Math 2 Honors worksheet on Law of Sines & Cosines with four triangle problems.

Let’s solve each triangle one by one using the Law of Sines and Law of Cosines, as needed. We’ll find missing sides and angles step by step.

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Triangle 1:

Given:
- Side AB = c = 24 (opposite angle C)
- Side BC = a = 19 (opposite angle A)
- Side AC = b = 26 (opposite angle B)

We need to find all three angles: ∠A, B, ∠C.

Since we have all three sides, use the Law of Cosines first to find one angle, then Law of Sines for others.

#### Step 1: Find ∠C (between sides a=19 and b=26, opposite side c=24)

Law of Cosines:
c² = a² + b² - 2ab·cos(C)
→ 24² = 19² + 26² - 2·19·26·cos(C)
→ 576 = 361 + 676 - 988·cos(C)
→ 576 = 1037 - 988·cos(C)
→ 988·cos(C) = 1037 - 576 = 461
→ cos(C) = 461 / 988 ≈ 0.4666
→ ∠C ≈ cos⁻¹(0.4666) ≈ 62.2°

#### Step 2: Use Law of Sines to find ∠A

Law of Sines:
sin(A)/a = sin(C)/c
→ sin(A)/19 = sin(62.2°)/24
→ sin(A) = 19 · sin(62.2°) / 24
→ sin(62.2°) ≈ 0.884
→ sin(A) ≈ 19 · 0.884 / 24 ≈ 16.796 / 24 ≈ 0.6998
→ ∠A ≈ sin⁻¹(0.6998) ≈ 44.4°

#### Step 3: Find ∠B

Sum of angles in triangle = 180°
∠B = 180° - ∠A - ∠C ≈ 180 - 44.4 - 62.2 = 73.4°

✔ Check with Law of Sines for ∠B:
sin(B)/b = sin(C)/c → sin(B)/26 = sin(62.2°)/24
→ sin(B) = 26 · 0.884 / 24 ≈ 22.984 / 24 ≈ 0.9577
→ ∠B ≈ sin⁻¹(0.9577) ≈ 73.3° — close enough (rounding error). Good.

So for Triangle 1:
m∠A ≈ 44.4°
m∠B ≈ 73.4°
m∠C ≈ 62.2°

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Triangle 2:

Given:
- Side AB = c = 22 (opposite angle C)
- Side BC = a = ? (opposite angle A)
- Side AC = b = 13 (opposite angle B)
- Angle B = 32.7°

We need to find: m∠A, m∠C, and side a.

Use Law of Sines first since we have an angle and its opposite side? Wait — we have angle B and side b? No! Actually, we have angle B and side AC = b = 13? Let’s label correctly.

Standard labeling:
Side opposite angle A is a = BC
Side opposite angle B is b = AC = 13
Side opposite angle C is c = AB = 22
Angle B = 32.7°

So we know:
angle B = 32.7°, side b = 13, side c = 22

We can use Law of Sines to find angle C:

sin(C)/c = sin(B)/b
→ sin(C)/22 = sin(32.7°)/13
→ sin(C) = 22 · sin(32.7°) / 13
→ sin(32.7°) ≈ 0.540
→ sin(C) ≈ 22 · 0.540 / 13 ≈ 11.88 / 13 ≈ 0.9138
→ ∠C ≈ sin⁻¹(0.9138) ≈ 66.0°

Now find angle A:
∠A = 180° - 32.7° - 66.0° = 81.3°

Now find side a (BC) using Law of Sines:

sin(A)/a = sin(B)/b
→ sin(81.3°)/a = sin(32.7°)/13
→ a = 13 · sin(81.3°) / sin(32.7°)
→ sin(81.3°) ≈ 0.988, sin(32.7°) ≈ 0.540
→ a ≈ 13 · 0.988 / 0.540 ≈ 12.844 / 0.540 ≈ 23.8

Check: Is this reasonable? Side opposite largest angle should be longest. Angle A = 81.3° is largest, so side a should be longest — yes, 23.8 > 22 > 13. Good.

