Law of Sines and Cosines; Area of Triangles Mixed Practice ... - Free Printable

Let’s solve each problem step by step. We’ll use the Law of Sines and Law of Cosines where appropriate, and for area problems, we’ll use the formula:
Area = (1/2)ab·sin(C)

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Section A: Find missing lengths to 2 decimal places

#### Problem 1)
Triangle ABC:
- Angle B = 83°
- Side c = AB = 5 (opposite angle C)
- Side a = BC = 4 (opposite angle A)
- Need to find side b = AC (opposite angle B)

We can use the Law of Cosines since we have two sides and the included angle? Wait — actually, angle B is between sides a and c? Let’s check:

In triangle ABC:
- Side a is opposite angle A → so side a = BC
- Side b is opposite angle B → so side b = AC
- Side c is opposite angle C → so side c = AB

Angle B is at vertex B, so it is between sides AB and BC → which are sides c and a.

So yes! We have two sides (a=4, c=5) and the included angle B=83°. So we can use Law of Cosines to find side b:

> b² = a² + c² - 2ac·cos(B)

Plug in:

b² = 4² + 5² - 2×4×5×cos(83°)
b² = 16 + 25 - 40×cos(83°)

Calculate cos(83°): ≈ 0.1219 (use calculator)

b² = 41 - 40×0.1219 = 41 - 4.876 = 36.124

b = √36.124 ≈ 6.01

✔ Final Answer for 1): 6.01

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#### Problem 2)
Triangle ABC:
- Angle A = 62°
- Angle B = 41°
- Side b = AC = 4 (opposite angle B)
- 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)

So:

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

✔ Final Answer for 2): 5.38

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

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 ≈ 8.18 × 0.6119 ≈ 5.00

✔ Final Answer for 3): 5.00

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#### Problem 4)
Triangle ABC:
- Angle C = 23°
- Side a = BC = 7.51 (opposite angle A)
- Side b = AC = 6.42 (opposite angle B)
- Need to find side c = AB (opposite angle C)

We have two sides and the included angle? Wait — angle C is at vertex C, so it is between sides AC and BC → which are sides b and a.

Yes! So again, 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 = 2×48.2142 = 96.4284
Multiply by cos(23°): 96.4284 × 0.9205 ≈ 88.76

c² = 97.6165 - 88.76 ≈ 8.8565

c = √8.8565 ≈ 2.98

✔ Final Answer for 4): 2.98

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Section B: Find missing angles to 3 significant figures

#### Problem 1) Find angle PQR

Triangle PQR:
- PQ = 2.3
- PR = 4.4
- QR = 6
- Need angle at Q → angle PQR

This is the angle between sides PQ and QR? Wait — let’s label properly.

Vertices: P, Q, R

Sides:
- Opposite P: QR = 6
- Opposite Q: PR = 4.4
- Opposite R: PQ = 2.3

We want angle at Q → that’s angle PQR.

We know all three sides → use Law of Cosines.

Formula for angle at Q:

> cos(Q) = (PQ² + QR² - PR²) / (2 × PQ × QR)

Wait — let’s be precise.

Standard Law of Cosines for angle at Q:

> cos(Q) = (side adjacent to Q)^2 + (other adjacent)^2 - (opposite)^2 / (2 × product of adjacents)

At vertex Q, the two sides forming the angle are QP and QR.

QP = 2.3, QR = 6, and the side opposite angle Q is PR = 4.4

So:

cos(Q) = (QP² + QR² - PR²) / (2 × QP × QR)
= (2.3² + 6² - 4.4²) / (2 × 2.3 × 6)

Calculate:

2.3² = 5.29
6² = 36
4.4² = 19.36

Numerator: 5.29 + 36 - 19.36 = 21.93

Denominator: 2 × 2.3 × 6 = 27.6

cos(Q) = 21.93 / 27.6 ≈ 0.7946

Q = arccos(0.7946) ≈ 37.4°

✔ Final Answer for 1): 37.4°

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#### Problem 2) Area of triangle DEF is 10. Find angle FDE

Triangle DEF:
- DE = 3.85
- DF = 7.63
- Area = 10
- Need angle at D → angle FDE

Angle at D is between sides DE and DF.

Area formula:
> Area = (1/2) × DE × DF × sin(angle D)

So:

10 = (1/2) × 3.85 × 7.63 × sin(D)

Calculate right side:

First, 3.85 × 7.63 ≈ 29.3755
Then × 1/2 = 14.68775

So:

10 = 14.68775 × sin(D)

sin(D) = 10 / 14.68775 ≈ 0.6808

D = arcsin(0.6808) ≈ 42.9°

But wait — sine is positive in both first and second quadrant. Could angle be obtuse?

The problem doesn’t specify, but since area is given and no other info, we assume acute unless told otherwise. Also, 42.9° makes sense with the side lengths.

Check: if angle were 180 - 42.9 = 137.1°, sin is same, but would that make sense? The sides are 3.85 and 7.63 — if angle is large, area might still be 10, but typically we take the acute one unless specified.

