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The Designers of a Water Park Are Creating a New θ\theta

Question 1

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The designers of a water park are creating a new slide and have sketched some preliminary drawings.The length of the ladder is a = 26 feet, and its angle of elevation is 60º (see figure) .
 The designers of a water park are creating a new slide and have sketched some preliminary drawings.The length of the ladder is a = 26 feet, and its angle of elevation is 60º (see figure) .     Find the angle of depression  \theta  from the top of the slide to the end of the slide at the ground in terms of the horizontal distance d the rider travels.   A)   \theta = \arctan \left( \frac { 22.52 } { d } \right)   B)   \theta = \arctan \left( \frac { 23.52 } { d } \right)   C)   \theta = \arctan \left( \frac { 24.52 } { d } \right)   D)   \theta = \arctan \left( \frac { 25.52 } { d } \right)   E)   \theta = \arctan \left( \frac { 26.52 } { d } \right)
Find the angle of depression θ\theta from the top of the slide to the end of the slide at the ground in terms of the horizontal distance d the rider travels.


A) θ=arctan(22.52d) \theta = \arctan \left( \frac { 22.52 } { d } \right)
B) θ=arctan(23.52d) \theta = \arctan \left( \frac { 23.52 } { d } \right)
C) θ=arctan(24.52d) \theta = \arctan \left( \frac { 24.52 } { d } \right)
D) θ=arctan(25.52d) \theta = \arctan \left( \frac { 25.52 } { d } \right)
E) θ=arctan(26.52d) \theta = \arctan \left( \frac { 26.52 } { d } \right)

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