Law of Sines and Cosines Word Problems Worksheet #2 Answer Key ... - Free Printable

Explanation:
Let’s solve each problem one by one, carefully checking the work.

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Problem 1:
Juan and Romella are 10 miles apart on a straight shoreline. A ship is out in the water.
- Angle between coastline and line from Juan to ship = 35°
- Angle between coastline and line from Romella to ship = 45°

We need the distance from Juan to the ship.

Let’s draw triangle J–R–S (J = Juan, R = Romella, S = Ship).
- Side JR = 10 miles (coastline).
- At J: angle between JR and JS is 35° → so angle at J in triangle JRS is 35°.
- At R: angle between RJ and RS is 45° → but note: RJ points *toward* J, while the coastline direction is the same line — so the interior angle at R is also 45°.
- Therefore, angle at ship S = 180° − 35° − 45° = 100°.

Now use Law of Sines:
\[
\frac{JS}{\sin(\angle R)} = \frac{JR}{\sin(\angle S)}
\]
So:
\[
\frac{x}{\sin 45^\circ} = \frac{10}{\sin 100^\circ}
\Rightarrow x = 10 \cdot \frac{\sin 45^\circ}{\sin 100^\circ}
\]

Compute:
- sin 45° ≈ 0.7071
- sin 100° ≈ sin(80°) (since sin(100°) = sin(80°)) ≈ 0.9848

So:
\[
x \approx 10 \cdot \frac{0.7071}{0.9848} \approx 10 \cdot 0.718 \approx 7.18 \text{ miles}
\]

Rounded to one decimal: 7.2 miles ✔
The student’s answer is correct.

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Problem 2:
Jack and Jill are on opposite sides of a 200-ft-wide canyon. Both see the trail guide at an angle of depression of 60°.

Angle of depression = angle below horizontal. So from Jack’s eye, line of sight down to guide makes 60° with horizontal. Same for Jill.

Assume Jack, Jill, and guide form an isosceles triangle:
- Horizontal distance between Jack and Jill = 200 ft.
- The guide is directly below the midpoint? Not necessarily — but if both angles of depression are equal (60°), and they’re at same height (same side of canyon, assumed level ground), then the guide must be centered horizontally — i.e., 100 ft from each.

Then, consider right triangle:
- Horizontal leg = 100 ft
- Angle at Jack = 60° (between horizontal and line of sight)
- So the line of sight (hypotenuse) = adjacent / cos(60°) = 100 / cos(60°) = 100 / 0.5 = 200 ft

Wait — but the student wrote “length to trail guide is 200” and said it's equilateral. Let’s verify:

If horizontal = 100 ft, vertical drop = ?
tan(60°) = opposite / adjacent = height / 100 → height = 100·√3 ≈ 173.2 ft
Then line-of-sight distance = √(100² + (100√3)²) = √(10,000 + 30,000) = √40,000 = 200 ft ✔

So yes — each person is 200 ft from the guide.
The student’s answer is correct.

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Problem 3:
Tents: Tom (T), Dick (D), Harry (H)
- TD = 153 ft
- TH = 201 ft
- DH = 175 ft
Find angle at Dick, i.e., angle ∠TDH (between TD and DH).

Use Law of Cosines:
For triangle with sides a, b, c, and angle C between sides a and b:
\[
c^2 = a^2 + b^2 - 2ab\cos C
\]

Here, angle at D is between TD and DH, so sides around angle D are TD = 153 and DH = 175; opposite side is TH = 201.

So:
\[
TH^2 = TD^2 + DH^2 - 2(TD)(DH)\cos(\angle D)
\]
\[
201^2 = 153^2 + 175^2 - 2(153)(175)\cos D
\]

Compute:
- 201² = 40,401
- 153² = 23,409
- 175² = 30,625
Sum = 23,409 + 30,625 = 54,034

So:
\[
40,401 = 54,034 - 2(153)(175)\cos D
\]
\[
2(153)(175) = 2 × 26,775 = 53,550
\]
Then:
\[
40,401 = 54,034 - 53,550 \cos D
\]
\[
53,550 \cos D = 54,034 - 40,401 = 13,633
\]
\[
\cos D = \frac{13,633}{53,550} \approx 0.2546
\]
Now find angle:
\[
D = \cos^{-1}(0.2546) \approx 75.2^\circ
\]

The student started the calculation correctly but didn’t finish — they left it as an expression. So the correct answer is about 75° (to nearest degree).

But since the question asks “what is the angle”, we give the value.

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Problem 4:
Three boats: J1, J2, J3.
From J1:
- Distance to J2 = 2 mi
- Distance to J3 = 4 mi
- Angle between lines of sight to J2 and J3 = 45°
Find distance between J2 and J3.

This is Law of Cosines again. Triangle with sides:
- a = J1J2 = 2
- b = J1J3 = 4
- included angle C = 45°
- side c = J2J3 = ?

\[
c^2 = a^2 + b^2 - 2ab\cos C = 2^2 + 4^2 - 2(2)(4)\cos 45^\circ
\]
\[
= 4 + 16 - 16 \cdot \frac{\sqrt{2}}{2} = 20 - 8\sqrt{2}
\]
Since √2 ≈ 1.4142,
8√2 ≈ 11.3137
So:
\[
c^2 ≈ 20 - 11.3137 = 8.6863
\]
\[
c ≈ \sqrt{8.6863} ≈ 2.947 \approx 2.95 \text{ miles}
\]

Student got x² = 8.686 and x = 2.94 — that’s fine (rounded to 2 decimals). Correct.

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Now, the user only asked to “solve the problem accurately” — and there are 4 problems. But the instruction says:
> ONLY THEN provide the final answer.
And Final Answer should be only the final correct answer.

But there are 4 answers. Since the prompt doesn’t specify which one, and the image contains 4 problems, likely the expectation is to give all four final answers clearly.

However, the format says:
Final Answer: [Provide only the final correct answer]

To resolve: In such multi-part homework, often the “problem” refers to the whole sheet — but the instruction says “the problem", singular. Looking back: the user says “Solve the problem accurately.” and uploaded a sheet with 4 numbered problems.

Given the ambiguity, and that the assistant must follow instructions strictly, and the Final Answer section expects *one* answer — but that would be incorrect if only one is given.

Re-examining the rules:
> CRITICAL: First work through the problem step by step, check your calculations carefully, and ONLY THEN provide the final answer.

It says “the problem”, but the input has 4. However, in many contexts, when a worksheet is given, the expectation is to solve all.

Also, the sample student work includes answers for all 4, and the assistant is to verify/correct.

I think the safest is to list all four final answers, as separate items, since that’s what’s needed for correctness.

But the format says “Final Answer: [only the final correct answer]” — singular.

Let me check: In prior similar tasks, when multiple questions are given, the model lists all answers in the Final Answer, e.g., "1. 7.2 mi, 2. 200 ft, 3. 75.2°, 4. 2.95 mi"

That is acceptable if clearly formatted.

I will provide all four, clearly labeled, as the final answer.

Final verification:
1. ≈ 7.2 miles
2. 200 feet
3. ≈ 75.2° (or 75° if rounding to nearest degree)
4. ≈ 2.95 miles

Since problem 3 didn’t specify rounding, and student left unfinished, we give precise: 75.2°.

But let’s compute more precisely:
13,633 / 53,550 = 0.254584...
cos⁻¹(0.254584) = using calculator: ≈ 75.22° → so 75.2° is good.

Final Answer:
1. 7.2 miles
2. 200 feet
3. 75.2 degrees
4. 2.95 miles