Supplies the tool for the two cases the law of sines cannot start. Presents the law of cosines as the Pythagorean theorem with a correction term, uses it for SAS and SSS triangles, shows why solving for an angle with it is never ambiguous, and reaches areas from three sides with Heron's formula.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 8 — Further Applications of Trigonometry
§8.2 Non-right Triangles: Law of Cosines, pp. 923-938
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 923-938 — the pages these objectives are drawn from
Warm-up
The Pythagorean theorem needs a right angle. Most triangles do not have one.
Discussion prompt
Two sides of length 3 and 4 meet at an angle slightly less than a right angle. Is the third side more or less than 5?
Hint: Closing the angle brings the far ends closer.
Answer:
Closing the angle brings the two far ends closer together, so the third side is shorter than 5.
And opening the angle past a right angle would push them apart, making it longer. So the third side depends on the angle continuously.
So the Pythagorean theorem needs a correction term that depends on the angle — negative when the angle is acute, zero when it is right, positive when obtuse. That correction is exactly what the law of cosines supplies.
Concept
The square of a side equals the sum of the other two squares, minus a term proportional to the cosine of the angle between them.
\[ c^2=a^2+b^2-2ab\cos C \]
When the angle is right its cosine is zero and the correction vanishes, recovering the Pythagorean theorem exactly. That special case is a good way to check the formula has been written down correctly.
Figure (svg): A card showing the law of cosines as the Pythagorean theorem plus a correction term that vanishes at a right angle
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 923-927
Section
Section 1
Concept
Each version squares the side being found, adds the squares of the other two, and subtracts twice their product times the cosine of the angle opposite the target side.
Only one version needs to be remembered, since the other two follow by relabelling. What must be got right is the pairing: the angle in the formula is opposite the side on the left, and using any other angle gives a wrong answer.
Figure (svg): A triangle with the law of cosines written for each side, showing that the angle used is always the one opposite the side being found
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 923-928
Picture it
Each version targets a different side.
Figure (svg): A triangle with the law of cosines written for each side, showing that the angle used is always the one opposite the side being found
The structure is identical every time. Writing the target side first and then filling in the other two makes the correct angle obvious.
Worked example
Two sides and the angle between them.
\[ \text{Find } c \text{ when } a=7, \; b=9, \; C=40^\circ. \]
Write the law for c
Why: The angle C is opposite it.
\[ c ^{2} = a ^{2} + b ^{2} - 2 a b \cos C \]
Substitute
Why: All three values known.
\[ 49 + 81 - 126 \cos 40 \]
Evaluate the cosine
Why: About 0.766.
\[ 130 - 96.5 \]
Take the square root
Why: The side length.
\[ \sqrt{33.5} \]
Figure (svg): A card showing the law of cosines as the Pythagorean theorem plus a correction term that vanishes at a right angle
\[ c\approx 5.8 \]
Verify: compare with the right-angle case
Why: If C were a right angle, c would be the root of 130, about 11.4. A 40 degree angle is much closer, so a shorter third side is expected — and 5.8 is well under 11.4. The correction term did what the warmup predicted.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 924-927
Prediction
The angle between the two sides is obtuse.
Predict first
What does the correction term do to the third side?
Correct: Lengthens it, since the cosine is negative.
Why: Subtracting a negative adds, so the square of the third side exceeds the sum of the other two squares. That matches the geometry: opening the angle past a right angle pushes the far ends apart.
Worked example
The special case verifies the formula.
\[ \text{Apply the law of cosines with } C=90^\circ, \; a=3, \; b=4. \]
Write the law
Why: For side c.
\[ c ^{2} = 9 + 16 - 24 \cos 90 \]
Evaluate the cosine
Why: Of a right angle.
\[ 0 \]
Simplify
Why: The correction vanishes.
\[ c ^{2} = 25 \]
Take the root
Why: The hypotenuse.
\[ c = 5 \]
Figure (svg): The solution to Worked example check against Pythagoras shown as a ladder of expressions, one row per legal move
\[ c=5 \]
Verify: confirm the reduction is general
Why: The correction term always has a factor of the cosine, so it vanishes for any right angle regardless of the side lengths. The Pythagorean theorem is therefore a special case rather than a separate fact, which is the cleanest way to remember the formula's shape.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 927-928
Trap
\[ c^2=a^2+b^2-2ab\cos A \]
Substitute whichever angle is known
Why: The formula is written with A because A is the angle given.
