Finding the distance from ceiling to each point in the clock and finding an equation with cosine or sine

Question

2) Clock #2: A clock has a radius of 6 inches. The center is 14.5 inches below the ceiling. Measure the distance from the ceiling to the tip of the hour hand at each hour beginning at 12 (which is t=0). For example, at t=0, d=8.5 inches. Complete the table and round your answers to the nearest hundreth if needed.

Answer 1

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.