SOLUTION: Admiral Al needs a ramp to perform a stunt at the Arlington High School pep rally. For the first 20 feet along the ground, the ramp must rise off of the ground at a constant rate.

Question 209396: Admiral Al needs a ramp to perform a stunt at the Arlington High School pep rally. For the first 20 feet along the ground, the ramp must rise off of the ground at a constant rate. After the construction crew had finished 5 feet of the horizontal distance, the track is 4 feet off the ground. After 20 feet of horizontal distance is constructed, Admiral Al needs to clear a 12 foot high barrier. Will he be able to do it?
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Admiral Al needs a ramp to perform a stunt at the Arlington High School pep rally. For the first 20 feet along the ground, the ramp must rise off of the ground at a constant rate. After the construction crew had finished 5 feet of the horizontal distance, the track is 4 feet off the ground. After 20 feet of horizontal distance is constructed, Admiral Al needs to clear a 12 foot high barrier. Will he be able to do it?
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Draw the picture of two overlapping right triangles.
The smaller triangle has a base of 5 and height of 4.
The larger has a base of 20 and a height of 12 ft.
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Find the angle of elevation of the smaller triangle.
tan^-1(4/5) = 38.66 degrees
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Using that angle solve for the height of the triangle when the base is 20.
tan(38.66) = height/20
height = 20*tan(38.66) = 20*0.8 = 16
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Al's going to make it as his height is 16 and the barrier is only 12 ft.
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Cheers,
Stan H.
Reply to stanbon@comcast.net
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
does not appear to be a problem.
the ratio of the vertical distance to the horizontal distance is 4/5
4/5 * 20 = 16 feet.
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he should clear the 12 foot barrier with 4 feet to spare.
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alternate way of proving this is tangent of angle ramp makes with the ground is 4/5 (opposite / adjacent)
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this makes the angle equal to 38.6599 degrees.
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if the horizontal distance is 20 feet, then the tan(38.699) degrees = o/a = o/20.
this means that o = 20 * tan(38.6599) = 16 feet.
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o means opposite side from the angle which is the vertical height.
a means adjacent side from the angle which is the horizontal distance.
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if you want to name the triangle, let A be where the construction starts, let B be 20 feet of horizontal distance from the starting point, let C be 16 feet of vertical distance directly above point B. the ramp is AC, the horizontal distance is AB, the vertical distance is BC.
the angle is angle BAC, otherwise known as angle A.
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the tangent of the angle will find you the vertical distance all along the way.
given 5 feet of horizontal distance from the starting point of A, the vertical distance is 5 * tan (38.6599) = 4 feet.
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