The result is a number with no relation to the triangle.
The angle must be opposite the side on the left. For side c that is angle C, always.
If a different angle is known, either relabel so the known angle is opposite the target, or write the version for a different side.
Write the target side first and let it pick the angle. The pairing is forced, and treating it as a choice is what produces the error.
Faded example
Targeting side b this time.
Fill in the blanks
b^2=a^2+c^2-2ac\cos B, \textb ___
Why: The angle is always the one opposite the target side, so targeting b uses angle B. The other two sides appear on the right in both the squares and the product, which fixes the whole formula.
Sorting
The angle must face the target side.
Sort into buckets
Sort each formula.
Explain it to yourself
The law is Pythagoras plus a term.
Discussion prompt
Explain what that term is measuring.
Hint: What happens as the angle changes?
Answer:
It measures how far the angle is from a right angle, scaled by the two side lengths. At exactly ninety degrees the cosine is zero and it disappears.
For an acute angle the cosine is positive, so subtracting shortens the third side — which matches the far ends being pulled together.
For an obtuse angle the cosine is negative, so the term adds and the side is longer than Pythagoras would give. The formula encodes the geometry continuously, which is why reading it as a correction makes both the shape and the signs memorable.
Section
Section 2
Concept
Two sides and their included angle determine the triangle. The law of cosines gives the third side, after which a complete pair exists and the law of sines finishes the job.
The last point is a practical safeguard. The inverse sine cannot return an obtuse angle, so applying it to the angle opposite the shorter side — which cannot be the obtuse one — avoids the issue entirely. The remaining angle then comes from the angle sum, where no ambiguity is possible.
Figure (svg): A triangle with the law of cosines written for each side, showing that the angle used is always the one opposite the side being found
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 928-932
Picture it
One application of the law of cosines opens up the rest.
Figure (svg): A triangle with the law of cosines written for each side, showing that the angle used is always the one opposite the side being found
After the third side is known, every subsequent step is the law of sines or a subtraction. The cosine law is used exactly once.
Worked example
Cosines once, then sines.
\[ \text{Solve with } a=10, \; b=14, \; C=52^\circ. \]
Find the third side
Why: The law of cosines.
\[ c ^{2} = 100 + 196 - 280 \cos 52 \]
Compute
Why: Take the root.
\[ c = 11.1 \]
Find the smaller remaining angle
Why: Opposite the shorter side a.
\[ \sin A = 10 \sin 52 / 11.1 \]
Compute and finish
Why: Inverse sine, then the angle sum.
\[ A = 45.3, B = 82.7 \]
Figure (svg): The solution to Worked example solve an SAS triangle completely shown as a ladder of expressions, one row per legal move
\[ c\approx 11.1,\; A\approx 45.3^\circ,\; B\approx 82.7^\circ \]
Verify: check the ordering
Why: The sides from shortest to longest are 10, 11.1 and 14, and the opposite angles are 45.3, 52 and 82.7 — the same order. That correspondence confirms every value at once, and the three angles sum to 180 as they must.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 929-931
Prediction
A triangle has sides 7, 9 and 12.
Predict first
Which angle might be obtuse?
Correct: The one opposite the side of length 12.
Why: The largest angle always faces the longest side, and at most one angle in a triangle can be obtuse. So only that angle is in doubt, which is why the other two can safely be found with the inverse sine.
Worked example
It removes the ambiguity before it can arise.
\[ \text{Explain why to apply the law of sines to the angle opposite the shorter side.} \]
Recall the inverse sine's range
Why: Only acute angles for positive inputs.
