Law of Sines and Cosines Notes and Worksheets - Lindsay Bowden - Free Printable

Let’s solve each problem step by step using the Law of Cosines.

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Section A — Finding side ‘a’

The Law of Cosines:
> a² = b² + c² - 2bc·cos(A)
where A is the angle between sides b and c, and a is the side opposite angle A.

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1) Triangle ABC:
- Angle A = 35°
- Side AB = 7 cm (this is side c, opposite C)
- Side AC = 9 cm (this is side b, opposite B)
- We want side BC = a (opposite angle A)

So:
a² = 7² + 9² - 2×7×9×cos(35°)
a² = 49 + 81 - 126×cos(35°)
cos(35°) ≈ 0.8192
a² = 130 - 126×0.8192 ≈ 130 - 103.2192 ≈ 26.7808
a ≈ √26.7808 ≈ 5.175 → 5.18 cm (to 3 sig figs)

✔ Check: 5.18² = 26.8324 — close enough to 26.7808 (rounding error acceptable).

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2) Triangle ABC:
- Angle A = 77°
- Side AB = 5 cm (side c)
- Side AC = 12 cm (side b)
- Want side BC = a

a² = 5² + 12² - 2×5×12×cos(77°)
a² = 25 + 144 - 120×cos(77°)
cos(77°) ≈ 0.2250
a² = 169 - 120×0.2250 = 169 - 27 = 142
a ≈ √142 ≈ 11.916 → 11.9 cm

✔ Check: 11.9² = 141.61 — very close to 142.

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3) Triangle ABC:
- Angle A = 61°
- Side AB = 4 cm (side c)
- Side AC = 6 cm (side b)
- Want side BC = a

a² = 4² + 6² - 2×4×6×cos(61°)
a² = 16 + 36 - 48×cos(61°)
cos(61°) ≈ 0.4848
a² = 52 - 48×0.4848 ≈ 52 - 23.2704 ≈ 28.7296
a ≈ √28.7296 ≈ 5.360 → 5.36 cm

✔ Check: 5.36² = 28.7296 — perfect match.

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Section B — Finding angle ‘x’

Now we rearrange the Law of Cosines to solve for an angle:

> cos(A) = (b² + c² - a²) / (2bc)
Then take arccos to get angle A.

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1) Triangle XYZ:
We want angle at X (labeled x).
Sides:
- Opposite X is YZ = 14 cm → this is side a
- Adjacent sides: XY = 8 cm, XZ = 10 cm → these are b and c

So:
cos(x) = (8² + 10² - 14²) / (2×8×10)
cos(x) = (64 + 100 - 196) / 160 = (-32)/160 = -0.2
x = arccos(-0.2) ≈ 101.537° → 102° (to 3 sig figs)

✔ Check: cos(102°) ≈ -0.2079 — close to -0.2 (rounding ok).

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2) Triangle XYZ:
Want angle at X (x).
Sides:
- Opposite X is YZ = 15 cm → side a
- Adjacent: XY = 29 cm, XZ = 21 cm → b and c

cos(x) = (29² + 21² - 15²) / (2×29×21)
= (841 + 441 - 225) / 1218 = 1057 / 1218 ≈ 0.8678
x = arccos(0.8678) ≈ 29.82° → 29.8°

✔ Check: cos(29.8°) ≈ 0.867 — good.

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3) Triangle XYZ:
Want angle at X (x).
Sides:
- Opposite X is YZ = 11.4 cm → side a
- Adjacent: XZ = 9.1 cm, XY = 14.08 cm → b and c

cos(x) = (9.1² + 14.08² - 11.4²) / (2×9.1×14.08)
Calculate:
9.1² = 82.81
14.08² = 198.2464
11.4² = 129.96
Numerator = 82.81 + 198.2464 - 129.96 = 151.0964
Denominator = 2×9.1×14.08 = 256.256
cos(x) = 151.0964 / 256.256 ≈ 0.5896
x = arccos(0.5896) ≈ 53.87° → 53.9°

✔ Check: cos(53.9°) ≈ 0.589 — correct.

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Section C — Mixed: find missing side or angle

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1) Triangle PQR:
Want side QR = x (opposite angle P)
Angle P = 28°
Side PQ = 16 cm (side r, opposite R)
Side PR = 11 cm (side q, opposite Q)

Use Law of Cosines:
x² = 16² + 11² - 2×16×11×cos(28°)
x² = 256 + 121 - 352×cos(28°)
cos(28°) ≈ 0.8829
x² = 377 - 352×0.8829 ≈ 377 - 310.7808 ≈ 66.2192
x ≈ √66.2192 ≈ 8.137 → 8.14 cm

✔ Check: 8.14² = 66.2596 — very close.

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2) Triangle DEF:
Want angle at D (x).
Sides:
- Opposite D is EF = 17 cm → side a
- Adjacent: DE = 15 cm, DF = 25 cm → b and c

cos(x) = (15² + 25² - 17²) / (2×15×25)
= (225 + 625 - 289) / 750 = 561 / 750 = 0.748
x = arccos(0.748) ≈ 41.58° → 41.6°

✔ Check: cos(41.6°) ≈ 0.748 — perfect.

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3) Triangle JKL:
Want side JL = x (opposite angle K)
Angle K = 54°
Side JK = 7.2 cm (side l, opposite L)
Side KL = 4.9 cm (side j, opposite J)

x² = 7.2² + 4.9² - 2×7.2×4.9×cos(54°)
x² = 51.84 + 24.01 - 70.56×cos(54°)
cos(54°) ≈ 0.5878
x² = 75.85 - 70.56×0.5878 ≈ 75.85 - 41.475 ≈ 34.375
x ≈ √34.375 ≈ 5.863 → 5.86 cm

✔ Check: 5.86² = 34.3396 — very close.

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Final Answers:

Section A:
1) 5.18 cm
2) 11.9 cm
3) 5.36 cm

Section B:
1) 102°
2) 29.8°
3) 53.9°

Section C:
1) 8.14 cm
2) 41.6°
3) 5.86 cm

Final Answer:
Section A: 1) 5.18, 2) 11.9, 3) 5.36
Section B: 1) 102, 2) 29.8, 3) 53.9
Section C: 1) 8.14, 2) 41.6, 3) 5.86