The law of cosines and its relationship to the Pythagorean theorem, solving triangles in the SAS and SSS cases, why the largest angle is found first, choosing between the two laws, and Heron's formula for the area from three sides.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 13 — Trigonometric Ratios and Functions
Apply the Law of Cosines
Objectives
Five outcomes. The two cases the law of sines cannot touch.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 889-893 — the lesson these objectives are drawn from
Warm-up
The law of sines needs an angle together with the side opposite it.
Discussion prompt
A triangle has sides 11 and 14 with a 34 degree angle between them. Try to start with the law of sines. What goes wrong?
Hint: Which angle is opposite a known side?
Answer:
The 34 degree angle is BETWEEN the two known sides, so the side opposite it is the unknown one. Every ratio in the law of sines is missing a piece.
\[ a^2 = b^2+c^2-2bc\cos A \]
The law of cosines needs no such pair. It relates all three sides to one angle, so it can start where the law of sines cannot — and it is simply the Pythagorean theorem with one correction term.
Concept
In any triangle, the square of a side equals the sum of the squares of the other two, minus twice their product times the cosine of the angle between them. When that angle is 90 degrees the cosine is zero and the Pythagorean theorem falls out.
law of cosines — For any triangle ABC, a squared equals b squared plus c squared minus 2bc cosine A, with matching formulas for b squared and c squared. It solves the SAS and SSS cases, which the law of sines cannot.
\[ a^2 = b^2+c^2-2bc\cos A \]
A negative cosine signals an obtuse angle, so the formula reports not only the size of an angle but which side of 90 degrees it lies on.
Figure (svg): Two columns comparing the law of sines with the law of cosines
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 889-890
Section
Section 1
Concept
Given two sides and the angle between them, the law of cosines produces the third side. From there the triangle has an angle with its opposite side, so the law of sines finishes it.
\[ b^2 = a^2+c^2-2ac\cos B \]
Each form squares the side whose letter matches the angle. Once you see that pattern, only one formula has to be remembered.
Figure (svg): The three forms of the law of cosines and its link to Pythagoras
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 889-889 — Law of Cosines
Picture it
The law of cosines, and what happens at a right angle.
Figure (svg): The three forms of the law of cosines and its link to Pythagoras
Setting the angle to 90 degrees kills the last term, leaving the Pythagorean theorem exactly. It is a special case of this law, not a separate fact.
Worked example
Example 1.
\[ \text{Solve } \triangle ABC \text{ with } a = 11, \; c = 14, \; B = 34^\circ. \]
Use the law of cosines for b
Why: One twenty-one plus 196 minus 308 cosine 34.
\[ b ^{2} = 61.7 \]
Take the positive root
Why: The root of 61.7.
\[ b = 7.85 \]
Use the law of sines for A
Why: Eleven sine 34 over 7.85.
\[ \sin A = 0.7836, A = 51.6 \]
Subtract for C
Why: One eighty minus 34 minus 51.6.
\[ C = 94.4 ^\circ \]
Figure (svg): A triangle solved from two sides and the angle between them
\[ b \approx 7.85, \; A \approx 51.6^\circ, \; C \approx 94.4^\circ \]
Verify: check the sides rank like their angles
Why: The angles are 34, 51.6 and 94.4 with opposite sides 7.85, 11 and 14 — the same order. The shortest side really does face the smallest angle, which is the free check available on every solved triangle.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 889-889
Fill the middle
Example 1.
Fill in the blanks
b^2 = 121+196-308\cos 34^\circ = 317-255.3 = 61.7
Why: About 61.7, so b is about 7.85. The correction term removed most of the 317, which is what a fairly small included angle does.
Worked example
Guided Practice 1.
\[ \text{Solve } \triangle ABC \text{ with } a = 8, \; c = 10, \; B = 48^\circ. \]
Use the law of cosines
Why: Sixty-four plus 100 minus 160 cosine 48.
\[ b ^{2} = 56.94 \]
Take the root
Why: The root of 56.94.
\[ b = 7.55 \]
Use the law of sines for A
Why: Eight sine 48 over 7.55.
\[ \sin A = 0.7878, A = 52.0 \]
Subtract for C
Why: One eighty minus 48 minus 52.0.
