Math worksheet for practicing the Law of Sines and Cosines, featuring problems to find missing lengths and angles in triangles.

Let’s solve each problem step by step. We’ll use the Law of Sines and Law of Cosines as needed, depending on what information is given.

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Section A: Find missing lengths (to 2 decimal places)

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Problem 1)
Triangle ABC:
Angle B = 83°, side c = AB = 5, side a = BC = 4 → find side b = AC.

We have two sides and the included angle → use Law of Cosines:

b² = a² + c² - 2ac·cos(B)
b² = 4² + 5² - 2·4·5·cos(83°)
b² = 16 + 25 - 40·cos(83°)

Calculate cos(83°):
cos(83°) ≈ 0.1219 (use calculator in degree mode)

So:
b² = 41 - 40 × 0.1219 = 41 - 4.876 = 36.124
b = √36.124 ≈ 6.01

✔ Answer for 1): 6.01

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Problem 2)
Triangle ABC:
Angle A = 62°, Angle B = 41°, side b = AC = 4 → find side a = BC.

First, find angle C:
Angle C = 180° - 62° - 41° = 77°

Now use Law of Sines:
a / sin(A) = b / sin(B)
→ a = b · sin(A) / sin(B)
a = 4 · sin(62°) / sin(41°)

sin(62°) ≈ 0.8829
sin(41°) ≈ 0.6561

a = 4 × 0.8829 / 0.6561 ≈ 3.5316 / 0.6561 ≈ 5.38

✔ Answer for 2): 5.38

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Problem 3)
Triangle ABC:
Angle B = 120°, Angle C = 28°, side b = AC = 8.18 → find side a = BC.

First, find angle A:
Angle A = 180° - 120° - 28° = 32°

Use Law of Sines:
a / sin(A) = b / sin(B)
→ a = b · sin(A) / sin(B)
a = 8.18 · sin(32°) / sin(120°)

sin(32°) ≈ 0.5299
sin(120°) = sin(60°) ≈ 0.8660

a = 8.18 × 0.5299 / 0.8660 ≈ 4.334 / 0.8660 ≈ 5.00

✔ Answer for 3): 5.00

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Problem 4)
Triangle ABC:
Side a = BC = 7.51, side b = AC = 6.42, Angle C = 23° → find side c = AB.

Two sides and included angle → use Law of Cosines:

c² = a² + b² - 2ab·cos(C)
c² = 7.51² + 6.42² - 2·7.51·6.42·cos(23°)

Calculate squares:
7.51² = 56.4001
6.42² = 41.2164
Sum = 97.6165

cos(23°) ≈ 0.9205
2·7.51·6.42 = 2 × 48.2142 = 96.4284
Multiply by cos(23°): 96.4284 × 0.9205 ≈ 88.76

So:
c² = 97.6165 - 88.76 ≈ 8.8565
c = √8.8565 ≈ 2.98

✔ Answer for 4): 2.98

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Section B: Find missing angles (to 3 significant figures)

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Problem 1) Triangle PQR:
Sides: PQ = 2.3, PR = 4.4, QR = 6 → find angle PQR (angle at Q).

We know all three sides → use Law of Cosines to find angle at Q.

In triangle PQR, angle at Q is between sides PQ and QR? Wait — let’s label carefully.

Actually, angle PQR is at vertex Q, so it’s between sides QP and QR.

But we are given:
- Side opposite P: QR = 6
- Side opposite Q: PR = 4.4
- Side opposite R: PQ = 2.3

Wait — better to think: To find angle at Q, we need the sides adjacent to Q and the side opposite.

