Law of Sines and Cosines - Riddle and Maze Worksheets | Law of ... - Free Printable

We are given a worksheet with 5 problems involving the Law of Cosines. Let’s solve each one step by step, carefully.

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Problem 1: Solve for the unknown in each triangle. Round to nearest hundredth.

#### A. Triangle with sides 22 m, 17 m, and angle 42° between them? Wait — check diagram description.
From the layout:
- Triangle A: side x (unknown), side 22 m, side 17 m, and angle 42° is *between* the 22 m and 17 m sides?
But looking at typical labeling: the angle shown (42°) is opposite side x? No — in triangle diagrams like this, the angle is usually placed at the vertex between the two labeled sides unless indicated otherwise.

Let me interpret each triangle based on standard conventions in such worksheets:

Triangle A: Two sides given: 22 m and 17 m, and the included angle is 42° → then x is the side *opposite* the 42° angle? Wait — no: if the angle is drawn between the two known sides, then it's the included angle, and x is the side opposite that angle.

Yes — in triangle A, the 42° angle is between the 22 m and 17 m sides, so x is the side opposite the 42° angle.

So use Law of Cosines:
\[
x^2 = 22^2 + 17^2 - 2(22)(17)\cos(42^\circ)
\]

Compute:
- \(22^2 = 484\)
- \(17^2 = 289\)
- Sum = 773
- \(2 \cdot 22 \cdot 17 = 748\)
- \(\cos(42^\circ) \approx 0.7431\) (using calculator in degree mode)

So:
\[
x^2 = 773 - 748 \cdot 0.7431 = 773 - 555.8388 = 217.1612
\]
\[
x = \sqrt{217.1612} \approx 14.74
\]

✔ So A: x ≈ 14.74 m

---

#### B. Triangle with sides 39 mm, 35 mm, 47 mm; find angle θ.
Angle θ is likely between sides 39 mm and 35 mm, opposite side 47 mm (since 47 is longest, and θ is at top, probably opposite 47). So use Law of Cosines to find angle opposite side 47:

\[
\cos\theta = \frac{39^2 + 35^2 - 47^2}{2 \cdot 39 \cdot 35}
\]

Compute:
- \(39^2 = 1521\)
- \(35^2 = 1225\)
- \(47^2 = 2209\)
- Numerator: \(1521 + 1225 - 2209 = 537\)
- Denominator: \(2 \cdot 39 \cdot 35 = 2730\)

So:
\[
\cos\theta = \frac{537}{2730} \approx 0.1967
\]
\[
\theta = \cos^{-1}(0.1967) \approx 78.65^\circ
\]

✔ B: θ ≈ 78.65°

---

#### C. Triangle with sides 13 cm, 9.4 cm, 7 cm; find angle θ.
Angle θ appears at the vertex between sides 9.4 cm and 7 cm, opposite side 13 cm (since 13 is longest). So again, use Law of Cosines for angle opposite 13 cm:

\[
\cos\theta = \frac{9.4^2 + 7^2 - 13^2}{2 \cdot 9.4 \cdot 7}
\]

Compute:
- \(9.4^2 = 88.36\)
- \(7^2 = 49\)
- \(13^2 = 169\)
- Numerator: \(88.36 + 49 - 169 = -31.64\)
- Denominator: \(2 \cdot 9.4 \cdot 7 = 131.6\)

\[
\cos\theta = \frac{-31.64}{131.6} \approx -0.2404
\]
\[
\theta = \cos^{-1}(-0.2404) \approx 103.91^\circ
\]

✔ C: θ ≈ 103.91°

---

#### D. Triangle with sides 23 m, 20 m, angle 47° between them? Unknown side x opposite 47°?
Again, angle 47° is likely between the 23 m and 20 m sides, and x is the side opposite that angle.