\[ C = 80.0 ^\circ \]
Figure (svg): The solution to Worked example another SAS triangle shown as a ladder of expressions, one row per algebraic move
\[ b \approx 7.55, \; A \approx 52.0^\circ, \; C \approx 80.0^\circ \]
Verify: check C with the law of cosines instead
Why: Cosine of C is 64 plus 56.94 minus 100, over 2 times 8 times 7.55, which is 20.94 over 120.8, giving 0.1734 and an angle of 80.0 degrees. The two routes agree, and using the law of cosines a second time is the safest way to check an angle found by sines.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 891-891
Trap
\[ b^2 = 11^2+14^2-2(11)(14)\cos 34^\circ \]
Subtract 2ac first, then multiply by the cosine
Why: The order of operations is misread.
\[ b^2 = (317-308)\cos 34^\circ = 7.46 \quad \text{(wrong)} \]
The cosine multiplies only the 2ac term. Order of operations puts that multiplication before the subtraction.
\[ b^2 = 317-308(0.829) = 317-255.3 \]
Multiply 2ac by the cosine, then subtract
Why: One term, computed whole.
\[ b^2 = 61.7 \;\Longrightarrow\; b \approx 7.85 \]
The wrong route gives b about 2.7, shorter than either given side, which is impossible for a triangle with a 34 degree included angle.
Matching
The squared side matches the angle.
Match the pairs
Why: The last row is the Pythagorean theorem appearing as a special case, since cosine of 90 degrees is zero. Remembering one form and swapping letters is easier than memorising three.
Sorting
Order of operations.
Sort into buckets
In b squared equals a squared plus c squared minus 2ac cosine B, sort each statement.
Computing the whole term 2ac cosine B as one number before subtracting removes the ambiguity, and it is worth writing that intermediate value down.
Prediction
Commit before reasoning.
Predict first
Solving for an angle you find cos B = -0.3854. What does the sign tell you?
Correct: That B is obtuse, since cosine is negative only between 90 and 180 degrees.
\[ \cos B < 0 \;\Longleftrightarrow\; 90^\circ < B < 180^\circ \]
Why: Across the angles of a triangle, from 0 to 180 degrees, the cosine is positive below 90 and negative above it. So the sign of the cosine reports the type of angle before the inverse cosine is even taken. This is a genuine advantage over the law of sines, whose sine is positive throughout that whole range and so tells you nothing about acute versus obtuse — which is exactly the source of the SSA ambiguity.
Section
Section 2
Concept
Given three sides, rearrange the law of cosines to solve for an angle. Find the angle opposite the LONGEST side first, so that the two remaining angles are guaranteed acute and the law of sines cannot go wrong.
\[ \cos B = \frac{a^2+c^2-b^2}{2ac} \]
A triangle can hold at most one non-acute angle, so once the largest is found the others must be acute and the inverse sine is safe.
Figure (svg): A triangle solved from three sides, largest angle first
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 890-890 — Solve a triangle for the SSS case
Picture it
Example 2: sides 12, 27 and 20.
Figure (svg): A triangle solved from three sides, largest angle first
The 27 is longest, so angle B is found first; it turns out obtuse, which guarantees the other two are not.
Worked example
Example 2.
\[ \text{Solve } \triangle ABC \text{ with } a = 12, \; b = 27, \; c = 20. \]
Pick the longest side
Why: Twenty-seven is b.
Solve the law of cosines for cosine B
Why: Seven twenty-nine minus 144 minus 400, over -480.
\[ \cos B = -0.3854 \]
Take the inverse cosine
Why: A negative cosine gives an obtuse angle.
\[ B = 112.7 ^\circ \]
Use the law of sines and subtract
Why: Sine A is 0.4100.
\[ A = 24.2, C = 43.1 \]
Figure (svg): A triangle solved from three sides, largest angle first
\[ A \approx 24.2^\circ, \; B \approx 112.7^\circ, \; C \approx 43.1^\circ \]
Verify: check the angles sum to 180
Why: Twenty-four point two plus 112.7 plus 43.1 is exactly 180. This is the cleanest possible check on an SSS solution, and it catches any misidentified angle immediately.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 890-890
Fill the middle
Example 2.
Fill in the blanks
\cos B = \frac-0.3854___ = \frac______ = ___
Why: About negative 0.3854. The negative sign says at once that B is obtuse, before the inverse cosine is taken.
Worked example
Guided Practice 2.
\[ \text{Solve } \triangle ABC \text{ with } a = 14, \; b = 16, \; c = 9. \]
Pick the longest side
Why: Sixteen is b.
