{2} = (\frac{1}{4}) a ^{2} b ^{2}(1 - \cos ^{2}) \]
Substitute the Law of Cosines
Why: The cosine becomes an expression in the three sides.
\[ \cos \gamma = \frac{a ^{2} + b ^{2} - c ^{2}}{2 a b} \]
Factor twice as a difference of squares
Why: Each factor rearranges into a semiperimeter difference.
\[ A ^{2} = s(s - a) (s - b) (s - c) \]
Figure (svg): The solution to Worked example the outline of the derivation shown as a ladder of expressions, one row per legal move
\[ A^2 = \frac{a^2b^2}{4}\left(1 - \cos^2\gamma\right) \;\longrightarrow\; s(s-a)(s-b)(s-c) \]
Verify: check one of the factor identities
Why: The semiperimeter minus a is the whole perimeter minus twice a, all over 2, which is b plus c minus a, over 2. That matches one of the factors produced by the difference of squares, confirming the last step of the derivation.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 914-915
Trap
\[ a = 4, b = 7, c = 5 \;\Longrightarrow\; A = \sqrt{16(16-4)(16-7)(16-5)} = \sqrt{19008} \]
Use the perimeter in every slot
Why: The perimeter is the obvious quantity and 16 is easy to compute.
But the formula is built from the semiperimeter, 8, not the perimeter, 16. The answer 137.9 is more than fourteen times too large for a triangle whose longest side is 7.
\[ s = \tfrac{1}{2}(4+7+5) = 8 \;\Longrightarrow\; A = \sqrt{8(4)(1)(3)} = 4\sqrt{6} \approx 9.80 \]
Halve the perimeter first, then subtract each side from that half
Why: The letter s always means the semiperimeter in this formula.
A sanity check catches this instantly: the area of a triangle cannot exceed half the product of its two longest sides, here half of 35, or 17.5. Any answer above that is wrong before it is checked in detail.
Prediction
You apply Heron's Formula to the lengths 2, 3 and 9 and one factor comes out negative.
Predict first
What does that tell you?
Correct: No triangle has those side lengths.
Why: The semiperimeter is 7, and 7 minus 9 is negative 2. A negative factor makes the product negative and the square root imaginary. Geometrically, 2 plus 3 is less than 9, so the two shorter sides cannot reach across the longest one — the triangle inequality fails. Heron's Formula detects impossible triangles automatically, which is a genuinely useful side effect.
Matching
Every factor in the formula has a geometric reading.
Match the pairs
Why: The semiperimeter minus a simplifies to half of b plus c minus a, which is positive exactly when b plus c exceeds a — the triangle inequality for that side. If it equals zero the three points are collinear and the area is zero, which the formula reports correctly.
Counterexample
A student proposes: two triangles with the same perimeter have the same area.
Discussion prompt
Give a counterexample, and say what perimeter alone does determine.
Hint: Try a very flat triangle and a very fat one.
Answer:
Take sides 6, 8, 10, with semiperimeter 12 and area 24. Now take 11, 11, 2, also perimeter 24, with semiperimeter 12 and area the square root of 12 times 1 times 1 times 10, about 10.95. Same perimeter, less than half the area.
Push it further: 11.9, 11.9, 0.2 has the same perimeter and an area of about 1.19. As the triangle flattens, the area approaches zero while the perimeter is unchanged, so perimeter places no positive lower bound on area at all.
What perimeter does determine is an upper bound: among all triangles of a given perimeter, the equilateral one encloses the most area. For perimeter 24 that is the 8-8-8 triangle, with area about 27.7 — larger than both examples above. This is the triangle case of the isoperimetric problem, and Heron's Formula is the natural tool for proving it.
Comparison
Fill the blanks from memory. Which law to use is decided before any arithmetic.
Comparison matrix
| Law of Sines | Law of Cosines | |
|---|---|---|
| cases handled | AAS, ASA, ASS | SAS and SSS |
| needs a matched pair | yes | no |
| inverse function used | arcsine, range 0 to 90 deg | arccosine, range 0 to 180 deg |
| can be ambiguous | yes, in the ASS case | never |
| identifies obtuse angles | no; sine is positive either way | yes; the cosine is negative |
The last two rows are the same fact stated twice. The arccosine can return the full range of triangle angles and the arcsine cannot, so only one of the two laws ever leaves a question open.
Pattern
One decision, then a fixed route.
Round once, at the end. Intermediate rounding is amplified by every squaring the formula performs.
OpenStax Algebra and Trigonometry 2e, §10.2 Non-right Triangles: Law of Cosines §10.2
Check
Choose the law.
Check your understanding
A triangle has a = 9, c = 14 and beta = 62 degrees. Which law finds the missing side, and why?
Answer: B
Why: The angle beta is between the two given sides, so its opposite side b is the missing one and no matched pair exists. This is Side-Angle-Side, and the version of the law with beta gives b directly as the square root of 81 plus 196 minus 252 cosine 62 degrees.
Check
The sign of the cosine.
Check your understanding
In a triangle with sides 7, 9 and 13, is the largest angle acute, right or obtuse?
