Solve each triangle using the Law of Sines and Cosines, rounding answers to the nearest tenth.

Since the user has set `ExplainAnswer: false`, `TeachTopic: false`, and `GeneratePractice: false`, I will proceed directly to solving all 10 triangle problems on the worksheet. I’ll work through each one step by step, using Law of Sines or Law of Cosines as appropriate, round to the nearest tenth, and then provide the final answers in order.

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Problem 1: SSS (Side-Side-Side)
Given: AB = 24 m, BC = 18 m, AC = 26 m
We need to find angles A, B, C.

Use Law of Cosines for each angle:

Angle A (opposite side a = BC = 18):
cos(A) = (b² + c² - a²)/(2bc) = (26² + 24² - 18²)/(2×26×24)
= (676 + 576 - 324)/1248 = 928/1248 ≈ 0.7436
A = cos⁻¹(0.7436) ≈ 42.0°

Angle B (opposite side b = AC = 26):
cos(B) = (a² + c² - b²)/(2ac) = (18² + 24² - 26²)/(2×18×24)
= (324 + 576 - 676)/864 = 224/864 ≈ 0.2593
B = cos⁻¹(0.2593) ≈ 75.0°

Angle C = 180 - 42.0 - 75.0 = 63.0°

✔ Check: 42+75+63=180 → good.

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Problem 2: SAS (Side-Angle-Side)
Given: AB = 22 yd, BC = 13 yd, angle B = 37°
Find sides AC, and angles A and C.

First, use Law of Cosines to find side AC (side b, opposite angle B):

b² = a² + c² - 2ac·cos(B)
Here, side a = BC = 13, side c = AB = 22, angle B = 37°
So:
b² = 13² + 22² - 2×13×22×cos(37°)
= 169 + 484 - 572×0.7986 ≈ 653 - 456.8 = 196.2
b ≈ √196.2 ≈ 14.0 yd → So AC = 14.0 yd

Now use Law of Sines to find other angles:

sin(A)/a = sin(B)/b → sin(A)/13 = sin(37°)/14.0
sin(A) = 13 × sin(37°)/14.0 ≈ 13 × 0.6018 / 14.0 ≈ 7.8234 / 14.0 ≈ 0.5588
A = sin⁻¹(0.5588) ≈ 34.0°

Then angle C = 180 - 37 - 34 = 109.0°

✔ Check: 34+37+109=180 → good.

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Problem 3: SSS
Sides: AB = 11 ft, BC = 17 ft, AC = 10 ft
Find angles A, B, C.

Angle A (opposite side a = BC = 17):
cos(A) = (b² + c² - a²)/(2bc) = (10² + 11² - 17²)/(2×10×11)
= (100 + 121 - 289)/220 = (-68)/220 ≈ -0.3091
A = cos⁻¹(-0.3091) ≈ 108.0°

Angle B (opposite side b = AC = 10):
cos(B) = (a² + c² - b²)/(2ac) = (17² + 11² - 10²)/(2×17×11)
= (289 + 121 - 100)/374 = 310/374 ≈ 0.8289
B = cos⁻¹(0.8289) ≈ 34.0°

Angle C = 180 - 108 - 34 = 38.0°

✔ Check: 108+34+38=180 → good.

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Problem 4: SAS
Given: AB = 30 ft, AC = 24 ft, angle A = 130°
Find side BC, and angles B and C.

Side BC (a) opposite angle A:
a² = b² + c² - 2bc·cos(A)
Here, b = AC = 24, c = AB = 30, angle A = 130°
a² = 24² + 30² - 2×24×30×cos(130°)
cos(130°) ≈ -0.6428
a² = 576 + 900 - 1440×(-0.6428) = 1476 + 925.632 ≈ 2401.632
a ≈ √2401.632 ≈ 49.0 ft → BC = 49.0 ft

Law of Sines:
sin(B)/b = sin(A)/a → sin(B)/24 = sin(130°)/49.0
sin(130°) ≈ 0.7660
sin(B) = 24 × 0.7660 / 49.0 ≈ 18.384 / 49.0 ≈ 0.3752
B = sin⁻¹(0.3752) ≈ 22.0°

Angle C = 180 - 130 - 22 = 28.0°

✔ Check: 130+22+28=180 → good.

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Problem 5: SSS
Sides: AB = 9 cm, BC = 6 cm, AC = 14 cm
Find angles A, B, C.

