Question 1184609: 2. An artillery observer in an airplane agrees that when he gets above the targe which is invisible to the guns he will level off at 10,000 feet elevation. If at that instant his angle of elevation as seen from the guns is 20° 40° , find the distance from the guns to the target.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the plane and the target and the gun placement form a right triangle.
the plane is 10,000 feet directly above the target.
the guns are x feet away from the target in a horizontal direction.
the angle of elevation from the guns to the plan is 20 degrees 40 minutes.
40 minutes is equal to 40/60 = 2/3 of a degrees.
the angle of elevation is (10 + 2/3) degrees between the guns and the plane.
tan(20 + 2/3) = opposite divided by adjacent = 10,000 / x
solve for x to get:
x = 10,000 / tan(20 + 1/3) = 26510.86671 feet.
that's the distance from the guns to the target in a horizontal direction on the ground.
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