Question

an 80-foot rope is attached to a boat 7 feet above the water level as shown and a boy jumps over the wake while wakeboarding. what is the distance between his hands and the water level? round to the nearest hundredth.

Explanation:

Step1: Find the angle θ

Since the angle between the rope and the vertical line of the boat is \(87^\circ\), the angle \(\theta\) between the rope and the horizontal line is \(90^\circ - 87^\circ=3^\circ\).

Step2: Calculate the vertical drop from the boat's rope attachment

We use the sine function for the right triangle formed by the rope (hypotenuse = 80 ft), the horizontal distance, and the vertical drop. The vertical drop \(d\) from the boat's rope attachment to the boy's hands is given by \(d = 80\times\sin(3^\circ)\).
First, calculate \(\sin(3^\circ)\approx0.0523\). Then \(d = 80\times0.0523 = 4.184\) ft.

Step3: Find the distance from hands to water

The boat is 7 ft above water, and the vertical drop from the boat's rope attachment to the boy's hands is \(4.184\) ft. So the distance from the boy's hands to the water is \(7 - 4.184 = 2.816\approx2.82\) ft? Wait, no, wait. Wait, actually, the angle: Wait, maybe I got the angle wrong. Wait, the angle between the rope and the vertical is \(87^\circ\), so the angle between the rope and the horizontal is \(3^\circ\), but the vertical component from the boat's rope point to the boy's hands: Wait, no, the boat is 7 ft above water. The rope is 80 ft long. The angle between the rope and the vertical (the 7 ft side) is \(87^\circ\), so the angle between the rope and the horizontal is \(3^\circ\). Wait, actually, the vertical distance from the rope's boat end to the boy's hands: Let's re - express. The triangle: the rope is the hypotenuse (80 ft), the angle between the rope and the vertical (the 7 ft line) is \(87^\circ\), so the vertical side from the boat's rope point to the boy's hands: Wait, no, the boat's height above water is 7 ft. The boy's hands are at a lower vertical level? Wait, no, the diagram: the boat is 7 ft above water, the rope is attached to the boat 7 ft above water, and the boy is on the water (wakeboarding). Wait, maybe the angle is with respect to the horizontal. Wait, the angle between the rope and the horizontal is \(\theta\), and the angle between the rope and the vertical is \(87^\circ\), so \(\theta=90 - 87 = 3^\circ\). Then the vertical distance from the rope's boat attachment to the boy's hands is \(80\times\sin(3^\circ)\), and the distance from the boy's hands to the water is \(7 - 80\times\sin(3^\circ)\)? Wait, no, maybe the other way. Wait, if the boat is 7 ft above water, and the rope is slanting down to the boy, then the vertical distance from the rope's boat end (7 ft above water) to the boy's hands is \(80\times\sin(3^\circ)\), so the boy's hands are \(7 - 80\times\sin(3^\circ)\) above water? Wait, no, that can't be, because if the angle is small, the vertical drop is small. Wait, let's calculate \(\sin(3^\circ)\): \(\sin(3^\circ)\approx0.052335956\). Then \(80\times\sin(3^\circ)\approx80\times0.052335956 = 4.18687648\) ft. Then the distance from the boy's hands to the water is \(7 - 4.18687648=2.81312352\approx2.81\) ft? Wait, but maybe I mixed up the angle. Wait, maybe the angle between the rope and the horizontal is \(87^\circ\)? Wait, the diagram shows a dashed horizontal line, and the rope makes an angle of \(87^\circ\) with the vertical (the 7 ft line). So the angle with the horizontal is \(3^\circ\). Alternatively, maybe the angle between the rope and the horizontal is \(87^\circ\), and the angle with the vertical is \(3^\circ\). Let's re - examine. If the angle between the rope and the horizontal is \(87^\circ\), then the vertical component is \(80\times\sin(87^\circ)\). \(\sin(87^\circ)\approx0.998629535\). Then \(80\times\sin(87^\circ)\approx80\times0.998629535 = 79.8903628\) ft. But that can't be, because the boat is only 7 ft above water. So my initial angle interpretation was wrong. Wait, the boat is 7 ft above water. The rope is attached to the boat 7 ft above water, and the boy is on the water. So the vertical distance from the rope's boat end to the water is 7 ft. The rope is 80 ft long. The angle between the rope and the vertical (the 7 ft line) is \(87^\circ\), so the angle between the rope and the horizontal is \(3^\circ\). The vertical distance from the rope's boat end to the boy's hands is \(80\times\cos(87^\circ)\)? Wait, no, cosine of the angle with the vertical. Let's use trigonometry correctly. Let's define the triangle: the rope is the hypotenuse (c = 80 ft). The vertical side (adjacent to the \(87^\circ\) angle) is \(a\), and the horizontal side is \(b\). So \(\cos(87^\circ)=\frac{a}{80}\), so \(a = 80\times\cos(87^\circ)\). \(\cos(87^\circ)\approx0.052335956\), so \(a = 80\times0.052335956 = 4.18687648\) ft. Then the distance from the boy's hands to the water is \(7 - 4.18687648 = 2.81312352\approx2.81\) ft? Wait, but that seems too small. Wait, maybe the angle is with respect to the horizontal, so the angle between the rope and the horizontal is \(87^\circ\), so the vertical component is \(80\times\sin(87^\circ)\), and the distance from the boy's hands to the water is \(80\times\sin(87^\circ)-7\)? Wait, that would be a large number, which doesn't make sense. Wait, the boat is 7 ft above water. The rope is attached to the boat 7 ft above water, and the boy is holding the rope. So the vertical distance from the boy's hands to the water is equal to the boat's height above water minus the vertical component of the rope from the boat's attachment point to the boy's hands. The vertical component of the rope from the boat's attachment point to the boy's hands is \(80\times\cos(87^\circ)\) (since the angle with the vertical is \(87^\circ\), so adjacent side is vertical). So \(80\times\cos(87^\circ)\approx4.19\) ft. Then \(7 - 4.19 = 2.81\) ft. Wait, but let's check with sine. If the angle with the horizontal is \(\theta = 3^\circ\), then the vertical component is \(80\times\sin(3^\circ)\approx4.19\) ft, same as above. So the distance from the boy's hands to the water is \(7 - 4.19 = 2.81\) ft (rounded to the nearest hundredth). Wait, but maybe I made a mistake in the angle. Wait, the problem says "the distance between his hands and the water level". The boat is 7 ft above water. The rope is 80 ft long. The angle between the rope and the vertical (the 7 ft line) is \(87^\circ\), so the vertical distance from the rope's boat end to the boy's hands is \(80\times\cos(87^\circ)\), and then the distance from hands to water is \(7 - 80\times\cos(87^\circ)\). Let's calculate \(\cos(87^\circ)\):

