A person is sitting on a park bench and sees a swing 100 feet away and a slide 60 f...

1. Identify the problem type: This problem involves finding the length of a side of a triangle when two sides and the included angle are known. This is a classic application of the Law of Cosines, so the problem type is trigonometry.

1. Visualize the problem: Imagine the park bench as a vertex of a triangle. The swing and the slide are the other two vertices. The distance from the bench to the swing is one side of the triangle (let's call it 'a'), the distance from the bench to the slide is another side (let's call it 'b'), and the angle between these two sides at the bench is given (let's call it 'C'). We need to find the distance between the swing and the slide, which is the third side of the triangle (let's call it 'c').

3. Assign variables:
- Let $a$ be the distance from the bench to the swing = 100 feet.
- Let $b$ be the distance from the bench to the slide = 60 feet.
- Let $C$ be the angle between the swing and the slide at the bench = $30^\circ$.
- Let $c$ be the distance between the swing and the slide (what we need to find).

4. Apply the Law of Cosines: The Law of Cosines states that for any triangle with sides $a$, $b$, and $c$, and the angle $C$ opposite side $c$, the following relationship holds:
$$c^2 = a^2 + b^2 - 2ab \cos(C)$$

5. Substitute the known values into the formula:
$$c^2 = (100)^2 + (60)^2 - 2(100)(60) \cos(30^\circ)$$

6. Calculate the squares of the distances:
- $a^2 = 100^2 = 10000$
- $b^2 = 60^2 = 3600$

7. Calculate the product term:
- $2ab = 2(100)(60) = 12000$

8. Find the cosine of the angle:
- $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660$

9. Substitute these values back into the Law of Cosines equation:
$$c^2 = 10000 + 3600 - 12000 \times \frac{\sqrt{3}}{2}$$
$$c^2 = 13600 - 6000\sqrt{3}$$

10. Calculate the numerical value of $c^2$:
$$c^2 \approx 13600 - 6000 \times 0.8660$$
$$c^2 \approx 13600 - 5196$$
$$c^2 \approx 8404$$

11. Find the distance $c$ by taking the square root of $c^2$:
$$c = \sqrt{8404}$$
$$c \approx 91.67 \text{ feet}$$

Correction: Let's re-evaluate the calculation. It seems there might be a mistake in the calculation or the options provided. Let's recheck the steps.

Re-calculating step 10:
$$c^2 = 13600 - 12000 \times \frac{\sqrt{3}}{2}$$
$$c^2 = 13600 - 6000\sqrt{3}$$
Using a calculator for $6000\sqrt{3} \approx 10392.3$
$$c^2 \approx 13600 - 10392.3$$
$$c^2 \approx 3207.7$$
$$c = \sqrt{3207.7} \approx 56.63 \text{ feet}$$

1. Compare the result with the given options: The calculated distance is approximately 56.63 feet. This is closest to option B, 57 feet.