Law of cosines — proof and applications: Exercises

Level 1 — Recognition

Here you only need to recognise which pieces go where. No cleverness yet.

Problem 1.1

In a triangle, sides $a = 5$ and $b = 7$ meet at an included angle $C = 90°$ . What is $c$ ?

Recall Solution 1.1

WHAT we know: two sides and the angle between them → this is SAS, and the angle is exactly a right angle. WHY Law of Cosines: it starts from SAS. But since $C = 90°$ and $cos 90° = 0$ , the correction term $- 2 ab cos C$ vanishes. $c^{2} = 5^{2} + 7^{2} - 2 (5) (7) cos 90° = 25 + 49 - 0 = 74$ $c = 74 \approx 8.60$ This is just Pythagoras Theorem — the special case where the correction dies. See Unit Circle and Cosine Values for why $cos 90° = 0$ .

Problem 1.2

A triangle has sides $a = 6$ , $b = 6$ , and included angle $C = 60°$ . Which formula do you write down (not solve) to find $c$ ? Then find $c$ .

Recall Solution 1.2

WHAT / WHY: SAS again → write the side form with the known angle $C$ in the cosine: $c^{2} = a^{2} + b^{2} - 2 ab cos C = 6^{2} + 6^{2} - 2 (6) (6) cos 60°$ $cos 60° = 0.5$ , so: $c^{2} = 36 + 36 - 72 (0.5) = 72 - 36 = 36 \Rightarrow c = 6$ Because all three sides are $6$ , this triangle is equilateral — a good sanity check.

Problem 1.3

You are given all three sides $a = 4$ , $b = 5$ , $c = 6$ and asked for the angle opposite the side of length 6. Which rearranged formula do you use, and which side is "the loner"?

Recall Solution 1.3

WHAT: three sides known → SSS. The angle opposite $c = 6$ is $C$ . WHY the rearranged form: $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ puts the loner side $c$ alone with the minus sign; $a, b$ frame the angle. $cos C = \frac{4 ^{2} + 5 ^{2} - 6 ^{2}}{2 ( 4 ) ( 5 )} = \frac{16 + 25 - 36}{40} = \frac{5}{40} = 0.125$ $C = arccos (0.125) \approx 82.82°$

Level 2 — Application

Now plug in and grind through the arithmetic carefully — signs matter.

Problem 2.1 (SAS)

$a = 8$ , $b = 5$ , $C = 60°$ . Find $c$ .

Recall Solution 2.1

$c^{2} = 8^{2} + 5^{2} - 2 (8) (5) cos 60° = 64 + 25 - 80 (0.5)$ $c^{2} = 89 - 40 = 49 \Rightarrow c = 7$ Clean whole number — the angle shrank $c$ below the $89 \approx 9.43$ you'd get with no correction.

Problem 2.2 (SAS, obtuse)

$a = 3$ , $b = 4$ , $C = 120°$ . Find $c$ .

Recall Solution 2.2

KEY: $120°$ lives in the second quadrant, so $cos 120° = - 0.5$ (negative). See Unit Circle and Cosine Values. $c^{2} = 3^{2} + 4^{2} - 2 (3) (4) (- 0.5) = 9 + 16 - (24) (- 0.5)$ $c^{2} = 25 + 12 = 37 \Rightarrow c = 37 \approx 6.08$ The two minus signs cancel: $- 2 ab cos C = - (24) (- 0.5) = + 12$ , which adds length. Compare to $c = 5$ if the angle were $90°$ .

Problem 2.3 (SSS)

$a = 7$ , $b = 8$ , $c = 9$ . Find angle $C$ .

Recall Solution 2.3

WHAT: three sides, no angles → SSS, and we want the angle $C$ opposite $c = 9$ . WHY the rearranged form: with no angle known, the side form $c^{2} = a^{2} + b^{2} - 2 ab cos C$ has $cos C$ buried inside it — so we solve it for $cos C$ first, which isolates the angle we want. The loner side $c = 9$ carries the minus; its neighbours $a = 7, b = 8$ frame the angle. $cos C = \frac{7 ^{2} + 8 ^{2} - 9 ^{2}}{2 ( 7 ) ( 8 )} = \frac{49 + 64 - 81}{112} = \frac{32}{112} = 0.2857$ WHY $arccos$ next: $arccos$ is the "which angle has this cosine?" undo-button, and on $0°$ – $180°$ (the full range of triangle angles) it returns exactly one answer — no ambiguity. $C = arccos (0.2857) \approx 73.40°$

Level 3 — Analysis

Here you decide which tool, reason about signs, or chase a hidden quantity.