So:
\[
x^2 = 23^2 + 20^2 - 2(23)(20)\cos(47^\circ)
\]

Compute:
- \(23^2 = 529\)
- \(20^2 = 400\)
- Sum = 929
- \(2 \cdot 23 \cdot 20 = 920\)
- \(\cos(47^\circ) \approx 0.6820\)

So:
\[
x^2 = 929 - 920 \cdot 0.6820 = 929 - 627.44 = 301.56
\]
\[
x = \sqrt{301.56} \approx 17.37
\]

✔ D: x ≈ 17.37 m

---

#### E. Triangle with sides 55 cm, 50 cm, angle 61° between them? Unknown side x opposite 61°?
Angle 61° is between sides 55 cm and 50 cm, so x is opposite 61°.

\[
x^2 = 55^2 + 50^2 - 2(55)(50)\cos(61^\circ)
\]

Compute:
- \(55^2 = 3025\)
- \(50^2 = 2500\)
- Sum = 5525
- \(2 \cdot 55 \cdot 50 = 5500\)
- \(\cos(61^\circ) \approx 0.4848\)

So:
\[
x^2 = 5525 - 5500 \cdot 0.4848 = 5525 - 2666.4 = 2858.6
\]
\[
x = \sqrt{2858.6} \approx 53.47
\]

✔ E: x ≈ 53.47 cm

---

#### F. Triangle with sides 4.9 m, 8.3 m, 9.1 m; find angle θ.
Angle θ is likely between sides 4.9 m and 8.3 m, opposite side 9.1 m (longest side), so use Law of Cosines:

\[
\cos\theta = \frac{4.9^2 + 8.3^2 - 9.1^2}{2 \cdot 4.9 \cdot 8.3}
\]

Compute:
- \(4.9^2 = 24.01\)
- \(8.3^2 = 68.89\)
- \(9.1^2 = 82.81\)
- Numerator: \(24.01 + 68.89 - 82.81 = 10.09\)
- Denominator: \(2 \cdot 4.9 \cdot 8.3 = 81.34\)

\[
\cos\theta = \frac{10.09}{81.34} \approx 0.1241
\]
\[
\theta = \cos^{-1}(0.1241) \approx 82.87^\circ
\]

✔ F: θ ≈ 82.87°

---

Problem 2: Solve for all missing sides and angles. Use proper variables.

#### A. ΔXYZ: x = 29 m, y = 15 m, ∠Z = 122°

Standard notation: side x = YZ (opposite X), side y = XZ (opposite Y), side z = XY (opposite Z). Angle Z is between sides x and y.

So we can find side z using Law of Cosines:
\[
z^2 = x^2 + y^2 - 2xy\cos Z = 29^2 + 15^2 - 2(29)(15)\cos(122^\circ)
\]

Compute:
- \(29^2 = 841\), \(15^2 = 225\), sum = 1066
- \(2 \cdot 29 \cdot 15 = 870\)
- \(\cos(122^\circ) = \cos(180-58) = -\cos(58^\circ) \approx -0.5299\)

So:
\[
z^2 = 1066 - 870 \cdot (-0.5299) = 1066 + 461.013 = 1527.013
\]
\[
z = \sqrt{1527.013} \approx 39.08 \text{ m}
\]

Now find angle X (opposite side x = 29):
Use Law of Sines:
\[
\frac{\sin X}{x} = \frac{\sin Z}{z} \Rightarrow \sin X = \frac{x \sin Z}{z}
\]
\[
\sin Z = \sin(122^\circ) \approx \sin(58^\circ) \approx 0.8480
\]
\[
\sin X = \frac{29 \cdot 0.8480}{39.08} = \frac{24.592}{39.08} \approx 0.6293
\]
\[
X = \sin^{-1}(0.6293) \approx 39.0^\circ
\]

Then angle Y = 180° − 122° − 39.0° = 19.0°

Check with Law of Sines for Y:
\[
\sin Y = \frac{y \sin Z}{z} = \frac{15 \cdot 0.8480}{39.08} = \frac{12.72}{39.08} \approx 0.3255 \Rightarrow Y \approx 19.0^\circ
\]

✔ So:
- z ≈ 39.08 m
- ∠X ≈ 39.00°
- ∠Y ≈ 19.00°

---

#### B. ΔGHI: g = 13 cm, h = 8 cm, i = 15 cm

Standard: side g = HI (opp G), h = GI (opp H), i = GH (opp I)

All sides known → find all angles.

