Thread: Math problem(s)

I know howmuch you all like math so I decided to ask your help on some of my shit.

It's about the hands of a clock. Like howmany degrees between the clock's hands at 5:30. This is easy though.

The question is this: The arms of the clock are symmetrical in the vertical line trough the middle. The time is close to 10:10 (10 past 10). What is the exact time? Exact means exact. Seconds or even miliseconds won't do.

What I already know:
The degrees from the "starting position" (= 12 o'clock) towards the small hand is calculated with this formula: S = 30t, with S = degrees of the small hand and t is the amount of hours that passed. So when 4 hours passed, the small hand is on the 4 which is 120 degrees from the starting position.

The degrees from the starting position to the big hand is calculated with this formula: B = 360t - 360*the hours that already passed. The big hand makes a full circle every hour, so when a full hour has passed you need to take 360 degrees off to get the right amount. For example: in the hour between 10 and 11 o'clock the formula will be B = 360t - 360*10 = 360t - 3600.

While I was writing all this down I already figured out how to solve this one, but I won't spoil it for anybody who actually likes this stuff. Edit... no I didn't figure it out. I thought I did. D:

This is part of a big project that I'm doing so when I run into any more bumps, I'll post it here to give you guys a try.

Have fun.

---------- Post added 20-06-2010 at 00:28 ---------- Previous post was 19-06-2010 at 20:26 ----------

Ok, I did solve it now. Next question I'm stuck on is very similar, but I'm not quite getting there.

The small hand is near the 8, and the big hand is so that it is symmetrical in the horizontal line trough the centre. That means it's little over 7:50. How late is it exactly? I could make all this a lot clearer with pictures but I cba to draw pictures and upload them. I will make those pictures if anybody who wants to solve this can't quite visualize what I'm saying though.