So for Triangle 2:
m∠A ≈ 81.3°
m∠C ≈ 66.0°
a ≈ 23.8

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Triangle 3:

Given:
- Side AB = c = ? (opposite angle C)
- Side BC = a = 22 (opposite angle A)
- Side AC = b = ? (opposite angle B)
- Angle C = 63°, Angle B = 79°

First, find angle A:
∠A = 180° - 63° - 79° = 38°

Now use Law of Sines to find sides b and c.

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

We know a = 22, ∠A = 38°

So:
22 / sin(38°) = b / sin(79°) = c / sin(63°)

Compute 22 / sin(38°):
sin(38°) ≈ 0.6157
→ 22 / 0.6157 ≈ 35.73

Now find b:
b = 35.73 · sin(79°)
sin(79°) ≈ 0.9816
→ b ≈ 35.73 · 0.9816 ≈ 35.1

Find c:
c = 35.73 · sin(63°)
sin(63°) ≈ 0.8910
→ c ≈ 35.73 · 0.8910 ≈ 31.8

Check: Largest angle is B=79°, so side b should be longest — yes, 35.1 > 31.8 > 22. Good.

So for Triangle 3:
m∠A = 38°
b ≈ 35.1
c ≈ 31.8

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Triangle 4:

Given:
- Side AB = c = 15 (opposite angle C)
- Side BC = a = 18 (opposite angle A)
- Side AC = b = ? (opposite angle B)
- Angle A = 86.3°

We need to find: m∠B, m∠C, and side b.

Use Law of Cosines to find side b first? Or Law of Sines?

Actually, we have two sides and included angle? Wait — angle A is between sides AB and AC? Let’s clarify.

Labeling:
Angle A is at vertex A, between sides AB and AC.
Sides adjacent to angle A: AB = c = 15, AC = b = ?
Opposite side to angle A is BC = a = 18.

Wait — that doesn’t match standard labeling. Let me reassign properly.

Standard:
In triangle ABC:
- Side a = BC (opposite angle A)
- Side b = AC (opposite angle B)
- Side c = AB (opposite angle C)

Given:
- a = BC = 18
- c = AB = 15
- angle A = 86.3°

We can use Law of Cosines to find side b (AC), because we have two sides and the included angle? Wait — angle A is NOT between sides a and c. Actually, angle A is between sides AB and AC, which are sides c and b. But we don’t know b yet.

Better approach: Use Law of Sines after finding another angle? But we only have one angle and two sides not including it.

Actually, we have side a, side c, and angle A — this is SSA case? But angle A is opposite side a, and we have side c too.

Use Law of Sines to find angle C:

sin(C)/c = sin(A)/a
→ sin(C)/15 = sin(86.3°)/18
→ sin(C) = 15 · sin(86.3°) / 18
→ sin(86.3°) ≈ 0.9979
→ sin(C) ≈ 15 · 0.9979 / 18 ≈ 14.9685 / 18 ≈ 0.8316
→ ∠C ≈ sin⁻¹(0.8316) ≈ 56.3°

Now find angle B:
∠B = 180° - 86.3° - 56.3° = 37.4°

Now find side b using Law of Sines:

sin(B)/b = sin(A)/a
→ sin(37.4°)/b = sin(86.3°)/18
→ b = 18 · sin(37.4°) / sin(86.3°)
→ sin(37.4°) ≈ 0.607, sin(86.3°) ≈ 0.9979
→ b ≈ 18 · 0.607 / 0.9979 ≈ 10.926 / 0.9979 ≈ 10.95

Check: Largest angle is A=86.3°, so side a=18 should be longest — yes, 18 > 15 > 10.95. Good.

So for Triangle 4:
m∠B ≈ 37.4°
m∠C ≈ 56.3°
b ≈ 10.95

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Final Answer:

Triangle 1:
m∠A ≈ 44.4°, m∠B ≈ 73.4°, m∠C ≈ 62.2°

Triangle 2:
m∠A ≈ 81.3°, m∠C ≈ 66.0°, a ≈ 23.8

Triangle 3:
m∠A = 38°, b ≈ 35.1, c ≈ 31.8

Triangle 4:
m∠B ≈ 37.4°, m∠C ≈ 56.3°, b ≈ 10.95