Solve for cosine B
Why: Two fifty-six minus 196 minus 81, over -252.
\[ \cos B = 0.0833 \]
Take the inverse cosine
Why: A small positive cosine.
\[ B = 85.2 ^\circ \]
Finish with sines and subtraction
Why: Sine A is 0.8720.
\[ A = 60.7, C = 34.1 \]
Figure (svg): The solution to Worked example another SSS triangle shown as a ladder of expressions, one row per algebraic move
\[ A \approx 60.7^\circ, \; B \approx 85.2^\circ, \; C \approx 34.1^\circ \]
Verify: notice this triangle has no obtuse angle
Why: The largest angle came out at 85.2 degrees, just under a right angle, so all three are acute. Finding the largest first was still worth doing: it is the only one that COULD have been obtuse, and knowing it is not makes the rest certain.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 891-891
Error analysis
A student solving a = 12, b = 27, c = 20 starts with angle A by the law of sines.
Annotate
On: \( \text{cannot start: no angle is known yet} \)
Had the student found A first by cosines and then used sines for B, the inverse sine would have returned 67.3 rather than the correct obtuse 112.7 — a silent error with no warning.
Fill the middle
Example 2.
Fill in the blanks
C = 180^\circ-24.2^\circ-112.7^\circ = 43.1^\circ
Why: Forty-three point one degrees. Subtracting for the last angle is always faster than a third application of either law.
Sorting
The largest angle faces the longest side.
Sort into buckets
Sort each side by whether its opposite angle should be found first.
The rule costs nothing and removes a whole class of silent error. It is worth applying even when the largest angle turns out to be acute after all.
Prediction
Commit before reasoning.
Predict first
Why is it safer to find the angle opposite the longest side before using the law of sines?
Correct: Because the remaining angles are then guaranteed acute, so the inverse sine returns the right one.
\[ \text{at most one angle } \ge 90^\circ, \text{ opposite the longest side} \]
Why: A triangle can hold at most one angle of 90 degrees or more, and that angle faces the longest side. Once it has been found by the law of cosines, whose negative cosines identify obtuse angles reliably, every remaining angle is acute — which is exactly the range the inverse sine returns. Reversing the order leaves you unable to tell whether an inverse sine's answer or its obtuse partner is the true one.
Section
Section 3
Concept
Ask whether you have a known angle together with the side opposite it. If yes, the law of sines has a complete ratio and can start. If no, only the law of cosines can begin.
\[ \text{AAS, ASA, SSA} \to \text{sines}; \quad \text{SAS, SSS} \to \text{cosines} \]
SAS and SSS each determine exactly one triangle, so neither is ever ambiguous. Only SSA can misbehave.
Figure (svg): A table matching each case of given information to the law that solves it
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 892-892 — Choosing a method
Picture it
Which law starts which case.
Figure (svg): A table matching each case of given information to the law that solves it
Three cases go to the law of sines and two to the law of cosines. Between them every determinable triangle is covered.
Worked example
Exercises 3 to 7 on page 892.
\[ \text{For SSS, ASA, SSA, SAS and AAS, say which law starts the solution.} \]
SSS
Why: No angle known at all.
ASA and AAS
Why: An angle is known and, after subtracting, so is every angle.
SSA
Why: The given angle is opposite one of the given sides.
SAS
Why: The given angle sits BETWEEN the two sides.
Figure (svg): A table matching each case of given information to the law that solves it
\[ \text{SSS, SAS} \to \cos; \qquad \text{ASA, AAS, SSA} \to \sin \]
Verify: notice what distinguishes SAS from SSA
Why: Both give two sides and one angle, and the only difference is where the angle sits. Included means SAS and the law of cosines with one guaranteed answer; opposite one of them means SSA and the law of sines with a possible ambiguity. One word changes both the method and the number of answers.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 892-892
Matching
Do you have an angle with its opposite side?
Match the pairs
Why: Three of the four give exactly one triangle. Only SSA can be ambiguous, and it is ambiguous precisely because the known angle is not between the known sides, leaving the triangle free to close two ways.
Worked example
Exercises 11 to 14 on page 892.
\[ \text{Which law for } B=25^\circ,a=8,c=6; \; A=103^\circ,b=15,c=24; \; a=18,b=28,c=13; \; a=38,b=31,c=35? \]
First: is B between a and c?