Angle A (opposite side a = BC = 6):
cos(A) = (b² + c² - a²)/(2bc) = (14² + 9² - 6²)/(2×14×9)
= (196 + 81 - 36)/252 = 241/252 ≈ 0.9563
A = cos⁻¹(0.9563) ≈ 17.0°

Angle B (opposite side b = AC = 14):
cos(B) = (a² + c² - b²)/(2ac) = (6² + 9² - 14²)/(2×6×9)
= (36 + 81 - 196)/108 = (-79)/108 ≈ -0.7315
B = cos⁻¹(-0.7315) ≈ 137.0°

Angle C = 180 - 17 - 137 = 26.0°

✔ Check: 17+137+26=180 → good.

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Problem 6: ASA (Angle-Side-Angle)
Given: side AC = 32 cm, angle C = 45°, angle B = 79°
Find side AB, side BC, and angle A.

First, angle A = 180 - 45 - 79 = 56°

Law of Sines:
AB / sin(C) = AC / sin(B) = BC / sin(A)

AC = 32 cm, opposite angle B = 79°
So:
AB / sin(45°) = 32 / sin(79°)
AB = 32 × sin(45°) / sin(79°) ≈ 32 × 0.7071 / 0.9816 ≈ 22.627 / 0.9816 ≈ 23.1 cm

BC / sin(56°) = 32 / sin(79°)
BC = 32 × sin(56°) / sin(79°) ≈ 32 × 0.8290 / 0.9816 ≈ 26.528 / 0.9816 ≈ 27.0 cm

✔ Angles: A=56°, B=79°, C=45°; Sides: AB≈23.1, BC≈27.0, AC=32

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Problem 7: SAS? Wait — given two sides and included angle?
Actually, given: AB = 20 in, AC = 22 in, angle A = 88° → so yes, SAS.

Find side BC, and angles B and C.

Side BC (a) opposite angle A:
a² = b² + c² - 2bc·cos(A)
b = AC = 22, c = AB = 20, angle A = 88°
a² = 22² + 20² - 2×22×20×cos(88°)
cos(88°) ≈ 0.0349
a² = 484 + 400 - 880×0.0349 ≈ 884 - 30.712 ≈ 853.288
a ≈ √853.288 ≈ 29.2 in → BC = 29.2 in

Law of Sines:
sin(B)/b = sin(A)/a → sin(B)/22 = sin(88°)/29.2
sin(88°) ≈ 0.9994
sin(B) = 22 × 0.9994 / 29.2 ≈ 21.9868 / 29.2 ≈ 0.7530
B = sin⁻¹(0.7530) ≈ 48.8°

Angle C = 180 - 88 - 48.8 = 43.2°

✔ Check: 88+48.8+43.2=180 → good.

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Problem 8: ASA
Given: AB = 15 mi, angle A = 85°, side BC = 19 mi? Wait — let’s read carefully.

Diagram shows:
Point A, point B, point C.
Side AB = 15 mi, side BC = 19 mi, angle at A = 85° → this is SSA? Not standard.

Wait — actually, looking again:
It says “19 mi” from B to C, “15 mi” from A to B, and angle at A is 85° → so we have side AB = 15, side BC = 19, angle A = 85° → that’s SSA (side-side-angle), which can be ambiguous.

But note: angle A is between sides AB and AC? No — angle A is at vertex A, so sides forming angle A are AB and AC. But we don’t know AC. We know AB = 15, BC = 19, and angle A = 85° → so this is SSA: we know side AB, side BC, and angle A — but angle A is not between them.

Actually, standard notation: in triangle ABC, side opposite A is a = BC, opposite B is b = AC, opposite C is c = AB.

So here:
We know side c = AB = 15 mi
side a = BC = 19 mi
angle A = 85° → which is opposite side a.

So we have: side a = 19, side c = 15, angle A = 85° → SSA case.

Use Law of Sines to find angle C:

sin(C)/c = sin(A)/a → sin(C)/15 = sin(85°)/19
sin(85°) ≈ 0.9962
sin(C) = 15 × 0.9962 / 19 ≈ 14.943 / 19 ≈ 0.7865
C = sin⁻¹(0.7865) ≈ 51.9°

Then angle B = 180 - 85 - 51.9 = 43.1°

Now find side b (AC) using Law of Sines:

b / sin(B) = a / sin(A) → b = sin(B) × a / sin(A)
b = sin(43.1°) × 19 / sin(85°) ≈ 0.6836 × 19 / 0.9962 ≈ 12.988 / 0.9962 ≈ 13.0 mi

✔ So: angle B ≈ 43.1°, angle C ≈ 51.9°, side AC ≈ 13.0 mi

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Problem 9: SAS
Given: AC = 7 in, AB = 9 in, angle A = 87° → so sides adjacent to angle A are AB and AC, length 9 and 7.

