Law of Cosines Worksheet 6 | PDF | Triangle | Angle - Free Printable

Let’s solve each problem step by step.

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Problem 1: Solve for the unknown in each triangle. Round to the nearest hundredth.

We’ll use the Law of Cosines:

> c² = a² + b² - 2ab·cos(C)

Where C is the angle between sides a and b, and c is the side opposite angle C.

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A. Given: two sides 17m, 22m, included angle 42°, find side x (opposite the 42° angle).

x² = 17² + 22² - 2·17·22·cos(42°)
x² = 289 + 484 - 748·cos(42°)
cos(42°) ≈ 0.7431
x² = 773 - 748·0.7431 ≈ 773 - 555.84 ≈ 217.16
x ≈ √217.16 ≈ 14.74 m

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B. Given: all three sides: 39mm, 47mm, 35mm. Find angle θ opposite 35mm.

Use Law of Cosines rearranged:

cos(θ) = (a² + b² - c²) / (2ab)

Here, let’s say:
a = 39, b = 47, c = 35 (side opposite θ)

cos(θ) = (39² + 47² - 35²) / (2·39·47)
= (1521 + 2209 - 1225) / 3666
= (2505) / 3666 ≈ 0.6833
θ = arccos(0.6833) ≈ 46.90°

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C. Given: sides 9.4cm, 7cm, 13cm. Find angle θ opposite 13cm.

cos(θ) = (9.4² + 7² - 13²) / (2·9.4·7)
= (88.36 + 49 - 169) / 131.6
= (-31.64) / 131.6 ≈ -0.2404
θ = arccos(-0.2404) ≈ 103.91°

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D. Given: sides 23m, 20m, included angle 47°, find side x (opposite 47°).

x² = 23² + 20² - 2·23·20·cos(47°)
= 529 + 400 - 920·cos(47°)
cos(47°) ≈ 0.6820
x² = 929 - 920·0.6820 ≈ 929 - 627.44 ≈ 301.56
x ≈ √301.56 ≈ 17.37 m

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E. Given: sides 55cm, 50cm, included angle 61°, find side x (opposite 61°).

x² = 55² + 50² - 2·55·50·cos(61°)
= 3025 + 2500 - 5500·cos(61°)
cos(61°) ≈ 0.4848
x² = 5525 - 5500·0.4848 ≈ 5525 - 2666.4 ≈ 2858.6
x ≈ √2858.6 ≈ 53.47 cm

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F. Given: sides 4.9m, 8.3m, 9.1m. Find angle θ opposite 9.1m.

cos(θ) = (4.9² + 8.3² - 9.1²) / (2·4.9·8.3)
= (24.01 + 68.89 - 82.81) / 81.34
= (10.09) / 81.34 ≈ 0.1240
θ = arccos(0.1240) ≈ 82.88°

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Problem 2: Solve for all missing sides and angles.

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A. ΔXYZ: x = 29m, y = 15m, ∠Z = 122°

Find side z first (opposite ∠Z):

z² = x² + y² - 2xy·cos(Z)
= 29² + 15² - 2·29·15·cos(122°)
= 841 + 225 - 870·cos(122°)
cos(122°) ≈ -0.5299
z² = 1066 - 870·(-0.5299) ≈ 1066 + 461.01 ≈ 1527.01
z ≈ √1527.01 ≈ 39.08 m

Now find other angles using Law of Sines or Law of Cosines.

Use Law of Sines: sin(X)/x = sin(Z)/z

sin(X) = x·sin(Z)/z = 29·sin(122°)/39.08
sin(122°) ≈ 0.8480
sin(X) ≈ 29·0.8480 / 39.08 ≈ 24.592 / 39.08 ≈ 0.6293
X ≈ arcsin(0.6293) ≈ 39.00°

Then Y = 180° - 122° - 39° = 19.00°

Check with Law of Sines: sin(Y)/y = sin(19°)/15 ≈ 0.3256/15 ≈ 0.0217
sin(Z)/z ≈ 0.8480/39.08 ≈ 0.0217 → matches.

So:
z ≈ 39.08 m, ∠X ≈ 39.00°, ∠Y ≈ 19.00°

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B. ΔGHI: g = 13cm, h = 8cm, i = 15cm

All sides known → find all angles.

Use Law of Cosines.

Angle G (opposite g=13):

cos(G) = (h² + i² - g²)/(2hi) = (64 + 225 - 169)/(2·8·15) = (120)/240 = 0.5
G = arccos(0.5) = 60.00°

Angle H (opposite h=8):

cos(H) = (g² + i² - h²)/(2gi) = (169 + 225 - 64)/(2·13·15) = (330)/390 ≈ 0.8462
H = arccos(0.8462) ≈ 32.20°

Angle I = 180 - 60 - 32.20 = 87.80°

Check: cos(I) = (g² + h² - i²)/(2gh) = (169+64-225)/(2·13·8) = (8)/208 ≈ 0.0385 → arccos ≈ 87.80° ✓

So:
∠G = 60.00°, ∠H ≈ 32.20°, ∠I ≈ 87.80°

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C. ΔMNO: n = 31m, o = 28m, ∠M = 62°

Find side m (opposite ∠M):

m² = n² + o² - 2no·cos(M)
= 31² + 28² - 2·31·28·cos(62°)
= 961 + 784 - 1736·cos(62°)
cos(62°) ≈ 0.4695
m² = 1745 - 1736·0.4695 ≈ 1745 - 815.05 ≈ 929.95
m ≈ √929.95 ≈ 30.50 m

Now find angles N and O.

