SOLUTION: The 12-foot bed of a dump truck loaded with heavy stone must rise to an angle of 36° before the stone will spill out. Approximately how high must the front of the bed rise (x) to u
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Question 363241: The 12-foot bed of a dump truck loaded with heavy stone must rise to an angle of 36° before the stone will spill out. Approximately how high must the front of the bed rise (x) to unload?
sin 36° ≈ 0.588, cos 36° ≈ 0.810, tan 36° ≈ 0.727
I have no clue how to work with Sine and Cosine.
How do I work this out?
You can put this solution on YOUR website! The 12-foot bed of a dump truck loaded with heavy stone must rise to an angle of 36° before the stone will spill out. Approximately how high must the front of the bed rise (x) to unload?
sin 36° ≈ 0.588.
Let a° represent the angle made by the bed of the truck when it is raised.
By definition (adjacent side) times sin a° = opposite side
(12 feet) times sin 36°= x feet
x = 12 sin 36°
x = 12(0.588)
x = 7.056 feet
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