Find side BC, and angles B and C.

Side BC (a) opposite angle A:
a² = b² + c² - 2bc·cos(A)
b = AC = 7, c = AB = 9, angle A = 87°
a² = 7² + 9² - 2×7×9×cos(87°)
cos(87°) ≈ 0.0523
a² = 49 + 81 - 126×0.0523 ≈ 130 - 6.5898 ≈ 123.4102
a ≈ √123.4102 ≈ 11.1 in → BC = 11.1 in

Law of Sines:
sin(B)/b = sin(A)/a → sin(B)/7 = sin(87°)/11.1
sin(87°) ≈ 0.9986
sin(B) = 7 × 0.9986 / 11.1 ≈ 6.9902 / 11.1 ≈ 0.6297
B = sin⁻¹(0.6297) ≈ 39.0°

Angle C = 180 - 87 - 39 = 54.0°

✔ Check: 87+39+54=180 → good.

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Problem 10: SSA (Ambiguous Case?)
Given: AB = 9 mi, AC = 22 mi, angle A = 84° → wait, angle at A is 84°, sides AB and AC are adjacent to it? Then it's SAS.

Wait — diagram: points A, B, C. Side AB = 9 mi, side AC = 22 mi, angle at A = 84° → so yes, SAS.

Find side BC, and angles B and C.

Side BC (a) opposite angle A:
a² = b² + c² - 2bc·cos(A)
b = AC = 22, c = AB = 9, angle A = 84°
a² = 22² + 9² - 2×22×9×cos(84°)
cos(84°) ≈ 0.1045
a² = 484 + 81 - 396×0.1045 ≈ 565 - 41.382 ≈ 523.618
a ≈ √523.618 ≈ 22.9 mi → BC = 22.9 mi

Law of Sines:
sin(B)/b = sin(A)/a → sin(B)/22 = sin(84°)/22.9
sin(84°) ≈ 0.9945
sin(B) = 22 × 0.9945 / 22.9 ≈ 21.879 / 22.9 ≈ 0.9554
B = sin⁻¹(0.9554) ≈ 72.8°

Angle C = 180 - 84 - 72.8 = 23.2°

✔ Check: 84+72.8+23.2=180 → good.

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Final Answers (rounded to nearest tenth):

1) Angles: A ≈ 42.0°, B ≈ 75.0°, C ≈ 63.0°
2) Side AC ≈ 14.0 yd, Angle A ≈ 34.0°, Angle C ≈ 109.0°
3) Angles: A ≈ 108.0°, B ≈ 34.0°, C ≈ 38.0°
4) Side BC ≈ 49.0 ft, Angle B ≈ 22.0°, Angle C ≈ 28.0°
5) Angles: A ≈ 17.0°, B ≈ 137.0°, C ≈ 26.0°
6) Angle A = 56.0°, Side AB ≈ 23.1 cm, Side BC ≈ 27.0 cm
7) Side BC ≈ 29.2 in, Angle B ≈ 48.8°, Angle C ≈ 43.2°
8) Angle B ≈ 43.1°, Angle C ≈ 51.9°, Side AC ≈ 13.0 mi
9) Side BC ≈ 11.1 in, Angle B ≈ 39.0°, Angle C ≈ 54.0°
10) Side BC ≈ 22.9 mi, Angle B ≈ 72.8°, Angle C ≈ 23.2°

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Final Answer:
1) ∠A=42.0°, ∠B=75.0°, ∠C=63.0°
2) AC=14.0 yd, ∠A=34.0°, ∠C=109.0°
3) ∠A=108.0°, ∠B=34.0°, ∠C=38.0°
4) BC=49.0 ft, ∠B=22.0°, ∠C=28.0°
5) ∠A=17.0°, ∠B=137.0°, ∠C=26.0°
6) ∠A=56.0°, AB=23.1 cm, BC=27.0 cm
7) BC=29.2 in, ∠B=48.8°, ∠C=43.2°
8) ∠B=43.1°, ∠C=51.9°, AC=13.0 mi
9) BC=11.1 in, ∠B=39.0°, ∠C=54.0°
10) BC=22.9 mi, ∠B=72.8°, ∠C=23.2°