Why: Yes, B is included.
Second: is A between b and c?
Why: Yes, A is included.
Third: what is given?
Why: All three sides.
Fourth: what is given?
Why: All three sides again.
Figure (svg): The solution to Worked example pick the law for four problems shown as a ladder of expressions, one row per algebraic move
\[ \text{SAS, SAS, SSS, SSS} \]
Verify: check no angle sits opposite a known side
Why: In the first, angle B is known and side b is not; in the third and fourth no angle is known at all. In every case the law of sines has no complete ratio to work from, which is precisely the test for needing the law of cosines.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 892-892
Trap
\[ B = 25^\circ, \; a = 8, \; c = 6 \]
Read it as two sides and an angle, so SSA
Why: The position of the angle is not checked.
\[ \frac{\sin B}{b} = \frac{\sin A}{8} \quad \text{(cannot start: b is unknown)} \]
Angle B sits BETWEEN sides a and c, so this is SAS. Side b, the one opposite B, is exactly what is missing.
\[ b^2 = 8^2+6^2-2(8)(6)\cos 25^\circ \]
Check where the angle sits
Why: Between the two sides means SAS and the law of cosines.
\[ b^2 \approx 12.98 \;\Longrightarrow\; b \approx 3.60 \]
Reading the position of the angle is the whole of the decision, and it takes one glance at the letters.
Sorting
Look for an angle with its opposite side known.
Sort into buckets
Sort each set of given information.
The fourth item is the ambiguous SSA case from Lesson 13.5, and the first and last are SAS. All three give two sides and an angle, so only the angle's position separates them.
Comparison
Fill the blanks. Two sides and an angle, twice.
Comparison matrix
| Question | SAS | SSA |
|---|---|---|
| Where the angle sits | between the two sides | opposite one of the two sides |
| Law to use | cosines | sines |
| Number of triangles | always exactly one | zero, one or two |
| Example | B = 25, a = 8, c = 6 | A = 40, a = 13, b = 16 |
SAS pins the triangle down because fixing an angle and both its arms leaves the third side no freedom. SSA leaves one arm swinging, which is where the second triangle comes from.
Prediction
Commit before reasoning.
Predict first
Why do two sides and the angle between them always give exactly one triangle?
Correct: Because fixing the angle and both its arms leaves the third vertex nowhere to move.
\[ \text{SAS: both arms fixed} \;\Longrightarrow\; \text{one triangle} \]
Why: Draw the angle, mark off the two given lengths along its arms, and the two endpoints are completely determined; joining them gives the one and only third side. In SSA the given angle has one arm of known length and one of unknown length, so the far end is free to slide, and the fixed opposite side may reach it in two places. The whole difference is whether both arms are pinned.
Section
Section 4
Concept
The area of a triangle with sides a, b and c is the square root of s times s minus a times s minus b times s minus c, where s is the semiperimeter, half the sum of the three sides.
\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}, \quad s = \tfrac{1}{2}(a+b+c) \]
No angle is needed, which fits the fact that three sides already determine the triangle completely.
Figure (svg): Heron's formula applied to a triangle of land
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 891-891 — Heron's Area Formula
Picture it
Example 4: sides of 170, 240 and 350 yards.
Figure (svg): Heron's formula applied to a triangle of land
The semiperimeter is 380, and the four factors 380, 210, 140 and 30 multiply to 335,160,000, whose root is about 18,300.
Worked example
Example 4.
\[ \text{Find the area of a triangle with sides } 170, \; 240 \text{ and } 350 \text{ yards.} \]
Find the semiperimeter
Why: Half of 170 plus 240 plus 350.
\[ s = 380 \]
Form the three differences
Why: Three eighty minus each side.
\[ 210, 140, 30 \]
Multiply all four
Why: Three eighty times 210 times 140 times 30.
\[ 335, 160, 000 \]
Take the square root
Why: The root of that product.
\[ \text{about } 18, 300 \]
Figure (svg): Heron's formula applied to a triangle of land
\[ \text{Area} \approx 18{,}300 \text{ yd}^2 \]
Verify: sanity-check against a rectangle
Why: A rectangle 170 by 240 would hold 40,800 square yards, and this triangle holds about 45 percent of that — plausible for a triangle whose longest side of 350 is close to the 410 that the other two sum to, making it rather flat. A very flat triangle encloses little area for its perimeter.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 891-891
Fill the middle
Example 4.
