A lighthouse that rises 49 ft above the surface of the water sits on a rocky cliff that extends 19 ft from its base. A sailor on the deck of a ship sights the top of the lighthouse at an angle of 30 degrees above the horizontal. If the sailor's eye level is 14 ft above the water, how far is the ship from the rocks?
2007-09-14 06:06:13 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Given the short distances involved, assume the Earth is flat. Therefore, we simply solve the right triangle formed by the sailor's eye, top of the lighthouse, and cliff at 5 ft. below the base of the lighthouse. Let x = r cos(30), which is the distance from the cliff (rocks) you are looking for; y = r sin(30) = 35 ft, which is the 30 ft lighthouse plus a 5 ft. bit of cliff that puts the bottom point of the triangle even with the sailor's eye level at 14 ft.
Solve for r = y/sin(30) = 2*35 = 70 ft. Then solve for x = r cos(30) = 70*cos(30). x is your distance from the rocks. I think the sailor, and the boat she is on, are in deep, dark trouble.
2007-09-14 06:29:26 · answer #1 · answered by oldprof 7 · 0⤊ 1⤋
answer is 46 ft
2016-05-25 12:20:55 · answer #2 · answered by Anonymous · 0⤊ 2⤋