Question

one skateboarding site recommends that the ideal kicker ramp for beginners have the side dimensions in the figure. to the nearest inch, how long would the base of the ramp be? this question was created by khan academy.

Explanation:

Step1: Identify the triangle type

The ramp's side is a right triangle with height \( h = 2 \) ft and angle \( \theta = 35^\circ \). We need to find the base \( b \).

Step2: Use trigonometric ratio

In a right triangle, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{b} \)? Wait, no: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \), here height is opposite, base is adjacent? Wait, no: the angle is at the base, so height is opposite, base is adjacent? Wait, no, let's correct: the angle given is \( 35^\circ \), the height is 2 ft (opposite side to the angle? Wait, no, the right angle is between the height and the base. So the triangle has angle \( 35^\circ \), height (opposite to the angle? No, wait: the angle at the base (the angle between the base and the hypotenuse) is \( 35^\circ \), so the height is the opposite side, and the base is the adjacent side? Wait, no: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \), where \( \theta = 35^\circ \), opposite side is height (2 ft), adjacent side is base \( b \)? Wait, no, that would be \( \tan(35^\circ) = \frac{2}{b} \), so \( b = \frac{2}{\tan(35^\circ)} \). Wait, no, maybe I mixed up. Wait, the height is 2 ft, the angle is \( 35^\circ \), so the angle between the base and the ramp (hypotenuse) is \( 35^\circ \), so the height is opposite to that angle, and the base is adjacent. So \( \tan(35^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{b} \), so \( b = \frac{2}{\tan(35^\circ)} \). Wait, no, that would be if the angle is at the base. Wait, maybe it's \( \tan(35^\circ) = \frac{\text{opposite}}{\text{adjacent}} \), where opposite is height, adjacent is base. So \( \tan(35^\circ) = \frac{2}{b} \), so \( b = \frac{2}{\tan(35^\circ)} \). Let's calculate that. First, convert 2 ft to inches: 2 ft = 24 inches. Wait, no, wait: the height is 2 ft, so we can work in feet and then convert to inches, or convert to inches first. Let's do feet first. \( \tan(35^\circ) \approx 0.7002 \). So \( b = \frac{2}{0.7002} \approx 2.856 \) feet. Now convert to inches: 2.856 feet 12 inches/foot ≈ 34.27 inches, which rounds to 34 inches. Wait, but maybe I had the ratio wrong. Wait, maybe it's \( \cot(35^\circ) = \frac{\text{adjacent}}{\text{opposite}} = \frac{b}{2} \), so \( b = 2 \cot(35^\circ) \). \( \cot(35^\circ) = \frac{1}{\tan(35^\circ)} \approx 1.4281 \), so \( b = 2 * 1.4281 \approx 2.856 \) feet, same as before. Then 2.856 feet 12 = 34.27 inches, so nearest inch is 34. Wait, but maybe the angle is with respect to the vertical? No, the diagram shows the angle at the base, 35 degrees, and height 2 ft. So the calculation is correct.

Wait, let's recheck: the ramp is a right triangle, height (vertical side) = 2 ft, angle at the base (between base and hypotenuse) = 35 degrees. So in the right triangle, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{height}}{\text{base}} \). So \( \tan(35^\circ) = \frac{2}{\text{base}} \), so \( \text{base} = \frac{2}{\tan(35^\circ)} \). Calculating \( \tan(35^\circ) \approx 0.7002 \), so \( \text{base} \approx \frac{2}{0.7002} \approx 2.856 \) feet. Convert to inches: 2.856 * 12 = 34.27 inches, so to the nearest inch, 34 inches.

Answer:

34 inches