Fill in the blanks
s = \tfrac380___(170+240+350) = \tfrac______(760) = ___
Why: Three hundred eighty. Every one of the three differences is then computed from this same number.
Worked example
Guided Practice 4, 5 and 6.
\[ \text{Find the area for sides } 8,11,5; \; 9,4,7; \; 23,15,12. \]
First: s is 12
Why: Differences 4, 1, 7.
First: evaluate
Why: The root of 336.
\[ \text{about } 18.3 \]
Second: s is 10
Why: Differences 1, 6, 3.
Third: s is 25
Why: Differences 2, 10, 13.
Figure (svg): The solution to Worked example three more areas shown as a ladder of expressions, one row per algebraic move
\[ \sqrt{336} \approx 18.3; \; \sqrt{180} \approx 13.4; \; \sqrt{6500} \approx 80.6 \]
Verify: check the first triangle's shape
Why: Sides 8, 11 and 5 have the two shorter summing to 13, only slightly more than 11, so the triangle is quite flat and its area of 18.3 is small for its perimeter of 24. A factor s minus a close to zero always signals a nearly degenerate triangle.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 891-891
Trap
\[ a=170, \; b=240, \; c=350 \]
Set s to the perimeter
Why: The word semiperimeter is skipped over.
\[ s = 760 \;\Longrightarrow\; \sqrt{760(590)(520)(410)} \approx 320{,}000 \]
That is more than seventeen times too large. The letter s means HALF the perimeter.
\[ s = \tfrac{1}{2}(760) = 380 \]
Halve the perimeter
Why: That is what semi means.
\[ \sqrt{380(210)(140)(30)} \approx 18{,}300 \]
A quick check: the area cannot exceed half the product of the two shorter sides, which here is 20,400.
Fill the middle
Guided Practice 6.
Fill in the blanks
s = 25: \; \sqrt80.6 = \sqrt___ \approx ___
Why: About 80.6 square units. The factor s minus a of 2 is small, so this triangle is somewhat flat and its area modest.
Ranking
Smallest first.
Put in order
Why: The areas are about 13.4, 18.3, 80.6, 103.9 and 18,300. The third and fourth have similar perimeters but different areas, because the fourth is closer to equilateral and so encloses more.
Prediction
Commit before reasoning.
Predict first
In Heron's formula, what does it mean if s minus a is very close to zero?
Correct: That side a is nearly as long as the other two combined, so the triangle is almost flat and its area almost zero.
\[ a = b+c \;\Longrightarrow\; s-a = 0 \;\Longrightarrow\; \text{Area} = 0 \]
Why: The semiperimeter s equals a exactly when a equals b plus c, which is the degenerate case where the triangle collapses onto a straight line. Heron's formula then returns zero, correctly. This makes the formula a diagnostic as well as a calculator: a factor near zero warns you that the triangle is nearly degenerate, and a negative factor would mean the three lengths cannot form a triangle at all.
Section
Section 5
Concept
Applied problems rarely announce their case. Identify which sides and angles are given, decide whether any angle sits opposite a known side, and choose the law accordingly.
\[ \cos B = \frac{a^2+c^2-b^2}{2ac} \]
Three measured lengths is the commonest applied situation, so SSS and Heron's formula appear far more often than the textbook exercise counts suggest.
Figure (svg): A triangle solved from three sides, largest angle first
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 890-891 — Use the law of cosines in real life
Picture it
Example 3: a dinosaur's step angle from three footprint distances.
Figure (svg): A triangle solved from three sides, largest angle first
Three distances and no angle at all is exactly SSS, so the law of cosines rearranged for a cosine is the tool.
Worked example
Example 3, and Guided Practice 3.
\[ \text{Footprints give } a = 155, b = 316, c = 197 \text{ cm. Find the step angle } B. \text{ Repeat for } 193, 335, 186. \]
Identify the case
Why: Three lengths, no angle.
