Worksheet with 10 geometry problems requiring the use of the Law of Cosines to find missing side lengths and angle measures in various triangles.

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

The Law of Cosines says:

> For any triangle with sides a, b, c and angle C opposite side c:
>
> c² = a² + b² - 2ab·cos(C)

We’ll use this to find missing sides or angles. Remember to round answers to the nearest tenth.

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Problem 1: Find AB

Triangle ABC:
- AC = 13
- BC = 29
- Angle C = 41°
→ We want side AB (opposite angle C)

Use Law of Cosines:
AB² = AC² + BC² - 2·AC·BC·cos(angle C)
AB² = 13² + 29² - 2·13·29·cos(41°)
AB² = 169 + 841 - 754·cos(41°)
cos(41°) ≈ 0.7547
AB² = 1010 - 754·0.7547 ≈ 1010 - 569.04 ≈ 440.96
AB ≈ √440.96 ≈ 21.0

✔ Final Answer for #1: 21.0

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Problem 2: Find BC

Triangle ABC:
- AB = 30
- AC = 21
- Angle A = 123°
→ We want side BC (opposite angle A)

BC² = AB² + AC² - 2·AB·AC·cos(angle A)
BC² = 30² + 21² - 2·30·21·cos(123°)
BC² = 900 + 441 - 1260·cos(123°)
cos(123°) ≈ -0.5446
BC² = 1341 - 1260·(-0.5446) = 1341 + 686.196 ≈ 2027.196
BC ≈ √2027.196 ≈ 45.0

✔ Final Answer for #2: 45.0

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Problem 3: Find BC

Triangle ABC:
- AC = 17
- AB = 28
- Angle A = 91°
→ We want side BC (opposite angle A)

BC² = 17² + 28² - 2·17·28·cos(91°)
BC² = 289 + 784 - 952·cos(91°)
cos(91°) ≈ -0.0175
BC² = 1073 - 952·(-0.0175) = 1073 + 16.66 ≈ 1089.66
BC ≈ √1089.66 ≈ 33.0

✔ Final Answer for #3: 33.0

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Problem 4: Find BC

Triangle ABC:
- AC = 9
- AB = 14
- Angle A = 17°
→ We want side BC (opposite angle A)

BC² = 9² + 14² - 2·9·14·cos(17°)
BC² = 81 + 196 - 252·cos(17°)
cos(17°) ≈ 0.9563
BC² = 277 - 252·0.9563 ≈ 277 - 240.9876 ≈ 36.0124
BC ≈ √36.0124 ≈ 6.0

✔ Final Answer for #4: 6.0

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Problem 5: Find AB

Triangle ABC:
- AC = 12
- BC = 13
- Angle C = 134°
→ We want side AB (opposite angle C)

AB² = 12² + 13² - 2·12·13·cos(134°)
AB² = 144 + 169 - 312·cos(134°)
cos(134°) ≈ -0.6947
AB² = 313 - 312·(-0.6947) = 313 + 216.7464 ≈ 529.7464
AB ≈ √529.7464 ≈ 23.0

✔ Final Answer for #5: 23.0

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Problem 6: Find AB

Triangle ABC:
- AC = 20
- BC = 22
- Angle C = 95°
→ We want side AB (opposite angle C)

AB² = 20² + 22² - 2·20·22·cos(95°)
AB² = 400 + 484 - 880·cos(95°)
cos(95°) ≈ -0.0872
AB² = 884 - 880·(-0.0872) = 884 + 76.736 ≈ 960.736
AB ≈ √960.736 ≈ 31.0

✔ Final Answer for #6: 31.0

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Problem 7: Find m∠A

Triangle ABC:
- AC = 9
- AB = 6
- BC = 14
→ We want angle A (between sides AB and AC)

Use Law of Cosines rearranged for angle:

cos(A) = (AB² + AC² - BC²) / (2·AB·AC)
cos(A) = (6² + 9² - 14²) / (2·6·9)
cos(A) = (36 + 81 - 196) / 108 = (-79)/108 ≈ -0.7315
A = arccos(-0.7315) ≈ 137.0°

✔ Final Answer for #7: 137.0°

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Problem 8: Find m∠B

Triangle ABC:
- AB = 22
- BC = ? → Wait, we have:
Actually, from diagram:
Side opposite B is AC = 17
Sides adjacent to B: AB = 22, BC = ? — wait, let’s label properly.