\(\cos(87^\circ)=\cos(90^\circ - 3^\circ)=\sin(3^\circ)\approx0.0523\)

So \(80\times\cos(87^\circ)=80\times0.0523 = 4.184\)

Then \(7 - 4.184 = 2.816\approx2.82\) (rounded to the nearest hundredth). Wait, maybe the correct approach is:

The angle between the rope and the vertical is \(87^\circ\), so the angle between the rope and the horizontal is \(3^\circ\). The vertical distance from the boy's hands to the boat's rope attachment is \(80\times\sin(3^\circ)\) (opposite side to the \(3^\circ\) angle). Then the distance from the boy's hands to the water is \(7 - 80\times\sin(3^\circ)\).

\(\sin(3^\circ)\approx0.052335956\)

\(80\times\sin(3^\circ)\approx4.18687648\)

\(7 - 4.18687648 = 2.81312352\approx2.81\) (rounded to the nearest hundredth). Wait, there is a slight difference due to more precise calculation of \(\sin(3^\circ)\). Let's use a calculator for more precision.

\(\sin(3^\circ)\):

Using a calculator, \(3^\circ\) in radians is \(3\times\frac{\pi}{180}=\frac{\pi}{60}\approx0.05235987756\)

\(\sin(0.05235987756)\approx0.05233595624\)

\(80\times0.05233595624 = 4.186876499\)

\(7 - 4.186876499 = 2.813123501\approx2.81\) when rounded to the nearest hundredth? Wait, 2.813123501 is closer to 2.81 or 2.82? The thousandth digit is 3, which is less than 5, so we round down to 2.81. But maybe I had the angle wrong. Wait, maybe the angle between the rope and the horizontal is \(87^\circ\), so the vertical component is \(80\times\sin(87^\circ)\), and the distance from the boy's hands to the water is \(80\times\sin(87^\circ)-7\). Let's calculate \(\sin(87^\circ)\):

\(\sin(87^\circ)=\sin(90^\circ - 3^\circ)=\cos(3^\circ)\approx0.998629535\)

\(80\times0.998629535 = 79.8903628\)

\(79.8903628 - 7 = 72.8903628\), which is impossible because the rope is only 80 ft long. So that can't be. So my initial angle interpretation was correct: the angle between the rope and the vertical is \(87^\circ\), so the angle with the horizontal is \(3^\circ\), and the vertical distance from the boy's hands to the water is \(7 - 80\times\sin(3^\circ)\approx2.81\) ft. Wait, but let's check the diagram again. The diagram shows a dashed horizontal line, the boat is 7 ft above water (vertical dashed line), the rope is 80 ft, and the angle between the rope and the vertical (7 ft line) is \(87^\circ\). So the vertical component from the boat's rope point to the boy's hands is \(80\times\cos(87^\circ)\) (adjacent to the \(87^\circ\) angle), so \(80\times\cos(87^\circ)\approx4.19\) ft, and then the distance from hands to water is \(7 - 4.19 = 2.81\) ft.

Answer:

\(2.81\) (or \(2.82\) depending on rounding precision, but with more precise calculation, it's approximately \(2.81\))