Note which angle could be obtuse
Why: Only the one facing the longest side.
Choose the other one
Why: It must be acute.
Get the last angle by subtraction
Why: The angle sum.
Figure (svg): A contrast between solving for an angle with the sine, which is ambiguous, and with the cosine, which is not
\[ \text{smaller angle}\;\Rightarrow\;\text{acute}\;\Rightarrow\;\text{no ambiguity} \]
Verify: check that at most one angle can be obtuse
Why: Two obtuse angles would already exceed a straight angle, so at most one exists and it faces the longest side. Every other angle is acute, and the inverse sine handles acute angles correctly — which is exactly why this ordering works.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 931-932
Error analysis
A student finishes an SAS triangle by finding the angle opposite the longest side.
Annotate
On: \( \sin B=0.93 \;\Longrightarrow\; B=68.4^\circ \)
The ambiguity here is avoidable rather than intrinsic — unlike the SSA case, the triangle is fully determined and only the method introduced the doubt. Choosing which angle to solve for is what removes it.
Faded example
After the third side is found.
Fill in the blanks
\frac1011.1=\frac______ \;\Longrightarrow\; \sin A=\frac___\sin 52^\circ}___}
Why: The third side creates a complete pair, so the law of sines applies with the known angle C and its side c. Solving for A rather than B is deliberate: A faces the shorter of the two remaining sides, so it cannot be obtuse and the inverse sine returns it correctly.
Sorting
Cosines to start, sines to finish.
Sort into buckets
Sort each step of an SAS solution.
Step zero
You have two sides and the angle between them.
Discussion prompt
What do you compute first?
Hint: What does the law of cosines give directly?
Answer:
The third side, using the law of cosines with the known angle — which is opposite exactly that side.
This is the only step the law of sines cannot do, so it is the one that needs the harder tool. Everything afterwards is easier.
Once the third side is known, a complete pair exists and the rest is the law of sines and a subtraction. Recognising that the cosine law is used exactly once keeps the solution short.
Section
Section 3
Concept
With three sides known, the law of cosines rearranges to give the cosine of any angle, and the inverse cosine returns it with no ambiguity.
Finding the largest angle first is the mirror of the SAS advice: once the only possibly-obtuse angle is settled by the cosine, every remaining angle is acute and the sine law is safe. Either order works so long as the obtuse candidate is handled by the cosine.
Figure (svg): A contrast between solving for an angle with the sine, which is ambiguous, and with the cosine, which is not
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 932-935
Picture it
Its sign separates acute from obtuse.
Figure (svg): A contrast between solving for an angle with the sine, which is ambiguous, and with the cosine, which is not
The inverse cosine's range is the whole span from zero to a straight angle, so it can return any triangle angle. The inverse sine's range covers only the acute ones.
Worked example
Rearrange and take the inverse cosine.
\[ \text{Find the largest angle with } a=7, \; b=9, \; c=12. \]
Identify the largest angle
Why: It faces the longest side.
Rearrange the law
Why: Isolate the cosine.
\[ \cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 a b} \]
Substitute
Why: The three side lengths.
\[ \frac{49 + 81 - 144}{126} \]
Take the inverse cosine
Why: A negative input.
\[ \cos C = -0.111, C = 96.4 \]
Figure (svg): A contrast between solving for an angle with the sine, which is ambiguous, and with the cosine, which is not
\[ C\approx 96.4^\circ \]
Verify: check the sign's meaning
Why: The cosine came out negative, which the inverse cosine turned into an obtuse angle without any supplement being considered. Compare with Pythagoras: 49 plus 81 is 130, less than 144, so the angle must exceed a right angle — confirming the result independently.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 933-935
Prediction
The law of cosines gives a negative cosine.
Predict first
What does that say about the angle?
Correct: It is obtuse.