Find ∠G (opposite g = 13):
\[
\cos G = \frac{h^2 + i^2 - g^2}{2hi} = \frac{8^2 + 15^2 - 13^2}{2 \cdot 8 \cdot 15}
= \frac{64 + 225 - 169}{240} = \frac{120}{240} = 0.5
\Rightarrow G = \cos^{-1}(0.5) = 60.00^\circ
\]

∠H (opposite h = 8):
\[
\cos H = \frac{g^2 + i^2 - h^2}{2gi} = \frac{169 + 225 - 64}{2 \cdot 13 \cdot 15}
= \frac{330}{390} = 0.84615
\Rightarrow H = \cos^{-1}(0.84615) \approx 32.20^\circ
\]

∠I = 180 − 60 − 32.20 = 87.80°

Check with Law of Cosines for I:
\[
\cos I = \frac{g^2 + h^2 - i^2}{2gh} = \frac{169 + 64 - 225}{2 \cdot 13 \cdot 8} = \frac{8}{208} = 0.03846
\Rightarrow I = \cos^{-1}(0.03846) \approx 87.80^\circ
\]

✔ So:
- ∠G = 60.00°
- ∠H ≈ 32.20°
- ∠I ≈ 87.80°

---

#### C. ΔMNO: n = 31 m, o = 28 m, ∠M = 62°

Standard: side n = MO (opp N), o = MN (opp O), m = NO (opp M)

Given: ∠M = 62°, sides adjacent to M are n and o? Wait — side n is opposite N, so side n = MO; side o = MN; so yes, sides MN and MO meet at M → so sides o and n include angle M.

So we can find side m (opposite ∠M) using Law of Cosines:
\[
m^2 = n^2 + o^2 - 2no\cos M = 31^2 + 28^2 - 2(31)(28)\cos(62^\circ)
\]

Compute:
- \(31^2 = 961\), \(28^2 = 784\), sum = 1745
- \(2 \cdot 31 \cdot 28 = 1736\)
- \(\cos(62^\circ) \approx 0.4695\)

So:
\[
m^2 = 1745 - 1736 \cdot 0.4695 = 1745 - 815.052 = 929.948
\]
\[
m = \sqrt{929.948} \approx 30.50 \text{ m}
\]

Now find ∠N (opposite side n = 31):
Use Law of Sines:
\[
\frac{\sin N}{n} = \frac{\sin M}{m} \Rightarrow \sin N = \frac{n \sin M}{m}
\]
\[
\sin M = \sin(62^\circ) \approx 0.8829
\]
\[
\sin N = \frac{31 \cdot 0.8829}{30.50} = \frac{27.3699}{30.50} \approx 0.8974
\]
\[
N = \sin^{-1}(0.8974) \approx 63.83^\circ
\]

Then ∠O = 180 − 62 − 63.83 = 54.17°

Check with Law of Sines for O:
\[
\sin O = \frac{o \sin M}{m} = \frac{28 \cdot 0.8829}{30.50} = \frac{24.7212}{30.50} \approx 0.8105
\Rightarrow O = \sin^{-1}(0.8105) \approx 54.17^\circ
\]

✔ So:
- m ≈ 30.50 m
- ∠N ≈ 63.83°
- ∠O ≈ 54.17°

---

Problem 3: Triangle sides 4 m, 11 m, 8 m. Find its angles (nearest tenth).

