Question 188391
The minute hand and the hour hand are each going in
a circle at a certain speed in degrees/hr
Between 2 and 3 o'clock, the hour hand covers 5 minutes
of the 60 minutes on the clock face
{{{5/60 = 1/12}}}, so if the whole face is 360 degrees
{{{(1/12)*360 = 30}}} degrees
When the hour hand moves 30 degrees, the minute and 
goes all the way around, or 360 degrees. That's 12 times
faster than the hour hand
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To solve the problem, I ask the question:
If the hour hand moves a certain number of degrees be
tween 2 and 3, what time is it when the minute hand  is
at the hour hand's position + 180 degrees? That would put
them directly opposite eachother.
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Let {{{t}}}= the elapsed time in hours after 2 o'clock when the hands
are opposite eachother
{{{360/1}}}= the rate of travel of the minute hand in degrees/ hr
{{{30/1}}} = the rate of travel of the hour hand in degrees/hr
--------------------
Reading from  12 and going clockwise, the hour hand is at the 2
which is {{{2/12 = 1/6}}} of the 360 degrees
{{{(1/6)*360 = 60}}} degrees
So, when the hands are opposite the hour hand is at
{{{60 + (30/1)*t}}}
I want the minute hand to be this position + 180 degrees.
I also know that {{{360/1}}}{ is the minute's hand rate of travel)
so, I can write 
{{{60 + (30/1)*t + 180 = (360/1)*t}}}
{{{330t = 240}}}
{{{t = 24/33}}}
{{{t = .7273}}}hrs
{{{t = .7273*60}}}min
{{{t = 43.64}}}min
and 
{{{.64*60 = 38}}}sec
The hands are opposite eachother at 43 min and 38 sec past 2 o'clock
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check answer:
In degrees the hour hand has moved 
{{{.7273*(30/1) = 21.82}}} degrees
In degrees, the minute hand has moved
{{{.7273*(360/1) = 261.82}}} degrees
{{{261.82 - 21.82 = 240}}} degrees
If I subtract the 60 degree headstart the hour hand had,
{{{240 - 60 = 180}}}