From diagram:
Points: A, B, C
Angle at A is 143°, sides: AB = 22, AC = 17 → so we want angle at B.

Wait — actually, in triangle ABC:
Given:
- Side opposite A is BC = ?
But we are given:
AB = 22, AC = 17, angle A = 143° → but question asks for angle B.

Better approach: Use Law of Cosines to find side BC first? Or directly find angle B?

Actually, we can use Law of Cosines for angle B if we know all three sides. But we don’t yet.

Wait — we know two sides and included angle? No — angle A is between AB and AC, so we can find side BC first.

Then use Law of Cosines again to find angle B.

Step 1: Find BC (side opposite angle A)

BC² = AB² + AC² - 2·AB·AC·cos(angle A)
BC² = 22² + 17² - 2·22·17·cos(143°)
BC² = 484 + 289 - 748·cos(143°)
cos(143°) ≈ -0.7986
BC² = 773 - 748·(-0.7986) = 773 + 597.3528 ≈ 1370.3528
BC ≈ √1370.3528 ≈ 37.02

Now, to find angle B:

In triangle ABC, angle B is between sides AB and BC.

So:

cos(B) = (AB² + BC² - AC²) / (2·AB·BC)
cos(B) = (22² + 37.02² - 17²) / (2·22·37.02)
Calculate numerator:
484 + 1370.4804 - 289 = 1565.4804
Denominator: 2·22·37.02 ≈ 1628.88
cos(B) ≈ 1565.4804 / 1628.88 ≈ 0.9611
B ≈ arccos(0.9611) ≈ 16.0°

✔ Final Answer for #8: 16.0°

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Problem 9: Find m∠A

Triangle ABC:
- AB = 28
- AC = 29
- Angle C = 52°
→ We want angle A.

First, find side BC using Law of Cosines (since we have two sides and included angle? Wait — angle C is between AC and BC? Let's see.

Actually, angle C is at vertex C, so between sides AC and BC. But we don’t know BC.

Wait — we know sides AC=29, AB=28, and angle C=52°. So angle C is NOT between the two known sides. That means we cannot directly apply Law of Cosines to find another side without more info.

Actually, we can use Law of Sines? But the worksheet is on Law of Cosines. Maybe we need to find side AB? Wait no — AB is given as 28.

Wait — let me re-read: “Find m∠A” — angle at A.

We know:
- Side opposite A is BC — unknown
- Side opposite B is AC = 29
- Side opposite C is AB = 28
- Angle C = 52°

So we can use Law of Sines? But instructions say Law of Cosines. Alternatively, we can find side BC first using Law of Cosines? But we don’t have two sides enclosing angle C.

Wait — angle C is between sides AC and BC. We know AC = 29, but not BC. So we can't compute BC directly.

Alternative plan: Use Law of Cosines to express relationship.

Actually, better: Use Law of Cosines formula for angle A.

We need all three sides to find an angle. So let’s find side BC first.

In triangle ABC, by Law of Cosines at angle C:

AB² = AC² + BC² - 2·AC·BC·cos(angle C)

Wait — that would be:

28² = 29² + BC² - 2·29·BC·cos(52°)

This is quadratic in BC — messy.

Alternatively, use Law of Sines since we have one angle and its opposite side.

Angle C = 52°, opposite side AB = 28
Side AC = 29 is opposite angle B
Side BC is opposite angle A

Law of Sines:
sin(A)/a = sin(B)/b = sin(C)/c

So:
sin(C)/AB = sin(A)/BC = sin(B)/AC

But we don’t know BC or angles A or B.

Set up:

sin(52°)/28 = sin(B)/29
→ sin(B) = 29·sin(52°)/28
sin(52°) ≈ 0.7880
sin(B) ≈ 29·0.7880/28 ≈ 22.852/28 ≈ 0.8161
B ≈ arcsin(0.8161) ≈ 54.7° or 125.3° — but since angle C is 52°, and sum must be 180°, both possible? Let’s check.