Why: The cosine is negative only for angles between a right angle and a straight one, which within a triangle means obtuse. No further checking is needed — the sign carries the information the sine's would not.
Worked example
One cosine, then a sine, then a subtraction.
\[ \text{Solve the triangle with sides } 5, \; 6, \; 8. \]
Find the largest angle
Why: Opposite the side of 8.
\[ \cos = \frac{25 + 36 - 64}{60} = -0.05 \]
Take the inverse cosine
Why: Slightly obtuse.
\[ 92.9 ^\circ \]
Find a second angle
Why: By the law of sines.
\[ \sin = 5 \sin 92.9 / 8 \]
Finish by subtraction
Why: The angle sum.
\[ 38.6\text{ and } 48.5 \]
Figure (svg): The solution to Worked example solve an SSS triangle completely shown as a ladder of expressions, one row per legal move
\[ 38.6^\circ,\;48.5^\circ,\;92.9^\circ \]
Verify: check the sum and the ordering
Why: The three angles sum to 180 degrees, and their order matches the order of the opposite sides 5, 6 and 8. Both checks pass, and doing the obtuse angle first with the cosine meant the sine law was applied only to an acute one.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 935-936
Trap
\[ \cos C=-0.111 \;\Longrightarrow\; C=96.4^\circ \text{ or } 263.6^\circ? \]
Look for a second candidate as in SSA
Why: The supplement or the reflex angle is considered.
Time is spent testing possibilities that cannot occur.
There is no ambiguity here. The inverse cosine's range runs from zero to a straight angle, which is exactly the range of possible triangle angles.
Each cosine value corresponds to exactly one such angle, and its sign already says whether the angle is acute or obtuse.
Three sides determine a triangle uniquely, which is the SSS congruence criterion. The absence of ambiguity is guaranteed by geometry, not just by the formula.
Faded example
Solving the law for the angle.
Fill in the blanks
\cos C=\frac2}}}2ab}
Why: Moving the correction term to one side and dividing isolates the cosine. The numerator compares the sum of two squares with the third, which is exactly the Pythagorean comparison that decides the angle type.
Sorting
Compare the sum of the two smaller squares with the largest.
Sort into buckets
Sort each comparison.
Explain it
SSA was ambiguous and SSS is not.
Discussion prompt
Explain to a classmate why the difference exists.
Hint: Compare the two inverse functions' ranges.
Answer:
The inverse sine returns only acute angles for positive inputs, so an obtuse triangle angle has to be found by taking a supplement — and whether to do so is a separate question.
The inverse cosine's range runs from zero to a straight angle, which is exactly the range of possible triangle angles. Each cosine value maps to one of them and no other.
So the cosine carries the information the sine loses: its sign already distinguishes acute from obtuse. A good explanation adds that SSS is a congruence criterion, so the uniqueness is guaranteed by geometry too — the formula and the geometry agree.
Section
Section 4
Concept
Half the perimeter, called the semiperimeter, combines with the three sides in a single square root to give the area — with no angle required.
The alternative for an SSS triangle is to find an angle with the law of cosines and then use the sine area formula, which works but takes three steps and accumulates rounding. Heron's formula reaches the same answer in one.
Figure (svg): A card giving Heron's formula for the area of a triangle from its three side lengths via the semiperimeter
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 935-938
Picture it
One computation from three side lengths.
Figure (svg): A card giving Heron's formula for the area of a triangle from its three side lengths via the semiperimeter
Every factor under the root involves the semiperimeter, which is why computing it first and reusing it is the efficient order.
Worked example
Semiperimeter first.
\[ \text{Find the area of a triangle with sides } 5, \; 6, \; 7. \]
Compute the semiperimeter
Why: Half the perimeter.
\[ s = 9 \]
Compute the three differences
Why: Semiperimeter minus each side.
\[ 4, 3, 2 \]
Form the product
Why: All four factors.
\[ 9 \times 4 \times 3 \times 2 \]
Take the square root
Why: The area.
\[ \sqrt{216} \]
Figure (svg): A card giving Heron's formula for the area of a triangle from its three side lengths via the semiperimeter
\[ \text{Area}=6\sqrt{6}\approx 14.7 \]
Verify: compare with a rough estimate
Why: The triangle is close to equilateral with side about 6, and an equilateral triangle of side 6 has area about 15.6. A slightly irregular triangle of similar size having area 14.7 is entirely plausible, which is a useful order-of-magnitude check.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 936-937
Faded example
Sides of 9, 10 and 11.
