Mixed practice worksheet on the Law of Sines and Cosines for calculating triangle areas and missing angles and sides.

Let’s work through each problem step by step.

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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 a = 4 (opposite angle A) → wait, actually in standard notation:
- side a is opposite angle A → but here it’s labeled as “a = 4” next to side BC? Let’s check the diagram.

Actually, looking at the triangle:

In triangle ABC:
- Side c = AB = 5 (since it’s between A and B)
- Side a = BC = 4 (between B and C)
- Angle at B = 83°
- We need to find side b = AC

This is SAS (side-angle-side): we have two sides and the included angle → use Law of Cosines.

Formula:
b² = a² + c² - 2ac·cos(B)

Plug in:
b² = 4² + 5² - 2×4×5×cos(83°)
b² = 16 + 25 - 40×cos(83°)
cos(83°) ≈ 0.1219
So:
b² = 41 - 40×0.1219 = 41 - 4.876 = 36.124
b = √36.124 ≈ 6.01

✔ Answer: 6.01

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Problem 2)
Triangle ABC:
- Angle A = 62°
- Angle B = 41°
- Side b = 4 (which is AC, opposite angle B)
We need to find side a = BC (opposite angle A)

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 ≈ 4 × 1.3457 ≈ 5.38

✔ Answer: 5.38

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Problem 3)
Triangle ABC:
- Angle B = 120°
- Angle C = 28°
- Side b = 8.18 (AC, opposite angle B)
Find side a = BC (opposite angle A)

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

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 ≈ 8.18 × 0.6119 ≈ 5.00

✔ Answer: 5.00

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Problem 4)
Triangle ABC:
- Side a = 7.51 (BC)
- Side b = 6.42 (AC)
- Angle C = 23°
Find side c = AB

Again, SAS → Law of Cosines:
c² = a² + b² - 2ab·cos(C)
c² = 7.51² + 6.42² - 2×7.51×6.42×cos(23°)
Calculate:
7.51² = 56.4001
6.42² = 41.2164
Sum = 97.6165
cos(23°) ≈ 0.9205
2×7.51×6.42×0.9205 ≈ 2×7.51×6.42×0.9205
First: 7.51×6.42 ≈ 48.2142
Then: 48.2142×0.9205 ≈ 44.38
Then ×2 = 88.76? Wait — no! The formula has only one 2×a×b×cos(C), so:

Actually:
2 × 7.51 × 6.42 × cos(23°)
= 2 × 7.51 × 6.42 × 0.9205
Compute step by step:
7.51 × 6.42 = 48.2142
48.2142 × 0.9205 ≈ 44.38
Then ×2? No — wait, the formula is:
c² = a² + b² - 2ab·cos(C) → so it’s minus 2ab·cos(C)

So:
2ab·cos(C) = 2 × 7.51 × 6.42 × 0.9205
= 2 × 44.38 ≈ 88.76? That can’t be right because a² + b² is only ~97.6, and subtracting 88 would give small c.

Wait — let me recalculate 7.51 × 6.42:

7.51 × 6.42:
7 × 6.42 = 44.94
0.51 × 6.42 ≈ 3.2742
Total ≈ 48.2142 → correct.

Now 48.2142 × 0.9205:
48.2142 × 0.9 = 43.39278
48.2142 × 0.0205 ≈ 0.988
Total ≈ 44.38 → yes.

Then 2 × 44.38? NO — the 2 is already included in the formula: it’s 2ab·cos(C), so we multiply ab by 2 and by cos(C).

So: 2 × 7.51 × 6.42 × 0.9205 = 2 × 48.2142 × 0.9205 = 96.4284 × 0.9205 ≈ ?

Better:
2 × 7.51 = 15.02
15.02 × 6.42 = 96.4284
96.4284 × 0.9205 ≈ let's compute:

96.4284 × 0.9 = 86.78556
96.4284 × 0.0205 ≈ 1.976
Total ≈ 88.76

So c² = 56.4001 + 41.2164 - 88.76 = 97.6165 - 88.76 = 8.8565
c = √8.8565 ≈ 2.98

✔ Answer: 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
QR = 6
RP = 4.4
Find angle PQR (angle at Q)

Use Law of Cosines:
For angle at Q, which is between sides PQ and QR, opposite side is PR = 4.4

So:
PR² = PQ² + QR² - 2·PQ·QR·cos(angle Q)
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)
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: 37.4°

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Problem 2)
Triangle DEF:
Area = 10
DE = 3.85
DF = 7.63
Find angle FDE (angle at D)

Formula for area with two sides and included angle:
Area = (1/2)·DE·DF·sin(angle D)
10 = (1/2) × 3.85 × 7.63 × sin(D)
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: 42.9°

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Problem 3)
Triangle LMN:
Angle L = 41°
Side LN = 51? Wait — labels:

From diagram:
- Side LM = 51 (from L to M)
- Side MN = 34 (from M to N)
- Angle at L = 41°
Find angle LMN (angle at M), given that angle LNM (angle at N) is obtuse.

First, use Law of Sines to find angle at N.

Law of Sines:
sin(N) / LM = sin(L) / MN
Wait — standard:
side opposite angle L is MN = 34
side opposite angle N is LM = 51
side opposite angle M is LN — not given.

So:
sin(N) / 51 = sin(41°) / 34
sin(N) = 51 × sin(41°) / 34
sin(41°) ≈ 0.6561
sin(N) = 51 × 0.6561 / 34 ≈ 33.4611 / 34 ≈ 0.98415
N = arcsin(0.98415) ≈ 80.0° or 180° - 80° = 100.0°

But the problem says angle LNM (angle at N) is obtuse → so N = 100.0°

Then angle M = 180° - 41° - 100° = 39.0°

✔ Answer: 39.0°

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Problem 4)
Y-Z-T is a straight line.
Triangle XYZ:
XY = 9.2
XZ = 4.3
YZ = 6.4
Find angle XZT (angle at Z, outside the triangle, since YZT is straight)

First, find angle XYZ (angle at Z inside triangle XYZ).

Use Law of Cosines in triangle XYZ to find angle at Z.

Sides:
XY = c = 9.2 (opposite angle Z)
XZ = b = 4.3 (opposite angle Y)
YZ = a = 6.4 (opposite angle X)

Angle at Z is between sides XZ and YZ, so:

XY² = XZ² + YZ² - 2·XZ·YZ·cos(angle Z)
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)
55.04·cos(Z) = 59.45 - 84.64 = -25.19
cos(Z) = -25.19 / 55.04 ≈ -0.4577
Z = arccos(-0.4577) ≈ 117.2°

This is the internal angle at Z in triangle XYZ.

Since YZT is a straight line, angle XZT is the supplement:
Angle XZT = 180° - 117.2° = 62.8°

✔ Answer: 62.8°

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Final Answers:

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

Section B:
1) 37.4°
2) 42.9°
3) 39.0°
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) 39.0°
4) 62.8°