But let’s see — the problem says “find angle FDE” — no restriction. However, in context of triangle, and since 42.9° gives reasonable shape, and 137.1° might make the third side very long — but mathematically both are possible.

Wait — the problem says “to 3 significant figures”, and doesn’t specify acute or obtuse. But in most cases like this, they expect the acute angle unless hint says otherwise.

Actually, let’s compute both:

arcsin(0.6808) ≈ 42.9° or 137.1°

Which one fits? Let’s think — if angle D is 137.1°, then the other two angles must sum to 42.9°, which is possible, but let’s see the side opposite — EF would be longer.

But since the problem doesn’t give more constraints, and area formula gives same value for both, we should consider if there’s a convention.

Looking back at the worksheet — in Section B Problem 3, they mention “if angle LNM is obtuse”, implying that sometimes you need to consider obtuse. Here, no such hint.

But let’s calculate the third side for both cases to see which is more reasonable.

Alternatively — perhaps the diagram implies acute? Since we don’t have diagram, and problem is from mixed practice, likely they expect the acute angle.

Moreover, 42.9° to 3 sig fig is fine.

But let me double-check calculation:

3.85 × 7.63 = let's compute exactly:

3.85 × 7.63:

3.85 × 7 = 26.95
3.85 × 0.63 = 2.4255
Total = 29.3755 → correct

Half is 14.68775

10 / 14.68775 = 0.6808 → correct

arcsin(0.6808) = ? Using calculator: sin⁻¹(0.6808) ≈ 42.9° (yes)

To 3 sig fig: 42.9°

✔ Final Answer for 2): 42.9°

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#### Problem 3) Find angle LMN if angle LNM is obtuse

Triangle LMN:
- Angle at L = 41°
- Side LM = 51 (opposite angle N)
- Side MN = 34 (opposite angle L)
- Need angle at M → angle LMN
- Given: angle at N (LNM) is obtuse

First, note: side opposite angle L is MN = 34
Side opposite angle N is LM = 51
Side opposite angle M is LN — unknown

We know angle L = 41°, and sides adjacent? Actually, we know two sides and a non-included angle — this is ambiguous case.

Specifically, we know:
- Angle L = 41°
- Side opposite angle L: MN = 34
- Side opposite angle N: LM = 51

So, use Law of Sines to find angle N.

> sin(N) / LM = sin(L) / MN

sin(N) / 51 = sin(41°) / 34

sin(41°) ≈ 0.6561

So:

sin(N) = 51 × 0.6561 / 34 ≈ 51 × 0.019297 ≈ wait, better:

51 / 34 = 1.5

So sin(N) = 1.5 × sin(41°) ≈ 1.5 × 0.6561 = 0.98415

Now, sin(N) = 0.98415

Possible angles: N = arcsin(0.98415) ≈ 80.0° or 180° - 80.0° = 100.0°

Given that angle LNM (which is angle at N) is obtuse → so N = 100.0°

Now, angle at M = 180° - angle L - angle N = 180 - 41 - 100 = 39.0°

✔ Final Answer for 3): 39.0°

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#### Problem 4) YZT is straight line. Find angle XZT

Points: X, Y, Z, T

YZT is straight line → so points Y, Z, T are colinear, with Z between Y and T? Or T-Z-Y? The diagram shows T--Z--Y, I assume.

From description:
- Triangle XYZ? With points X, Y, Z
- Then T is on extension of YZ beyond Z? Since YZT straight, and angle XZT is asked.

Given:
- XY = 9.2
- XZ = 4.3
- YZ = 6.4
- YZT straight → so angle at Z in triangle XYZ is angle XZY, and angle XZT is the adjacent angle on the straight line.

So, first find angle at Z in triangle XYZ → angle XZY.

Then, since YZT is straight, angle XZT = 180° - angle XZY.

So, in triangle XYZ:
- Sides: XY = 9.2 (opposite angle Z)
- XZ = 4.3 (opposite angle Y)
- YZ = 6.4 (opposite angle X)

We want angle at Z → angle XZY.

Use Law of Cosines:

> cos(Z) = (XZ² + YZ² - XY²) / (2 × XZ × YZ)

Plug in:

XZ = 4.3, YZ = 6.4, XY = 9.2

cos(Z) = (4.3² + 6.4² - 9.2²) / (2 × 4.3 × 6.4)

Calculate:

4.3² = 18.49
6.4² = 40.96
9.2² = 84.64

Numerator: 18.49 + 40.96 - 84.64 = 59.45 - 84.64 = -25.19

Denominator: 2 × 4.3 × 6.4 = 2 × 27.52 = 55.04

cos(Z) = -25.19 / 55.04 ≈ -0.4577

Z = arccos(-0.4577) ≈ 117.2° (since cosine negative, obtuse)

Then angle XZT = 180° - 117.2° = 62.8°

✔ Final Answer for 4): 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°