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OG-12 Pg 173 Problem no.147

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OG-12 Pg 173 Problem no.147

by theachiever » Mon Nov 26, 2012 10:59 pm
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In the figure above ,V represents an observation point at one end of a pool.From V,an object that is actually located on the bottom of the pool at the point R appears to be at point S.If VR=10 feet,what is the distance RS in feet between the actual position and the perceived position of the object?

A.10-5root3
B.10-5root2
C.2
D.2 1/2
E.4
[/img]
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by Bill@VeritasPrep » Mon Nov 26, 2012 11:43 pm
If VR = 10, and V to the bottom of the pool (call that point P for simplicity's sake) is 5 feet, we can set up a right triangle with VR as the hypotenuse. We can then solve for PR, the other leg of the triangle:

PR^2 + VP^2 = VR^2

PR^2 + 5^2 = 10^2

PR^2 + 25 = 100

PR^2 = 75 = 5sqrt3

The length of RS is the difference in the length of PS and the length of PR. We already know that PS is 10 feet, so PR must be 10 - 5sqrt3.
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