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You are lying on the bottom of a swimming pool filled 15 ft deep with water (ftwater = 1.33). What is the diameter of the "hole” at the water surface through which you can see out of the pool?
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Answered by
bobpursley
sinTheta*4/3=1
sinTheta=.75
theta=arcsin.75=48.6 deg
radius=15*tan48.6=17 ft
diameter= 34 feet
check my work.
sinTheta=.75
theta=arcsin.75=48.6 deg
radius=15*tan48.6=17 ft
diameter= 34 feet
check my work.
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