Use Law of Sines:

sin(N)/n = sin(M)/m → sin(N) = n·sin(M)/m
sin(N) = 31·sin(62°)/30.50
sin(62°) ≈ 0.8829
sin(N) ≈ 31·0.8829 / 30.50 ≈ 27.37 / 30.50 ≈ 0.8974
N ≈ arcsin(0.8974) ≈ 63.80°

Then O = 180 - 62 - 63.80 = 54.20°

Check with Law of Sines: sin(O)/o = sin(54.2°)/28 ≈ 0.811 / 28 ≈ 0.02896
sin(M)/m ≈ 0.8829 / 30.50 ≈ 0.02895 → close enough.

So:
m ≈ 30.50 m, ∠N ≈ 63.80°, ∠O ≈ 54.20°

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Problem 3: Triangle with sides 4m, 11m, 8m. Find its angles.

Label: a=4, b=11, c=8

Find angle A (opposite a=4):

cos(A) = (b² + c² - a²)/(2bc) = (121 + 64 - 16)/(2·11·8) = 169/176 ≈ 0.9602
A ≈ arccos(0.9602) ≈ 16.2°

Angle B (opposite b=11):

cos(B) = (a² + c² - b²)/(2ac) = (16 + 64 - 121)/(2·4·8) = (-41)/64 ≈ -0.6406
B ≈ arccos(-0.6406) ≈ 129.8°

Angle C = 180 - 16.2 - 129.8 = 34.0°

Check: cos(C) = (a² + b² - c²)/(2ab) = (16+121-64)/(2·4·11) = 73/88 ≈ 0.8295 → arccos ≈ 34.0° ✓

Angles: 16.2°, 129.8°, 34.0°

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Problem 4: Ship leaves port at 1pm north at 30 mph. At 3pm, turns to bearing N 20° E. How far from port at 4pm?

From 1pm to 3pm: 2 hours → distance = 30 × 2 = 60 miles north

From 3pm to 4pm: 1 hour → distance = 30 × 1 = 30 miles on bearing N 20° E

Bearing N 20° E means 20° east of north → so direction is 20° from vertical toward right.

We can model this as a triangle:

Side A: 60 miles straight north
Side B: 30 miles at 20° from north (so angle between them is 20°)
We want the third side (distance from port).

Actually, the angle between the two paths is 20°, because first leg is due north, second is 20° east of north → so included angle is 20°.

Wait — actually, when the ship turns, it changes direction. The angle between the two legs is 20°.

So we have two sides: 60 and 30, included angle 20°? No!

Actually, the first leg is north. The second leg is N 20° E — which is 20° from north toward east. So the turn is 20° to the right. Therefore, the angle *between* the two paths is 180° - 20° = 160°? Let me think carefully.

Imagine:

- From port, go north 60 miles to point A.
- From A, go 30 miles on bearing N 20° E — that means you’re going 20° east of north, so your direction is 20° from the original north direction.

So the angle between the two vectors (from port to A, and from A to B) is 180° - 20° = 160°? Actually, no.

If you're facing north, then turn 20° to the right (east), your new direction is 20° from north. The angle between your old direction (north) and new direction (N20°E) is 20°. But in the triangle formed by port, A, and B, the angle at A is the supplement of that? Let's draw mentally.

Actually, the triangle has vertices: Port (P), Point A (after 2h), Point B (after 3h).

Vector PA is north. Vector AB is N20°E. So the angle at A between PA and AB is 180° - 20° = 160°, because PA is coming from south to north, and AB is going northeast — so the internal angle at A is 160°.

Yes.

So in triangle PAB:

PA = 60 miles
AB = 30 miles
Angle at A = 160°
Find PB (distance from port to final position).

Use Law of Cosines:

PB² = PA² + AB² - 2·PA·AB·cos(angle at A)
But angle at A is 160°, and cos(160°) = -cos(20°) ≈ -0.9397

PB² = 60² + 30² - 2·60·30·cos(160°)
= 3600 + 900 - 3600·(-0.9397)
= 4500 + 3382.92 ≈ 7882.92
PB ≈ √7882.92 ≈ 88.79 miles

Round to nearest unit: 89 miles

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Problem 5: Area of triangle with sides 12cm, 5cm, 13cm.

Notice: 5-12-13 is a Pythagorean triple! 5² + 12² = 25 + 144 = 169 = 13² → right triangle!

So area = (1/2) × leg1 × leg2 = (1/2) × 5 × 12 = 30 cm²

Alternatively, use Heron’s formula to verify:

s = (5+12+13)/2 = 15
Area = √[s(s-a)(s-b)(s-c)] = √[15(10)(3)(2)] = √[900] = 30 ✓

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Final Answers:

1. A. 14.74 m
B. 46.90°
C. 103.91°
D. 17.37 m
E. 53.47 cm
F. 82.88°

2. A. z ≈ 39.08 m, ∠X ≈ 39.00°, ∠Y ≈ 19.00°
B. ∠G = 60.00°, ∠H ≈ 32.20°, ∠I ≈ 87.80°
C. m ≈ 30.50 m, ∠N ≈ 63.80°, ∠O ≈ 54.20°

3. Angles: 16.2°, 129.8°, 34.0°

4. 89 miles

5. 30 cm²

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Final Answer:
1. A. 14.74 m, B. 46.90°, C. 103.91°, D. 17.37 m, E. 53.47 cm, F. 82.88°
2. A. z ≈ 39.08 m, ∠X ≈ 39.00°, ∠Y ≈ 19.00°; B. ∠G = 60.00°, ∠H ≈ 32.20°, ∠I ≈ 87.80°; C. m ≈ 30.50 m, ∠N ≈ 63.80°, ∠O ≈ 54.20°
3. 16.2°, 129.8°, 34.0°
4. 89 miles
5. 30 cm²