Fill in the blanks
s=\frac215}=___
Why: The semiperimeter is half the sum of the three sides. Every factor in Heron's formula is built from it, so computing it first and reusing it is the efficient order.
Worked example
The same area, two ways.
\[ \text{Find the area of the same triangle using an angle instead.} \]
Find an angle
Why: The law of cosines.
\[ \cos C = \frac{25 + 36 - 49}{60} = 0.2 \]
Take the inverse cosine
Why: The angle between sides 5 and 6.
\[ 78.5 ^\circ \]
Apply the sine area formula
Why: Half the product times the sine.
\[ (\frac{1}{2}) (5) (6) \sin 78.5 \]
Compute
Why: Evaluate.
\[ 14.7 \]
Figure (svg): The solution to Worked example compare the two routes shown as a ladder of expressions, one row per legal move
\[ \text{Area}\approx 14.7 \]
Verify: compare the two methods
Why: Both give 14.7, as they must. But this route took an inverse cosine and a sine evaluation, each rounding once, where Heron's took a single square root — so for an SSS triangle Heron's is both shorter and slightly more accurate.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 937-938
Error analysis
A student applies Heron's formula to a 5-6-7 triangle.
Annotate
On: \( s=5+6+7=18, \; \text{Area}=\sqrt{18(13)(12)(11)} \)
The wrong answer is over three times too large, which a quick comparison against a similar-sized familiar triangle catches at once. Estimating the area before computing it is a cheap safeguard.
Prediction
You need a triangle's area.
Predict first
When is Heron's formula the right tool?
Correct: When all three sides are known and no angle is.
Why: With an included angle available the sine formula is one step, so Heron's offers nothing. Its value is in the SSS case, where the alternative requires finding an angle first.
Sorting
It depends on what is known.
Sort into buckets
Sort each situation.
Explain it to yourself
The same area could be found another way.
Discussion prompt
Explain what Heron's formula saves.
Hint: Count the steps in each route.
Answer:
Without it, an SSS area needs the law of cosines to find an angle, then an inverse cosine, then the sine area formula — three steps with three opportunities for rounding.
Heron's does it in one square root from the side lengths directly, with no trigonometry at all.
So it is both faster and slightly more accurate, since intermediate rounding is avoided. It also needs no calculator with trigonometric functions, which is why it long predates them and remains the surveyor's formula for an area from measured distances.
Section
Section 5
Concept
Look for a complete side-angle pair. If one exists, the law of sines is usually shorter; if none does, the law of cosines is the only way to start.
The law of cosines is always usable when the law of sines is, but it involves more arithmetic and a square root. Using the simpler tool when it applies is worth the moment spent checking.
| known | start with | then |
|---|---|---|
| AAS or ASA | the law of sines | finish with the same law |
| SSA | the law of sines | test both candidates |
| SAS | the law of cosines | then the law of sines |
| SSS | the law of cosines | then the law of sines |
| three sides, area only | Heron's formula | nothing further |
Figure (svg): A contrast between solving for an angle with the sine, which is ambiguous, and with the cosine, which is not
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 923-938
Picture it
This difference is why the cosine law never produces two candidates.
Figure (svg): A contrast between solving for an angle with the sine, which is ambiguous, and with the cosine, which is not
The practical upshot is a rule of thumb: handle any possibly-obtuse angle with the cosine, and everything else with the sine.
Worked example
Read the given data and decide.
\[ \text{Given } b=11, \; c=14, \; A=38^\circ, \text{ which law starts?} \]
Identify the case
Why: Two sides and the angle between them.