Let sides: a = 4, b = 8, c = 11 (largest side = 11 ⇒ largest angle opposite it)

Find angle C (opposite side c = 11):
\[
\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{16 + 64 - 121}{2 \cdot 4 \cdot 8} = \frac{-41}{64} = -0.640625
\]
\[
C = \cos^{-1}(-0.640625) \approx 130.0^\circ
\]

Angle A (opposite a = 4):
\[
\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{64 + 121 - 16}{2 \cdot 8 \cdot 11} = \frac{169}{176} \approx 0.9602
\]
\[
A = \cos^{-1}(0.9602) \approx 16.2^\circ
\]

Angle B = 180 − 130.0 − 16.2 = 33.8°

Check with Law of Cosines for B:
\[
\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{16 + 121 - 64}{2 \cdot 4 \cdot 11} = \frac{73}{88} \approx 0.8307
\Rightarrow B = \cos^{-1}(0.8307) \approx 33.8^\circ
\]

✔ Angles:
- ≈ 16.2°, 33.8°, 130.0°

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Problem 4: Ship problem

- Leaves port at 1 pm, goes north at 30 mph.
- At 3 pm (2 hours later), changes course to bearing N 20° E.
- We want distance from port at 4 pm (i.e., 1 hour after turning).

So:
- From 1–3 pm: 2 hrs × 30 mph = 60 miles north.
- From 3–4 pm: 1 hr × 30 mph = 30 miles on bearing N 20° E.

Bearing N 20° E means 20° east of north → direction vector:
- North component: 30 cos(20°)
- East component: 30 sin(20°)

Compute:
- cos(20°) ≈ 0.9397 → north component = 30 × 0.9397 ≈ 28.191 mi
- sin(20°) ≈ 0.3420 → east component = 30 × 0.3420 ≈ 10.26 mi

Total north displacement = 60 + 28.191 = 88.191 mi
East displacement = 10.26 mi

Distance from port = √(88.191² + 10.26²)

Compute:
- 88.191² ≈ 7777.66
- 10.26² ≈ 105.27
- Sum ≈ 7882.93
- √7882.93 ≈ 88.79 → round to nearest unit: 89 miles

✔ Answer: 89 miles

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Problem 5: Area of triangle with sides 12 cm, 5 cm, 13 cm

Check if right triangle: 5² + 12² = 25 + 144 = 169 = 13² → yes! Right triangle with legs 5 and 12.

Area = (1/2) × base × height = 0.5 × 5 × 12 = 30 cm²

✔ Area = 30 cm²

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Now compile final answers per problem:

1.
A. 14.74
B. 78.65°
C. 103.91°
D. 17.37
E. 53.47
F. 82.87°

2.
A: z = 39.08 m, ∠X = 39.00°, ∠Y = 19.00°
B: ∠G = 60.00°, ∠H = 32.20°, ∠I = 87.80°
C: m = 30.50 m, ∠N = 63.83°, ∠O = 54.17°

3. Angles: 16.2°, 33.8°, 130.0°

4. 89 miles

5. 30 cm²

But the user asked to provide only the final answer — and since there are multiple parts, likely they expect each answer listed clearly.

However, instruction says:
> Final Answer: [Provide only the final correct answer — this should come AFTER all explanations and reasoning.]

Since there are multiple questions, and no specific one singled out, but the format suggests we give all final answers in a clean list.

Let me format the final answers as requested — only the answers, no explanation.

Final Answer:
1. A. 14.74 m; B. 78.65°; C. 103.91°; D. 17.37 m; E. 53.47 cm; F. 82.87°
2. A. z = 39.08 m, ∠X = 39.00°, ∠Y = 19.00°; B. ∠G = 60.00°, ∠H = 32.20°, ∠I = 87.80°; C. m = 30.50 m, ∠N = 63.83°, ∠O = 54.17°
3. 16.2°, 33.8°, 130.0°
4. 89 miles
5. 30 cm²