Gregg O. answered • 06/23/16

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circular trig is your friend here. We'll think about the problem using the point of attachment of the clock hands as the origin.

Facing the clock directly, the vertical displacement of the hour hand from the origin is given as r*sin(x), where x is the angle the hour hand makes from the x-axis, and r is the length of the hour hand (The problem has a problem, in assuming the radius of the clock is equal to this length).

The ceiling is located directly above the clock; imagine a drawing of a circle, with a horizontal line placed somewhere above it. That line represents the ceiling. The center of the circle is 14.5 inches below this line. What is the distance from the ceiling to an arbitrary point on the circle? Such a distance only requires y-components to compute. In fact, it is 14.5 - y. y is given by r*sin(x), from circular trig.

Since x is our angle with respect to the x-axis, 3 o'clock is x=0. 9 o'clock is x=pi. What is 2 o'clock? 1 o'clock? Figure out the angular spacing between clock points, and how the numbers on the face of the clock relate to the angle x.

Then you'll have solved the problem.

Gregg O.

Hi again Hannah. Let's imagine the hour positions as regular marks along the perimeter of the circle. How many of these marks are there? 12. This divides the circle into 12 regular partitions. Imagine an angle with vertex at the center of the clock, and edges intersecting the circle at consecutive numbers (11 and 12, 12 and 1, etc.). What is the measure of this angle? 360/12 = 30 degrees. Starting at 3 o'clock, each hour mark, when rotating counterclockwise is an increment of 30 degrees. So the height of the 2 o'clock mark is sin(30) above the origin; the height of the 1 o'clock mark is sin (2*30) units above the origin. Follow this pattern to determine heights relative to the origin. From there, you can find the distance to the horizontal line above the origin.

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06/23/16

Hannah L.

06/23/16