Law of cosines — proof and applications: Worked examples

This page is a drill: we hunt down every kind of triangle the Law of Cosines can meet — every sign of $cos$ , every degenerate collapse, the limiting cases, a real word problem, and an exam-style twist. If you have not yet seen the formula built from scratch, read the parent first: Law of Cosines — proof.

Recall The one formula we keep reusing

$c^{2} = a^{2} + b^{2} - 2 ab cos C ⟺ cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ Left form: given two sides and the angle between them, find the far side. Right form: given all three sides, find any angle. The angle $C$ always sits opposite the lone side $c$ .

The scenario matrix

Every triangle problem this topic throws lands in exactly one of these cells. We will hit each one.

#	Cell class	What is special	Sign of $cos$	Example
1	SAS, acute angle	correction subtracts	$cos C > 0$	Ex 1
2	SAS, right angle	correction $= 0$ (Pythagoras)	$cos 90° = 0$	Ex 2
3	SAS, obtuse angle	correction adds	$cos C < 0$	Ex 3
4	SSS, find an angle	use rearranged form	any	Ex 4
5	Degenerate $C \to 0°$	triangle flattens, $c=	a-b	$
6	Degenerate $C \to 180°$	triangle flattens, $c = a + b$	$cos 180° = - 1$	Ex 5
7	Impossible SSS (no triangle)	$cos$ falls outside $[- 1, 1]$	—	Ex 6
8	Word problem (real distances)	model then apply	any	Ex 7
9	Exam twist: SSS then Law of Sines	find largest angle safely	any	Ex 8

Figure — Law of cosines — proof and applications

Ex 1 — Cell 1: SAS with an acute angle

Worked example Two sides

$6$ and $9$ , angle $40°$ between them Find the far side $c$ when $a = 6$ , $b = 9$ , $C = 40°$ .

Forecast: The angle is less than $90°$ , so the correction subtracts. Guess: $c$ is less than the Pythagorean $6^{2} + 9^{2} = 117 \approx 10.8$ . Pick a number before reading on.

Step 1 — Confirm $C$ is between the two given sides. Why this step? The left form only works if the known angle is the included one. Here $40°$ is between the $6$ and the $9$ . ✓

Step 2 — Substitute. $c^{2} = 6^{2} + 9^{2} - 2 (6) (9) cos 40°$ Why this step? Drop the SAS inputs straight into the definition.

Step 3 — Evaluate. With $cos 40° \approx 0.7660$ : $c^{2} = 36 + 81 - 108 (0.7660) = 117 - 82.73 = 34.27$ $c = 34.27 \approx 5.85$ Why this step? The correction $- 82.73$ is a big negative, so $c$ collapses well below $10.8$ .

Verify: $5.85 < 10.8$ — matches the forecast (acute ⇒ shrink). Also $c \approx 5.85$ is close to but below $b = 9$ ; a narrow hinge gives a short gap. ✓

Ex 2 — Cell 2: SAS at exactly $90°$ (correction dies)

Ex 3 — Cell 3: SAS with an obtuse angle (correction adds)

Ex 4 — Cell 4: SSS, find an angle

Worked example All three sides known:

$a = 5$ , $b = 7$ , $c = 10$ Find angle $C$ (opposite the $10$ ).

Forecast: The side $10$ is the longest, so $C$ is the largest angle. A triangle with one side dominating usually has an obtuse biggest angle. Predict $C > 90°$ .

Step 1 — Choose the rearranged form. $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ Why this step? We have three sides, no angle — the left form can't start. Solving the definition for $cos C$ turns three lengths into one cosine.

Step 2 — Put the lone side $c$ with the minus. $cos C = \frac{5 ^{2} + 7 ^{2} - 1 0 ^{2}}{2 ( 5 ) ( 7 )} = \frac{25 + 49 - 100}{70} = \frac{- 26}{70} = - 0.3714$ Why this step? $C$ is opposite $c = 10$ , so $c^{2}$ is the term that gets subtracted; $a, b$ are the neighbours in the denominator.