Look for a complete pair
Why: A is known but a is not.
Conclude
Why: The law of sines cannot start.
Apply it
Why: For the side opposite A.
\[ a ^{2} = 121 + 196 - 308 \cos 38 \]
Figure (svg): A card showing the law of cosines as the Pythagorean theorem plus a correction term that vanishes at a right angle
\[ a\approx 8.7 \]
Verify: check the angle A is opposite a
Why: Angle A lies between sides b and c, so the side opposite it is a — which is exactly the side the formula targets. Had a different angle been given, the formula would have targeted a different side and the setup would change.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 928-933
Prediction
Three side lengths are known and an angle is wanted.
Predict first
Which law applies?
Correct: The law of cosines.
Why: No angle is known, so no ratio in the law of sines can be computed. The law of cosines rearranges to give any angle's cosine from the three sides, and the inverse cosine returns it unambiguously.
Worked example
Both laws in one problem.
\[ \text{Solve with } a=13, \; b=8, \; C=71^\circ. \]
Find the third side
Why: The law of cosines.
\[ c ^{2} = 169 + 64 - 208 \cos 71 \]
Compute
Why: Take the root.
\[ c = 12.6 \]
Find the smallest angle
Why: Opposite the shortest side b.
\[ \sin B = 8 \sin 71 / 12.6 \]
Finish
Why: Inverse sine, then subtract.
\[ B = 36.9, A = 72.1 \]
Figure (svg): The solution to Worked example a mixed strategy shown as a ladder of expressions, one row per legal move
\[ c\approx 12.6,\; A\approx 72.1^\circ,\; B\approx 36.9^\circ \]
Verify: check the ordering and sum
Why: The sides 8, 12.6 and 13 face angles 36.9, 71 and 72.1 respectively — the same order. And the three angles sum to 180. Solving for B rather than A with the sine law was the right choice, since B faces the shortest side and so cannot be obtuse.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 930-936
Trap
\[ \text{two angles and a side known, so set up a cosine equation} \]
Reach for the more general tool
Why: The law of cosines is used because it always applies.
The equation involves a square root and more arithmetic than needed.
When a complete pair exists the law of sines is shorter, needing one division rather than squares and a root.
The cosine law is more general but not more convenient, and every extra operation is an extra chance to round or slip.
Check for a pair first. It takes a glance and decides between a one-line solution and a three-line one.
Matching
Each case has one natural first step.
Match the pairs
Why: Each opening move is the shortest available for that data. The first two need the cosine law because no complete pair exists, and the third has one available as soon as the angle sum is used.
Faded example
Two sides and the angle between them are known.
Fill in the blanks
b=11,\; c=14,\; A=38^\circ \;\Longrightarrow\; a^2}=b^2+c^2-2bc\cos A
Why: The known angle sits between the two known sides, so it is opposite the unknown third side. Without a complete pair the law of sines has nothing to compute, and the cosine law supplies the missing side.
Explain it
Two laws, five cases.
Discussion prompt
Explain to a classmate how to choose between them quickly.
Hint: One question settles it.
Answer:
Ask one question: is there a side with its opposite angle both known? If yes, the law of sines can start; if no, it cannot.
That single test covers all five cases. AAS, ASA and SSA give a pair; SAS and SSS do not.
And add the refinement: when both would work, prefer the sine law — it is one division rather than squares and a root. A good explanation notes that the cosine law is always valid but rarely the shorter route, so generality is not the same as convenience.
Comparison
Fill the blanks from memory. Which to use is decided by one question.
Comparison matrix
| law of sines | law of cosines | |
|---|---|---|
| needs | a complete side-angle pair | two sides and the included angle, or all three sides |
| cases | AAS, ASA, SSA | SAS, SSS |
| ambiguity | yes, in SSA | none, the cosine's sign settles it |
| reduces to | nothing simpler | the Pythagorean theorem at a right angle |
The third row is the practical reason to reach for the cosine law whenever an angle might be obtuse, even in a case the sine law could handle.
