Quote:
Originally Posted by Honus
Sorry to come to this discussion so late, but I am confused.
Do we agree that the distance from the observer to the boat is the hypotenuse?
And that the calculation goes as follows:
distance = 300/sin(15.9) = 1095' ?
Is that correct?
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not entirely.
you need to estimate without a calculator, using the trig tables.
distance is about 1088 @ 16*