If a solid semicylindrical block (a blockletter D) is placed on its curved side on a horizontal table, the top (the flat side) will be parallel to the table.
The table is tilted by an angle, a, perpendicular to the axis of the cylinder. What angle will the top of the block form with the table now?
Assume angle a is small enough to prevent the block from slipping or tipping over.
(In reply to
Solution by Bractals)
I think in
BC sin(a) = CA sin(B)
the B needs to be lowercased, as I think you are referring to the law of sines applied to triangle ABC, where angle BAC = b. Below, I've made this change and added some intermediate steps in the combining of the equations, that make it easier for me to understand.
Then this reads:
BA sin(b) = BA sin(BAD) = BD = BC sin(BCD) = BC sin(a+b)
BC sin(a) = CA sin(b)
Combining these two we get
sin(a+b) = BD/BC
= BA sin(b)/(CA sin(b)/sin(a))
= BA sin(a)/CA
sin(a+b) = sin(a)/(CA/BA) = sin(a)/h
or
b = arcsin(sin(a)/h)  a

Posted by Charlie
on 20060117 10:07:25 