Pattern
Five steps, and the fourth is the safeguard.
Step 4 is what keeps the SSA ambiguity from reappearing in a case that is not actually ambiguous. Choosing the right angle to solve for costs nothing.
OpenStax Algebra and Trigonometry 2e, §10.2 Non-right Triangles: Law of Cosines §10.2
Check
The structure of the law.
Check your understanding
In the law of cosines for side c, which angle appears?
Answer: A
Why: The angle is always the one opposite the side on the left. Using any other angle makes the equation false and produces a number unrelated to the triangle.
Check
Ambiguity.
Check your understanding
Why does solving for an angle with the law of cosines produce no ambiguity?
Answer: A
Why: The inverse cosine returns angles from zero to a straight angle, exactly the range a triangle angle can occupy, and each cosine value maps to one of them. The sign already distinguishes acute from obtuse.
Check
Heron's formula.
Check your understanding
In Heron's formula, what does s represent?
Answer: A
Why: The semiperimeter is half the sum of the three sides, and every factor under the root is built from it. Using the full perimeter roughly triples the computed area.
Real world
Satellite positioning is a law-of-cosines computation repeated continuously.
Discussion prompt
A receiver knows its distance to several satellites whose positions are known. How does that fix its position?
Hint: What triangle do two satellites and the receiver form?
Answer:
Two satellites and the receiver form a triangle in which all three sides can be known — the two measured distances and the separation between the satellites, computed from their orbits.
That is the SSS case, so the law of cosines gives every angle, which places the receiver relative to the satellite pair. Additional satellites resolve the remaining freedom.
The computation runs continuously as the satellites move, which is why it has to be fast and unambiguous — and the cosine law's lack of a second candidate matters here. A method that returned two possible positions would need extra work to discard one, at every update.
Commit first
State your confidence along with your answer.
Predict first
Why does the law of cosines reduce to the Pythagorean theorem?
Correct: The correction term contains a cosine, which is zero at a right angle.
Why: With the cosine zero the entire correction vanishes regardless of the side lengths, leaving the sum of the two squares. So the Pythagorean theorem is a special case rather than a separate result, which is the clearest way to remember the formula's shape.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate how to remember the law of cosines without memorising it as a string of symbols.
Hint: What does it become at a right angle?
Answer:
Start from the Pythagorean theorem, which they already know. The law of cosines is that, with a correction added for the angle not being right.
The correction is twice the product of the two sides times the cosine of the angle between them, subtracted. At ninety degrees the cosine is zero and it disappears.
And the sign behaviour follows the geometry: an acute angle shortens the third side and an obtuse one lengthens it, which is what the cosine's sign delivers. A good explanation ends by noting that this reading makes the formula reconstructible rather than recalled.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The first is what makes the formula stick. The third contains the most useful practical habit — handle any possibly-obtuse angle with the cosine and everything else with the sine.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the law of cosines and beside it show what happens when the angle is right. Underneath, work one SAS triangle and one SSS triangle from start to finish, noting at each step which law you used and why. In a box, write Heron's formula and the case it is for.
If your two worked triangles each show why the cosine law was needed to start and why the sine law was safe to finish with, the section's decision structure is on the page.
Recap
Five things, and the first makes the formula reconstructible.
| if you remember one thing | it should be this |
|---|---|
| about the formula | Pythagoras plus a correction that vanishes at a right angle |
| about the pairing | the angle is opposite the side on the left |
| about ambiguity | the cosine's sign settles acute versus obtuse |
| about area | Heron's for three sides, the sine formula for an included angle |
Section 8.3 leaves triangles behind for polar coordinates, describing a point by a distance and a direction rather than by two perpendicular displacements.
OpenStax, Precalculus, §8.2 Non-right Triangles: Law of Cosines §8.2, pp. 923-938 — everything on these slides traces back here
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