Question 89162
A pilot was flying a small plane towards an oil spill at an altitude of 10,000 ft. He found that the near edge of the spill had an angle of depression on 58 degrees and the far edge of the spill had an angle of depression of 44 degrees. What is the width of the spill? (Find the horizontal distances away from the near edge and from the far edge. Then subtract these two amounts to get the distance across.) 
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Draw the figure with the plane at 10,000 ft above the ground.
Draw the oil spill as a circle 3 or 4 inches to the right of the plane
Draw a line from the plane to the nearest point on the circle; the angle
that line makes with the ground is 58 degrees.
Draw a line from the plane to the farthest point on the circle; the angle
that line makes with the ground is 44 degrees.
Draw a diameter of the circle connecting the two points on the ground.
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You now have two right triangles.
If you know some trig you know the base of the 
smaller right triangle = 10000/tan58 = 6248.69 ft.
and the base of the larger triangle is 10000/ = 10355.30 ft
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So the width of the spill is the d6ifference or about 4106.61 ft.
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Cheers,
stan H.