The captain drops the 3- foot- long anchor and lets out 100 feet of anchor rope. The boat floats away from where the anchor was dropped until the anchor rope becomes straight due to the tide. The captain uses the yardstick and determines that a total 27 inches of the anchor rope is above the water. Using the inclinometer, the captain determines that the angle of depression of the anchor rope is 42 degrees. The captain knows the water is parallel to the bay floor. ( Elevation and Depression)
How do I determine the distance, to the nearest inch, from point p, the place where the anchor rope leaves the boat, to the water line?
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Using Trig with the right triangle that is formed by the rope and the water we have a hypotenuse of 27 inches and the angle where the rope leaves the boat of 48 degrees(This is 90 - the depression angle they gave us)
The Cos of 48 degrees = Adjacent/hypotenuse = (distance from p to the water line)/27
So distance from p to the water line= 27(Cos 48) = 18.067 inches or 18 inches to the nearest inch.
It's hard for me to answer this right now, but I would suggest starting with a picture. "draw" a boat on the water with a rope taught to the bay floor, and label everything you know. I would then represent what you want to find out with the variable x, or something else you like better, and then go from there. If I were able to draw you a picture I would, but that's kind of hard to do this way.