Question #162500

A fountain of water is located at the center of a circular pool. Not wishing 

to get his feet wet, a student walks around the pool and measures its 

circumference to be 16.5 m. Next, the student stands at the edge of the pool 

and uses a protractor to gauge the angle of elevation at the bottom of the 

fountain to be 55.0°. How high is the fountain?


1
Expert's answer
2021-02-11T08:55:12-0500

Since the circumference is L=16.5mL = 16.5 m, the radius of the pool is:


r=L2πr = \dfrac{L}{2\pi}

From the right triangle in which one leg is radus, and another leg is the fountain, the height of the fountain is:


h=rtanθ=Ltanθ2πh = r\tan\theta = \dfrac{L\tan\theta}{2\pi}

where θ=55°\theta = 55\degree is the angle measured by the student. Thus, obtain:


h=16.5tan55°2π3.75mh =\dfrac{16.5\cdot \tan55\degree}{2\pi} \approx 3.75m

Answer. 3.75m.


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Chrishan
10.02.21, 09:55

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