SOLUTION: Suppose an angle sweeps out an arc length of 2.34 inches on a circle with a radius of 2.5 inches. Assume the initial ray of the angle is in standard position (passing through the

Question 1002009: Suppose an angle sweeps out an arc length of 2.34 inches on a circle with a radius of 2.5 inches.
Assume the initial ray of the angle is in standard position (passing through the 3 o'clock position) and the terminal ray swept counterclockwise, what is the horizontal displacement to the right of the vertical diameter (in radius lengths) of the arc's ending point?
_____ radius lengths
Please explain
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
It helps to draw this
The arc is (2.34/2.5) radians or 0.936 radians.
This is 53.628 degrees.
The horizontal displacement is the sine of the complement (36.372 degrees) divided by the hypotenuse, which is the radius of 2.5. The triangle has hypotenuse 2.5, the leg leading to the 12 o'clock position is the vertical displacement, and the leg leading into the first quadrant is the horizontal displacement.
The horizontal displacement is 1*sin {(pi/2)-0.936}=1* sin 0.634 radians, or 0.593 radian horizontal displacement. This would be 1.48 inches.
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