Question 973143
Two ships are out in a storm in the ocean.
 The captain of the ship that is farthest out sees a lighthouse on shore.
 The angle of elevation in looking to the top of the lighthouse is 4 degrees.
 The light house is 100 m in height.
 From a second ship that is closer to the lighthouse the angle of elevation looking up at the tower is 6 degrees.
 Use this information to determine how far apart the two ships 


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100
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_________a__________6\______________4\______
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You find the leg of each right triangle using the law of sines
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The small triangle, find the 3rd angle 180 - 90 - 6 = 94
{{{a/sin(94)}}} = {{{100/sin(6)}}}
{{{a/.99756}}} = {{{100/.10453}}}
cross multiply
.10453a = 100*.99756
a = {{{99.756/.10453}}}
a = 954.3 m from the lighthouse is the nearby ship
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The larger triangle, find the 3rd angle 180 - 90 - 4 = 96
{{{A/sin(96)}}} = {{{100/sin(4)}}}
{{{A/.99452}}} = {{{100/.069756}}}
cross multiply
.069756A = 100*.99452
A = {{{99.452/.069756}}}
A = 1430.1 m from the lighthouse is the further ship
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The distance between the ships: 1430.1 - 954.3 = 475.8 meters
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You better check my math here.