Problem 3.1 (Classify without solving fully)

Sides $a = 5$ , $b = 6$ , $c = 10$ . Without finding every angle, decide whether the angle opposite the longest side is acute, right, or obtuse.

Recall Solution 3.1

WHY this works: the sign of $cos C$ tells you the angle type. $cos C > 0 \Rightarrow$ acute, $= 0 \Rightarrow$ right, $< 0 \Rightarrow$ obtuse. The largest angle sits opposite the largest side $c = 10$ . $cos C = \frac{5 ^{2} + 6 ^{2} - 1 0 ^{2}}{2 ( 5 ) ( 6 )} = \frac{25 + 36 - 100}{60} = \frac{- 39}{60} = - 0.65$ Negative → the angle is obtuse. (Numerically $C = arccos (- 0.65) \approx 130.54°$ .) A slick shortcut: compare $a^{2} + b^{2}$ with $c^{2}$ . Here $25 + 36 = 61 < 100$ , so the correction must be positive → obtuse. This generalises Pythagoras Theorem.

Problem 3.2 (Which law?)

You know angle $A = 40°$ , its opposite side $a = 6$ , and another side $b = 9$ . Can you start with the Law of Cosines? What should you use?

Recall Solution 3.2

Analyse the given data: you have a side, its opposite angle, and one more side — that's SSA, the ambiguous case, not SAS. The Law of Cosines' SAS start needs the angle between the two known sides; here $A$ is not between $a$ and $b$ . Correct tool: the Law of Sines: $\frac{s i n A}{a} = \frac{s i n B}{b} \Rightarrow sin B = \frac{9 s i n 40°}{6} = \frac{9 ( 0.6428 )}{6} = 0.9642$ $B \approx 74.62°$ or its supplement $105.38°$ — two possible triangles (ambiguous). See Solving Triangles. (You could force Law of Cosines by treating the unknown side as a variable and solving a quadratic — that quadratic's two roots are exactly the two ambiguous triangles.)

Problem 3.3 (Find a side length via a full solve)

Given $a = 10$ , $c = 13$ , and the angle $B = 95°$ between them, find $b$ .

Recall Solution 3.3

SAS with the included angle $B$ . Use the $b$ -form: $b^{2} = a^{2} + c^{2} - 2 a c cos B = 100 + 169 - 2 (10) (13) cos 95°$ $cos 95° \approx - 0.08716$ (obtuse, so negative): $b^{2} = 269 - 260 (- 0.08716) = 269 + 22.66 = 291.66$ $b = 291.66 \approx 17.08$

Level 4 — Synthesis

Combine the Law of Cosines with another idea in the same problem.

Problem 4.1 (Cosines then Sines)

In a triangle, $a = 7$ , $b = 10$ , and the included angle $C = 44°$ . Find side $c$ , then find angle $A$ .

Recall Solution 4.1

Stage 1 — Law of Cosines for the missing side $c$ (SAS start): $c^{2} = 7^{2} + 1 0^{2} - 2 (7) (10) cos 44° = 49 + 100 - 140 (0.71934)$ $c^{2} = 149 - 100.71 = 48.29 \Rightarrow c \approx 6.949$ Stage 2 — now we have side $c$ and its opposite angle $C$ , so Law of Sines can finish: $\frac{s i n A}{a} = \frac{s i n C}{c} \Rightarrow sin A = \frac{7 s i n 44°}{6.949} = \frac{7 ( 0.69466 )}{6.949} = 0.6998$ $A = arcsin (0.6998) \approx 44.40°$ Sanity check: $A$ is opposite $a = 7$ , the shorter of the two given sides, so $A$ should be smaller than $B$ — and indeed $B = 180° - 44° - 44.40° = 91.60°$ is the largest angle, opposite the largest side $b = 10$ . ✓

Problem 4.2 (Diagonal of a parallelogram)

A parallelogram has adjacent sides $8$ and $5$ meeting at an angle of $70°$ . Find the length of both diagonals.

Recall Solution 4.2

WHY two answers: a parallelogram's two diagonals cut across angles that are supplementary — one triangle has the $70°$ corner, the other has $180° - 70° = 110°$ . Short diagonal (across the $70°$ corner): $d_{1}^{2} = 8^{2} + 5^{2} - 2 (8) (5) cos 70° = 64 + 25 - 80 (0.34202)$ $d_{1}^{2} = 89 - 27.36 = 61.64 \Rightarrow d_{1} \approx 7.85$ Long diagonal (across the $110°$ corner, $cos 110° = - 0.34202$ ): $d_{2}^{2} = 64 + 25 - 80 (- 0.34202) = 89 + 27.36 = 116.36 \Rightarrow d_{2} \approx 10.79$ Check (parallelogram law): $d_{1}^{2} + d_{2}^{2} = 61.64 + 116.36 = 178 = 2 (8^{2} + 5^{2}) = 2 (89)$ . ✓

Level 5 — Mastery

Full multi-step problems; connect to structure and other theorems.