Solve for cosine B
Why: Three sixteen squared minus the other two squares, over -2ac.
\[ \cos B = -0.6062 \]
Take the inverse cosine
Why: A negative cosine gives an obtuse angle.
\[ B = 127.3 ^\circ \]
Repeat with the second set
Why: Cosine B is -0.5624.
\[ B = 124.2 ^\circ \]
Figure (svg): The solution to Worked example a dinosaur's step angle shown as a ladder of expressions, one row per algebraic move
\[ B \approx 127.3^\circ; \qquad B \approx 124.2^\circ \]
Verify: interpret the numbers
Why: The closer the step angle is to 180, the more efficiently the animal walked, so 127.3 degrees is a moderately efficient gait. The second set gives 124.2, slightly less efficient. Both being obtuse is what the negative cosines predicted before either inverse was taken.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 890-891
Fill the middle
Example 3.
Fill in the blanks
\cos B = -0.6062 \;\Longrightarrow\; B \approx 127.3^\circ
Why: About 127.3 degrees. The negative cosine guaranteed an obtuse answer before the inverse was taken.
Worked example
Exercise 44 on page 893.
\[ \text{Three cities are } 18, \; 23 \text{ and } 8 \text{ miles apart. Find the area they enclose.} \]
Identify the case
Why: Three lengths, area wanted.
Find the semiperimeter
Why: Half of 18 plus 23 plus 8.
\[ s = 24.5 \]
Form the differences
Why: Twenty-four point five minus each.
\[ 6.5, 1.5, 16.5 \]
Multiply and take the root
Why: Twenty-four point five times 6.5 times 1.5 times 16.5.
Figure (svg): Heron's formula applied to a triangle of land
\[ \sqrt{24.5(6.5)(1.5)(16.5)} \approx 62.8 \]
Verify: notice the small factor
Why: The factor 1.5 comes from 24.5 minus 23, and it is small because 18 plus 8 is 26, only three more than 23. The three cities are close to being in a straight line, so they enclose much less area than their 49 mile perimeter might suggest.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 893-893
Error analysis
A student solving for cos B from a = 12, b = 27, c = 20 writes:
Annotate
On: \( \cos B = \frac{b^2-a^2-c^2}{2ac} = \frac{185}{480} = 0.3854 \)
The safer rearrangement puts the two squares of the adjacent sides first, subtracts the square of the opposite side, and divides by a positive 2ac. Both routes give the same answer, and only one invites a sign slip.
Matching
What is given decides the tool.
Match the pairs
Why: The third is the only one where an angle sits opposite a known side, so it is the only one the law of sines can start. The other three all begin with the law of cosines or its area consequence.
Comparison
Fill the blanks. Three sides given either way.
Comparison matrix
| Question | An angle wanted | The area wanted |
|---|---|---|
| Tool | law of cosines | Heron's formula |
| First computation | a cosine | the semiperimeter |
| Final step | an inverse cosine | a square root |
| Needs any angle? | no | no |
Neither needs an angle to start, which is the practical meaning of SSS determining the triangle completely: everything about it is already implied by the three lengths.
Prediction
Commit before reasoning.
Predict first
Three side lengths determine every angle and the area with no further information. Why?
Correct: Because a triangle with three fixed side lengths is rigid; there is no way to flex it.
\[ \text{SSS} \;\Longrightarrow\; \text{one triangle, all angles fixed} \]
Why: A four-sided frame with fixed side lengths can be pushed out of shape, but a triangle cannot — the three lengths leave no freedom at all. That rigidity is why bridges and roof trusses are built from triangles, and mathematically it is why SSS determines a unique triangle and therefore every angle and the area. The law of cosines and Heron's formula are simply the arithmetic that extracts those determined values.
Comparison
Fill the blanks. Together they cover everything.
Comparison matrix
| Question | Law of sines | Law of cosines |
|---|---|---|
| Statement | a/sin A = b/sin B = c/sin C | a^2 = b^2 + c^2 - 2bc cos A |
| Cases | AAS, ASA, SSA | SAS, SSS |
| Needs | an angle with its opposite side | no such pair |
| Ambiguity | possible in SSA | never |
The law of cosines is the more powerful of the two but the messier to use, so the usual strategy is to use it exactly once to break into the triangle and then finish with the law of sines.
Pattern
Classify, then choose.
A negative cosine means an obtuse angle. A sine gives no such information, which is why the largest angle is found by cosines.
OpenStax Algebra and Trigonometry 2e, §10.2 Non-right Triangles: Law of Cosines §10.2
Check
The angle is between the two sides.
Check your understanding
In triangle ABC, a = 11, c = 14 and B = 34 degrees. What is b?
Answer: A
Why: b^2 = 121 + 196 - 2(11)(14) cos 34, which is about 61.7.