Answer: C
Why: The largest angle is opposite the side of length 13, and its cosine has numerator 49 plus 81 minus 169, which is negative 39. A negative cosine means an obtuse angle, about 106.6 degrees. The sign of the numerator settled it without any arccosine.
Check
Heron's Formula.
Check your understanding
What is the area of the triangle with sides 5, 5 and 6?
Answer: A
Why: The semiperimeter is 8, and the differences are 3, 3 and 2. The product 8 times 3 times 3 times 2 is 144, whose square root is 12. This isosceles triangle has base 6 and height 4, giving half of 6 times 4, which is also 12.
Real world
Three cell towers are at known positions. A phone reports its distance from each of them, computed from signal travel time. The distances are 3.2 km, 4.7 km and 5.1 km, and the tower positions are fixed and known.
Discussion prompt
Explain what the Law of Cosines contributes here, why the answer is unambiguous, and what happens if the three measured distances are slightly inconsistent.
Hint: The tower separations are known, so what triangles do you actually have?
Answer:
Each pair of towers plus the phone forms a triangle in which all three sides are known: the two measured distances and the tower separation. That is SSS, so the Law of Cosines gives every angle of that triangle, and the angles fix the phone's bearing from each tower.
The answer is unambiguous because SSS determines a triangle rigidly and every angle is found with the arccosine, which cannot return a wrong branch. There is no ambiguous case to worry about, which is why trilateration is preferred over methods that measure angles.
If the measurements are slightly inconsistent — and they always are, since signal timing has noise — the three circles do not meet at a single point but bound a small region. Heron's Formula on that region's triangle is a direct measure of the inconsistency: an area near zero means the fix is tight, and a growing area means the measurements are disagreeing.
This is exactly how GPS reports its accuracy. The receiver is not just solving for a position, it is quantifying how badly the equations fail to agree, and the geometry of this lesson is what turns that disagreement into a number.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
You compute cosine gamma and get a value of negative 0.6. What do you know immediately?
Correct: Gamma is obtuse.
\[ \arccos(-0.6) \approx 126.87^\circ \]
Why: The cosine is negative exactly for angles between 90 and 180 degrees, so a negative value identifies an obtuse angle with no further work. There is no second candidate, because the cosine is one-to-one on the range of possible triangle angles. And nothing is impossible: a cosine of negative 0.6 gives a perfectly ordinary angle of about 126.9 degrees.
Explain it
They have memorised both laws but pick between them by trial and error, writing one down and abandoning it when it fails.
Discussion prompt
In no more than five sentences, give them a rule that decides which law to use before writing anything.
Hint: What does the Law of Sines need that the other does not?
Answer:
A usable answer: scan the given data for an angle whose opposite side is also given. If you find such a pair, the Law of Sines can start; if you do not, it cannot, and the Law of Cosines is the only option.
That scan takes two seconds and is never wrong. Two angles plus any side always yields a pair, since the third angle comes free from the angle sum. Two sides with the angle between them, and three sides with no angle, never yield one — and those are precisely SAS and SSS.
Worth adding: when in doubt the Law of Cosines is the safer guess, because it works whenever it applies and never leaves a second answer hidden.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Choosing between the laws is fixed by the matched-pair scan. SAS is fixed by using the version whose angle is the given one and rooting the whole right side. SSS is fixed by finding the angle opposite the longest side first and reading the sign of its cosine. Heron is fixed by writing s equals half the perimeter before anything else. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page draw a triangle in standard position with the angle at the origin and one side along the x-axis, label all three vertices with their coordinates, and write the four-line derivation of the Law of Cosines beside it. Underneath, write both forms of the law, the one solved for a side and the one solved for a cosine, and note in one line why the Pythagorean Theorem is a special case. In the middle of the page draw two columns headed Law of Sines and Law of Cosines, listing which cases each handles and whether each can be ambiguous, and write beneath them the one question that decides between them. In the bottom left, solve completely the triangle with sides 4, 7 and 5, finding the largest angle first and noting its sign. In the bottom right, write Heron's Formula with the semiperimeter defined, and apply it to the same triangle. Finally, circle the single property of the cosine function that makes this law free of ambiguity.
The circled property is that the cosine is negative for obtuse angles and positive for acute ones, so its sign identifies the angle type. The arccosine's range covers every possible triangle angle, which is why nothing is ever left undecided.
Recap
Five things, and the first is the one that decides every problem in the chapter.
| If the question says | Your first move is |
|---|---|
| Two sides and the angle between | Law of Cosines with that angle |
| Three sides, no angles | Law of Cosines, largest angle first |
| Is the triangle obtuse? | Compare the longest side squared against the other two |
| Find the area, three sides given | Compute the semiperimeter, then Heron |
| An angle might be obtuse | Use the cosine form, never the arcsine |
Every triangle that can be determined at all is now solvable, by one law or the other. The chapter turns next to polar coordinates, where the same circular functions describe position rather than shape.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.3 The Law of Cosines §11.3, pp. 910-916 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Trigonometry — $55/session, free consultation.