Step 3 — Undo cosine with $arccos$ . $C = arccos (- 0.3714) \approx 111.8°$ Why this step? $arccos$ answers "which angle has this cosine?" A negative cosine's angle lands between $90°$ and $180°$ — an obtuse angle, exactly as forecast.

Verify: Plug back: $cos 111.8° \approx - 0.371$ , and $- 0.3714$ came out. Consistent. And $C > 90°$ matches the "longest side ⇒ obtuse" prediction. ✓

Ex 5 — Cells 5 & 6: the two degenerate collapses

Worked example Squash the triangle flat both ways

Keep $a = 6$ , $b = 9$ and push $C$ to the two extremes $0°$ and $180°$ .

Forecast: At $C = 0°$ the two sticks fold onto each other pointing the same way — the gap should be the difference $∣9 - 6∣ = 3$ . At $C = 180°$ they point opposite — the gap should be the sum $9 + 6 = 15$ . Predict $c = 3$ and $c = 15$ .

Step 1 — Case $C \to 0°$ , with $cos 0° = 1$ . $c^{2} = 36 + 81 - 2 (6) (9) (1) = 117 - 108 = 9 \Rightarrow c = 3$ Why this step? At the maximum $cos = 1$ the correction is largest negative, shrinking $c$ to its smallest possible value $∣ a - b ∣$ .

Step 2 — Case $C \to 180°$ , with $cos 180° = - 1$ . $c^{2} = 36 + 81 - 2 (6) (9) (- 1) = 117 + 108 = 225 \Rightarrow c = 15$ Why this step? At the most negative $cos = - 1$ the correction is largest positive, stretching $c$ to its biggest possible value $a + b$ .

Verify: $c = 3 = ∣9 - 6∣$ and $c = 15 = 9 + 6$ — exactly the triangle inequality boundaries: any real triangle with these two sides has $3 < c < 15$ . The Law of Cosines reproduces those limits. ✓

Ex 6 — Cell 7: impossible SSS (no such triangle)

Worked example Do these three lengths form a triangle?

$a = 2$ , $b = 3$ , $c = 9$ . Try to find angle $C$ .

Forecast: $2 + 3 = 5$ , which is less than $9$ — the two short sides can't even reach across. Predict: no triangle exists, and the formula should signal this.

Step 1 — Compute $cos C$ anyway. $cos C = \frac{2 ^{2} + 3 ^{2} - 9 ^{2}}{2 ( 2 ) ( 3 )} = \frac{4 + 9 - 81}{12} = \frac{- 68}{12} = - 5.667$ Why this step? Let the algebra run and watch what value pops out.

Step 2 — Test against the legal range of cosine. Why this step? Cosine of a real angle is always between $- 1$ and $1$ (it's an $x$ -coordinate on the unit circle). Here $- 5.667 < - 1$ , which is impossible.

Verify: $arccos (- 5.667)$ is undefined — the calculator errors out. The Law of Cosines has correctly reported "these sides cannot close into a triangle," matching the triangle-inequality failure $2 + 3 < 9$ . ✓ This is your built-in impossibility detector: if $∣ cos C ∣ > 1$ , no triangle.

Ex 7 — Cell 8: a real-world word problem

Worked example Two hikers leave the same camp

Hiker P walks $8 km$ on a bearing, Hiker Q walks $5 km$ on a different bearing. The angle between their two paths is $110°$ . How far apart are they?

Forecast: The paths splay wide ( $110°$ is obtuse), so they end up far apart — expect more than the $8^{2} + 5^{2} \approx 9.4$ km a right angle would give.

Step 1 — Draw the model: it's a triangle. Why this step? Camp is the shared vertex; the two path lengths $8$ and $5$ are two sides; the angle at the camp between them is $110°$ . The gap between the hikers is the far side $c$ — a clean SAS problem.

