Question

solving a real-world problem
a playground slide is 14.5 feet long with an 8.5-foot tall ladder.
what is the measure of the angle that the slide makes with the ground, x, to the nearest degree?
62°
36°
54°

Explanation:

Step1: Identify the triangle type

The slide, ladder, and ground form a right triangle. The slide is the hypotenuse (\(c = 14.5\) ft), the ladder is the opposite side to angle \(x\) (\(a = 8.5\) ft). We use the sine function: \(\sin(x)=\frac{\text{opposite}}{\text{hypotenuse}}\).

Step2: Calculate \(\sin(x)\)

\(\sin(x)=\frac{8.5}{14.5}\approx0.5862\)

Step3: Find \(x\)

Take the inverse sine: \(x = \arcsin(0.5862)\approx36^\circ\) (to nearest degree). Wait, no, wait: Wait, the opposite side to angle \(x\) is the ladder? Wait, no, the ladder is vertical (height 8.5 ft), the slide is the hypotenuse (14.5 ft). So angle \(x\) is between slide and ground. So \(\sin(x)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{8.5}{14.5}\). Wait, but let's recalculate: \(\frac{8.5}{14.5}\approx0.5862\). \(\arcsin(0.5862)\approx36^\circ\)? Wait, no, \(\sin(36^\circ)\approx0.5878\), which is close. Wait, but wait, maybe I mixed up opposite and adjacent. Wait, the ladder is the opposite side to angle \(x\), so \(\sin(x)=\frac{8.5}{14.5}\). Let's compute \(\arcsin(8.5/14.5)\):

\(8.5\div14.5\approx0.5862\)

\(\arcsin(0.5862)\approx36^\circ\)? Wait, no, \(\sin(36^\circ)\approx0.5878\), which is very close. So the angle is approximately \(36^\circ\)? Wait, but wait, maybe I made a mistake. Wait, let's check \(\sin(36^\circ)\approx0.5878\), \(\sin(35^\circ)\approx0.5736\), so 0.5862 is between 35 and 36, closer to 36. Wait, but the options include 36°, 54°, 62°. Wait, maybe I mixed up sine and cosine. Wait, if the ladder is the adjacent side? No, the ladder is vertical, so it's opposite to angle \(x\) (the angle between slide and ground). Wait, no, the right angle is between ladder and ground. So the triangle has: right angle at the base of the ladder, ladder (8.5 ft) is one leg (opposite to angle \(x\)), ground is adjacent leg, slide (14.5 ft) is hypotenuse. So angle \(x\) is at the bottom of the slide, between slide and ground. So \(\sin(x)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{8.5}{14.5}\). So \(x=\arcsin(8.5/14.5)\approx36^\circ\). Wait, but let's check \(\cos(x)=\frac{\text{adjacent}}{\text{hypotenuse}}\), but we don't know adjacent. Alternatively, \(\cos(x)=\frac{\text{adjacent}}{14.5}\), but adjacent is \(\sqrt{14.5^2 - 8.5^2}=\sqrt{210.25 - 72.25}=\sqrt{138}\approx11.75\). Then \(\cos(x)=11.75/14.5\approx0.8097\), which is \(\cos(36^\circ)\approx0.8090\), which is very close. So that's correct. So angle \(x\) is approximately \(36^\circ\). Wait, but earlier when I calculated \(\sin(x)\), I got 0.5862, and \(\sin(36^\circ)\approx0.5878\), which is almost the same. So the correct answer is \(36^\circ\).

Wait, but let's re-express:

In right triangle, \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\)

Opposite side to \(\theta = x\) is the height of the ladder: 8.5 ft

Hypotenuse is the length of the slide: 14.5 ft

So \(\sin(x)=\frac{8.5}{14.5}\approx0.5862\)

Then \(x = \arcsin(0.5862)\approx36^\circ\) (since \(\sin(36^\circ)\approx0.5878\), which is very close, considering rounding).

Answer:

\(36^\circ\) (the option with \(36^\circ\))