Question

1. performance task teo designed a skateboard ramp.

ramp a has a right triangle mpn with angle at m being 18°, height np = 6 ft. ramp b has a right triangle osr with hypotenuse or = 10 ft.
part a list the sides of ramp a in order from shortest to longest.
part b list the angles of ramp b from smallest to largest, and explain how you know.
part c ramp b cannot be steeper than 45°. is it possible to build ramp b so that sr is shorter than 6 ft? explain.

Explanation:

Part A

Step1: Find the third angle of ramp A

In right - triangle \(MNP\) (ramp A), one angle is \(18^{\circ}\) and one is \(90^{\circ}\). Using the angle - sum property of a triangle (\(A + B + C=180^{\circ}\)), the third angle \(\angle NMP=180-(90 + 18)=72^{\circ}\)

Step2: Apply the side - angle relationship

In a triangle, the side opposite the smaller angle is shorter. The sides of right - triangle \(MNP\) are: \(NP = 6\) (opposite \(18^{\circ}\)), \(MP\) (opposite \(72^{\circ}\)), and \(MN\) (opposite \(90^{\circ}\))
So, the order from shortest to longest is \(NP

Part B

Step1: Find the third angle of ramp B

In right - triangle \(QSR\) (ramp B), let the angles be \(\angle QSR = 90^{\circ}\), \(\angle R\) and \(\angle O\). Using the angle - sum property of a triangle (\(A + B + C = 180^{\circ}\)), if we assume the height \(QS\) is equal to \(NP = 6\) (since \(NP\) and \(QS\) are heights of the ramps in the parallel - top situation). Using \(\sin R=\frac{QS}{QR}\), \(\sin R=\frac{6}{10}=0.6\), so \(\angle R=\sin^{- 1}(0.6)\approx37^{\circ}\), and \(\angle O=180-(90 + 37)=53^{\circ}\)

Step2: Order the angles

The order of angles from smallest to largest is \(\angle R<\angle O<\angle QSR\)

Part C

Step1: Use the tangent function

In right - triangle \(QSR\), \(\tan R=\frac{QS}{SR}\). If \(QS = 6\) (assuming the height is fixed as in the parallel - top structure of the ramps). If \(SR<6\), then \(\tan R=\frac{6}{SR}>1\) (since if \(SR = 6\), \(\tan R = 1\) (when \(R = 45^{\circ}\)).
Since \(y = \tan x\) is an increasing function for \(01\), then \(R>45^{\circ}\)
So, it is not possible to build ramp B so that \(SR\) is shorter than \(6\) ft because if \(SR<6\) ft, then the angle \(R\) (the angle of the ramp) will be greater than \(45^{\circ}\)

Answer:

• Part A: \(NP
• Part B: \(\angle R<\angle O<\angle QSR\) (because in right - triangle \(QSR\), using angle - sum property and trigonometric relations as shown above)
• Part C: No. Because if \(SR<6\) ft, using \(\tan R=\frac{QS}{SR}\) (\(QS = 6\) assumed from parallel - top structure), \(\tan R>1\) which implies \(R>45^{\circ}\)