If B = 54.7°, then A = 180 - 52 - 54.7 = 73.3°
If B = 125.3°, then A = 180 - 52 - 125.3 = 2.7° — possible but unlikely given side lengths.

Check which makes sense: side AC = 29 is longer than AB = 28, so angle B should be larger than angle C? Angle C is 52°, so angle B > 52° — both options satisfy.

But side BC is opposite angle A — if A is small, BC is small; if A is large, BC is large.

Actually, let’s use Law of Cosines properly.

We can write:

For angle A:

cos(A) = (AB² + AC² - BC²)/(2·AB·AC)

But we don’t know BC.

Instead, let’s find BC using Law of Cosines at angle C:

AB² = AC² + BC² - 2·AC·BC·cos(C)

28² = 29² + x² - 2·29·x·cos(52°) where x = BC

784 = 841 + x² - 58x·0.6157 [since cos(52°)≈0.6157]

784 = 841 + x² - 35.7106x

Bring all to one side:

x² - 35.7106x + 841 - 784 = 0
x² - 35.7106x + 57 = 0

Solve quadratic:

Discriminant D = (35.7106)^2 - 4·1·57 ≈ 1275.24 - 228 = 1047.24
√D ≈ 32.36

x = [35.7106 ± 32.36]/2

x1 = (35.7106 + 32.36)/2 ≈ 68.07/2 ≈ 34.035
x2 = (35.7106 - 32.36)/2 ≈ 3.35/2 ≈ 1.675

Now, which one makes sense? If BC = 1.675, then sides are 28, 29, 1.675 — almost degenerate, angle at A would be very small. If BC = 34.035, then it’s a normal triangle.

Check with Law of Sines: if BC = 34.035, then angle A opposite it:

sin(A)/34.035 = sin(52°)/28
sin(A) = 34.035 * 0.7880 / 28 ≈ 26.82 / 28 ≈ 0.9579
A ≈ arcsin(0.9579) ≈ 73.3° — matches our earlier calculation.

If BC = 1.675, sin(A) = 1.675*0.7880/28 ≈ 1.32/28 ≈ 0.047, A≈2.7° — also mathematically possible, but looking at the diagram (though we shouldn’t rely on it), it looks like a regular triangle, so likely 73.3°.

Moreover, in problem 9, the triangle is drawn with angle C=52°, sides 29 and 28, so angle A should be around 70-80 degrees.

So we take A ≈ 73.3°

But let’s verify using Law of Cosines with BC=34.035:

cos(A) = (AB² + AC² - BC²)/(2·AB·AC)
= (28² + 29² - 34.035²)/(2·28·29)
= (784 + 841 - 1158.38)/(1624)
= (1625 - 1158.38)/1624 = 466.62/1624 ≈ 0.2873
A = arccos(0.2873) ≈ 73.3° — yes.

So answer is 73.3°

✔ Final Answer for #9: 73.3°

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Problem 10: Find m∠C

Triangle ABC:
- BC = 29
- AB = 24
- Angle B = 101°
→ We want angle C.

First, find side AC using Law of Cosines (since we have two sides and included angle? Angle B is between AB and BC.

Yes! So side opposite angle B is AC.

AC² = AB² + BC² - 2·AB·BC·cos(angle B)
AC² = 24² + 29² - 2·24·29·cos(101°)
AC² = 576 + 841 - 1392·cos(101°)
cos(101°) ≈ -0.1908
AC² = 1417 - 1392·(-0.1908) = 1417 + 265.5936 ≈ 1682.5936
AC ≈ √1682.5936 ≈ 41.02

Now, to find angle C:

Angle C is between sides BC and AC.

So:

cos(C) = (BC² + AC² - AB²) / (2·BC·AC)
cos(C) = (29² + 41.02² - 24²) / (2·29·41.02)
Numerator: 841 + 1682.6404 - 576 = 1947.6404
Denominator: 2·29·41.02 ≈ 2379.16
cos(C) ≈ 1947.6404 / 2379.16 ≈ 0.8186
C ≈ arccos(0.8186) ≈ 35.0°

✔ Final Answer for #10: 35.0°

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Final Answers:

1) 21.0
2) 45.0
3) 33.0
4) 6.0
5) 23.0
6) 31.0
7) 137.0°
8) 16.0°
9) 73.3°
10) 35.0°