Problem 5.1 (Solve the triangle completely, SSS)

Sides $a = 6$ , $b = 9$ , $c = 13$ . Find all three angles and confirm they sum to $180°$ .

Recall Solution 5.1

Find the largest angle first — opposite $c = 13$ — using arccos (no ambiguity): $cos C = \frac{6 ^{2} + 9 ^{2} - 1 3 ^{2}}{2 ( 6 ) ( 9 )} = \frac{36 + 81 - 169}{108} = \frac{- 52}{108} = - 0.48148$ $C = arccos (- 0.48148) \approx 118.78°$ Negative cosine → obtuse, as expected for the side longest by a clear margin. Angle $B$ (opposite $9$ ): $cos B = \frac{6 ^{2} + 1 3 ^{2} - 9 ^{2}}{2 ( 6 ) ( 13 )} = \frac{36 + 169 - 81}{156} = \frac{124}{156} = 0.79487$ $B = arccos (0.79487) \approx 37.30°$ Angle $A$ by the sum (cheapest, exact): $A = 180° - 118.78° - 37.30° = 23.92°$ Check: $23.92 + 37.30 + 118.78 = 180.00°$ . ✓ (For a deeper cross-check, verify $\frac{s i n A}{a} = \frac{s i n C}{c}$ via Law of Sines.)

Problem 5.2 (Law of Cosines as a dot product)

Two vectors $u = (3, 4)$ and $v = (5, 0)$ start at the same point. Find the angle between them (a) with the Dot Product formula, and (b) by building the triangle whose third side is $v - u$ and applying the Law of Cosines. Show the two answers match.

Recall Solution 5.2

(a) Dot product. $u \cdot v = ∣ u ∣∣ v ∣ cos θ$ , so: $cos θ = \frac{u \cdot v}{∣ u ∣ ∣ v ∣} = \frac{( 3 ) ( 5 ) + ( 4 ) ( 0 )}{3 ^{2} + 4 ^{2} 5 ^{2} + 0 ^{2}} = \frac{15}{( 5 ) ( 5 )} = 0.6$ $θ = arccos (0.6) \approx 53.13°$ (b) Triangle sides. The two vectors have lengths $a = ∣ u ∣ = 5$ and $b = ∣ v ∣ = 5$ ; the opposite side is $v - u = (5 - 3, 0 - 4) = (2, - 4), c = 2^{2} + (- 4)^{2} = 20 \approx 4.472$ Law of Cosines for the enclosed angle $θ$ : $cos θ = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab} = \frac{25 + 25 - 20}{2 ( 5 ) ( 5 )} = \frac{30}{50} = 0.6$ Same $0.6$ , same $53.13°$ . This is why the parent note says the Dot Product "is the Law of Cosines in disguise": expand $∣ v - u ∣^{2} = ∣ u ∣^{2} + ∣ v ∣^{2} - 2 u \cdot v$ and the middle term $- 2 u \cdot v = - 2∣ u ∣∣ v ∣ cos θ$ is the correction term.

Problem 5.3 (Reverse engineering — find a side from a cosine constraint)

A triangle has sides $a = 10$ , $b = 14$ , and its included angle $C$ satisfies $cos C = 0.25$ . Find $c$ , and find the angle $C$ in degrees.

Recall Solution 5.3

Straight substitution — the cosine is given directly, no need to know $C$ first: $c^{2} = 1 0^{2} + 1 4^{2} - 2 (10) (14) (0.25) = 100 + 196 - 280 (0.25)$ $c^{2} = 296 - 70 = 226 \Rightarrow c = 226 \approx 15.03$ And the angle itself: $C = arccos (0.25) \approx 75.52°$ Acute (positive cosine), and $c$ is the longest side — consistent, since $C$ is the largest angle here.

Connections

Law of cosines — proof and applications — the parent this drills.
Pythagoras Theorem — the $C = 90°$ collapse used in 1.1 and 3.1.
Law of Sines — the finisher in 4.1 and the correct tool in the SSA of 3.2.
Dot Product — unmasked as the Law of Cosines in 5.2.
Solving Triangles — the decision tree behind every "which law?" choice.
Unit Circle and Cosine Values — why $cos$ turns negative past $90°$ (2.2, 3.1, 5.1).

Exercises — Law of cosines — proof and applications