Step 2 — Apply the left form with $cos 110° \approx - 0.3420$ . $c^{2} = 8^{2} + 5^{2} - 2 (8) (5) cos 110° = 64 + 25 - 80 (- 0.3420)$ Why this step? Obtuse ⇒ negative cosine ⇒ the correction adds separation.

Step 3 — Finish. $c^{2} = 89 + 27.36 = 116.36 \Rightarrow c = 116.36 \approx 10.79 km$ Why this step? Root, then attach the unit: kilometres, because every length input was in km.

Verify: $10.79 > 9.4$ km — the obtuse splay pushed them apart more than a right angle would, as forecast. Units consistent (km in ⇒ km out). ✓

Ex 8 — Cell 9: exam twist (SSS, then finish with Law of Sines the safe way)

Worked example Find

all three angles of $a = 7$ , $b = 8$ , $c = 13$ A classic exam trap.

Forecast: $13$ is much the longest side, so its opposite angle $C$ is the big one — probably obtuse. The other two are acute.

Step 1 — Find the largest angle first, with the Law of Cosines. $cos C = \frac{7 ^{2} + 8 ^{2} - 1 3 ^{2}}{2 ( 7 ) ( 8 )} = \frac{49 + 64 - 169}{112} = \frac{- 56}{112} = - 0.5$ $C = arccos (- 0.5) = 120°$ Why this step? Always pin down the biggest (possibly obtuse) angle with Cosines, because $arccos$ handles negative values and returns the correct obtuse angle directly — no ambiguity.

Step 2 — Now switch to the Law of Sines for a remaining angle. $\frac{s i n A}{a} = \frac{s i n C}{c} \Rightarrow sin A = \frac{7 s i n 120°}{13} = \frac{7 ( 0.8660 )}{13} = 0.4663$ $A = arcsin (0.4663) \approx 27.8°$ Why this step? With one angle known, the Law of Sines is quicker. And $A$ is opposite the shortest side, so it is guaranteed acute — the notorious ambiguous SSA trap can't bite here.

Step 3 — Get $B$ by the angle sum. $B = 180° - 120° - 27.8° = 32.2°$ Why this step? Angles of a triangle sum to $180°$ — the cheapest way to close the last angle.

Verify: Check $B$ independently via Cosines: $cos B = \frac{7 ^{2} + 1 3 ^{2} - 8 ^{2}}{2 ( 7 ) ( 13 )} = \frac{49 + 169 - 64}{182} = \frac{154}{182} = 0.8462 \Rightarrow B \approx 32.2°.$ Both methods agree, and $120° + 27.8° + 32.2° = 180°$ . ✓

The trap avoided: If you'd used the Law of Sines to find the obtuse angle $C$ , $arcsin$ would have returned $60°$ (its acute twin) and you'd have missed $120°$ entirely. Cosines-first for the big angle is the safe habit.

Recall check

Recall What signals "impossible triangle" in an SSS calculation?

When $cos C$ comes out with $∣ cos C ∣ > 1$ — cosine can never exceed $1$ in size, so no real angle (and no triangle) exists. Equivalent to the triangle inequality failing. ::: correct Why compute the largest angle first in a full SSS solve? ::: Only that angle may be obtuse; $arccos$ returns it correctly, whereas $arcsin$ would give its acute twin and lose the obtuse solution. At $C = 0°$ and $C = 180°$ , what does $c$ become? ::: $∣ a - b ∣$ and $a + b$ — the triangle-inequality boundaries.

Connections

Parent: proof & applications
Pythagoras Theorem — the $C = 90°$ cell (Ex 2).
Law of Sines — finishes the SSS solve safely (Ex 8).
Solving Triangles — decision tree for which law, which order.
Unit Circle and Cosine Values — why $cos$ flips sign past $90°$ (governs the whole matrix).
Dot Product — same formula in vector clothing.