A lighthouse that rises h1 = 47.0 ft above the surface of the water sits on a rocky cliff that extends d = 17.4 ft from its base, as shown in the figure below.
A sailor on the deck of a ship sights the top of the lighthouse at an angle of theta = 29.3 degree above the horizontal. If the sailor's eye level is h2 = 15.1 ft above the water, how far is the ship from the rocks?
well if d is horizontal, subtract that 17.4 at the end
the light is 47 - 15.1 = 31.9 ft above our horizontal line of sight
horizontal distance to light = 31.9 tan 29.3
d from rocks = that - 17.4
When I did that My answer came to be .501 but it was marked wrong. Was that the answer I was supposed to get?
I am sure that is wrong. You are not half a foot from the rocks.
whoops typo
31.9 / tan 29.3 upside down
31.9/tan 29.3 = 56.84
56.84 - 17.4 = 39.4
Okay this time my answer came to be 39.4 m but it still says incorrect. Did I not do my calculations correctly?
Do we have the picture right? Is subtracting 17.4 correct ?
yes
Then I am at a loss.
Hmmm. I get
(47.0-15.1)/(x-17.4) = tan 29.3 = 0.56
x = 74.36
As my final answer do I use ft or m ?
I put the lighthouse on the cliff, Steve put it out in the water. Which is it in your picture?
You whole problem is in feet, so I assume you would answer in feet unless your class has a rule saying always reply in SCI units.
Sorry you were right my unit was wrong, thanks!
I put the lighthouse so its peak was 47 ft above the water (as stated), regardless of the cliff.
I put the lighthouse behind the rocks. You put it ahead of the rocks. Could be either. (Boston Light versus Deer Island light)
Tan29.3 = (47-15.1)/(17.4+d2)
(17.4+d2)*Tan29.3 = 47-15.1 = 31.9
(17.4+d2) = (47-15.1)/Tan29.3 = 56.9
d2 = 56.9-17.4 = 39.5 Ft.