Check
Find the largest angle first.
Check your understanding
In triangle ABC, a = 17, b = 29 and c = 14. What is angle B?
Answer: A
Why: cos B = (289 + 196 - 841)/(2 times 17 times 14) = -0.7479, so B is obtuse.
Check
Halve the perimeter first.
Check your understanding
What is the area of a triangle with sides 21, 16 and 13?
Answer: A
Why: s = 25, and the area is the root of 25(4)(9)(12) = 10,800.
Real world
A surveyor needs the distance across a lake. Standing at a point on shore, she measures 480 metres to one landing and 350 metres to the other, with an angle of 118 degrees between the two lines of sight.
Discussion prompt
How far apart are the landings, and how much area does the triangle enclose?
Hint: Two sides and the angle between them.
Answer:
\[ d^2 = 480^2+350^2-2(480)(350)\cos 118^\circ \]
\[ = 230400+122500-336000(-0.4695) = 352900+157{,}752 \]
\[ d = \sqrt{510{,}652} \approx 715 \text{ m} \]
\[ \text{Area} = \tfrac{1}{2}(480)(350)\sin 118^\circ \approx 74{,}150 \text{ m}^2 \]
The landings are about 715 metres apart, enclosing about 74,150 square metres.
Watch what the obtuse angle did. Its cosine is negative, so the correction term was ADDED rather than subtracted, making the third side longer than either given one — which is right, since it faces the largest angle. Had the angle been 62 degrees instead, the same two sides would have given only about 440 metres. That is the law of cosines earning its keep: one measurement of an angle, and a distance no one can walk becomes a calculation.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the Pythagorean theorem a separate fact from the law of cosines, or a special case of it?
Correct: A special case: at 90 degrees the cosine is zero and the correction term vanishes.
\[ \cos 90^\circ = 0 \;\Longrightarrow\; a^2 = b^2+c^2 \]
Why: Substituting A equal to 90 degrees into a squared equals b squared plus c squared minus 2bc cosine A kills the last term outright, leaving a squared equals b squared plus c squared. So the law of cosines contains the Pythagorean theorem, and the correction term measures exactly how far from a right angle the triangle is: negative cosines make the third side longer than Pythagoras would predict, positive ones shorter.
Explain it
They know the Pythagorean theorem and think it only works for right triangles.
Discussion prompt
In four sentences or fewer, explain what the law of cosines does.
Hint: Start from Pythagoras and add something.
Answer:
For a right triangle, a squared plus b squared equals c squared. For any other triangle that is not quite true, and the law of cosines says exactly how far off it is.
You subtract twice the product of the two sides times the cosine of the angle between them. At 90 degrees that cosine is zero, so the correction disappears and you are back to Pythagoras.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For order of operations, compute the whole 2bc cosine A term as one number before subtracting. For the largest angle, find the one opposite the longest side so the rest are guaranteed acute. For choosing a law, ask whether any known angle sits opposite a known side. For Heron's formula, s means half the perimeter, so compute it and write it down before anything else.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a law of cosines page. Top left: write all three forms of the law, label which side is squared and which angle it matches, and show in one line how setting the angle to 90 degrees returns the Pythagorean theorem. Top right: solve one SAS triangle completely, marking clearly which step used cosines and which used sines. Middle: solve one SSS triangle, starting with the angle opposite the longest side, and write one sentence beside it on why that order matters. Bottom left: draw the decision table for all five cases, and beside it write the single question that decides between the two laws. Bottom right: compute one area by Heron's formula, showing the semiperimeter and all four factors before the root, and note what a small factor would have meant.
If any SSS solution on your page found a small angle first and then used the law of sines, redo it: that order can return the wrong angle with no warning.
Recap
Five things, and every solvable triangle now covered.
| If you see | Then |
|---|---|
| An angle between two known sides | SAS: law of cosines for the third side |
| Three sides | SSS: law of cosines for the largest angle first |
| An angle opposite a known side | Law of sines |
| A negative cosine | The angle is obtuse |
| Three sides and an area wanted | Heron's formula, with s half the perimeter |
| A right angle | The correction term vanishes and Pythagoras returns |
That closes Chapter 13. Chapter 14 takes the six functions built here and graphs them, turning the unit circle into the waves that model everything periodic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.6 Apply the Law of Cosines §13.6, pp. 889-893 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.