Adjacent to angle Q: sides QP and QR → lengths 2.3 and 6
Opposite to angle Q: side PR = 4.4

Law of Cosines:

(PR)² = (QP)² + (QR)² - 2·QP·QR·cos(angle PQR)

So:

4.4² = 2.3² + 6² - 2·2.3·6·cos(Q)

19.36 = 5.29 + 36 - 27.6·cos(Q)
19.36 = 41.29 - 27.6·cos(Q)

Rearrange:

27.6·cos(Q) = 41.29 - 19.36 = 21.93
cos(Q) = 21.93 / 27.6 ≈ 0.7946

Q = arccos(0.7946) ≈ 37.4°

✔ Answer for 1): 37.4°

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Problem 2) Triangle DEF: Area = 10 square units.
Sides: DE = 3.85, DF = 7.63 → find angle FDE (angle at D).

Area formula with two sides and included angle:

Area = (1/2)·DE·DF·sin(angle FDE)

10 = (1/2)·3.85·7.63·sin(D)

First, compute 3.85 × 7.63 = ?

3.85 × 7.63:
= 3.85 × 7 + 3.85 × 0.63
= 26.95 + 2.4255 = 29.3755

Then:
10 = (1/2) × 29.3755 × sin(D)
10 = 14.68775 × sin(D)
sin(D) = 10 / 14.68775 ≈ 0.6808

D = arcsin(0.6808) ≈ 42.9°

✔ Answer for 2): 42.9°

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Problem 3) Triangle LMN:
Given: Angle L = 41°, side LM = 51, side MN = 34 → find angle LMN (angle at M), but note: angle LNM is obtuse.

Hint says: First find angle LNM, then do 180 minus that answer before finding LMN.

Wait — actually, hint says: “Find angle LNM first but do 180 minus the answer before then finding LMN”

That seems confusing. Let me re-read.

Actually, probably means: Use Law of Sines to find angle at N (LNM), but since it's obtuse, you get an acute angle from calculator, so subtract from 180 to get actual obtuse angle. Then use angle sum to find angle at M.

Let’s try:

In triangle LMN:
Angle at L = 41°
Side opposite L: MN = 34
Side opposite N: LM = 51
Side opposite M: LN = unknown

Use Law of Sines to find angle at N:

sin(N)/LM = sin(L)/MN
→ sin(N)/51 = sin(41°)/34

sin(41°) ≈ 0.6561

So:
sin(N) = 51 × 0.6561 / 34 ≈ 33.4611 / 34 ≈ 0.98415

N = arcsin(0.98415) ≈ 79.8° — but this is acute.

But problem says angle LNM is obtuse → so actual angle N = 180° - 79.8° = 100.2°

Now, angle at M = 180° - angle L - angle N = 180 - 41 - 100.2 = 38.8°

✔ Answer for 3): 38.8°

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Problem 4) YZT is straight line. Triangle XYZ:
XY = 9.2, XZ = 4.3, YZ = 6.4 → find angle XZT.

Note: YZT is straight → so angle XZY and angle XZT are supplementary (add to 180°).

So if we find angle XZY (inside triangle), then angle XZT = 180° - angle XZY.

In triangle XYZ, find angle at Z (angle XZY).

Sides:
XZ = 4.3, YZ = 6.4, XY = 9.2

Use Law of Cosines at angle Z:

(XY)² = (XZ)² + (YZ)² - 2·XZ·YZ·cos(angle XZY)

9.2² = 4.3² + 6.4² - 2·4.3·6.4·cos(Z)

84.64 = 18.49 + 40.96 - 55.04·cos(Z)
84.64 = 59.45 - 55.04·cos(Z)

Rearrange:

55.04·cos(Z) = 59.45 - 84.64 = -25.19
cos(Z) = -25.19 / 55.04 ≈ -0.4577

Z = arccos(-0.4577) ≈ 117.2°

So angle XZY = 117.2° → then angle XZT = 180° - 117.2° = 62.8°

✔ Answer for 4): 62.8°

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Final Answer:
Section A:
1) 6.01
2) 5.38
3) 5.00
4) 2.98

Section B:
1) 37.4°
2) 42.9°
3) 38.8°
4) 62.8°