SOLUTION: At what time between 2 and 3 o'clock are the hands of the clock opposite each other? Thank you for your help..=)

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Question 188391: At what time between 2 and 3 o'clock are the hands of the clock opposite each other?

Thank you for your help..=)

Found 2 solutions by J2R2R, josmiceli:
Answer by J2R2R(94) About Me  (Show Source):
You can put this solution on YOUR website!
Let us start at 12:00 when the two hands are together.

The hour hand will pass through A degrees.

The minute hand will pass through 720 degrees (2 hours) + (180 + A) degrees to be opposite the hour hand.

Since the minute hand travels 12 times as fast as the hour hand,

12A = 720 + 180 + A

11A = 900

A = 81 + 9/11

Which is 21 + 9/11 into the third hour since the first two hours take up 60 degrees.

The hour hand will have passed through 21 + 9/11 degrees since 2 o’clock, it is twice this amount in minutes past 2 o’clock since the hour hand takes 60 minutes to pass through 30 degrees (2 minutes per degree).

Therefore the time is 43 + 7/11 minutes past 2 o’clock.

We can verify this by calculating the proportions of the hour the hands have moved through in degrees with respect to an hour.

i.e. (21 + 9/11)/30 = (180 + 60 + 21 + 9/11)/360 = 8/11

180 is to be opposite, 60 is for the two hours and 21 + 9/11 is A.

Time is 2:43:38 + 2/11 of a second.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
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%2F60+=+1%2F12, so if the whole face is 360 degrees
%281%2F12%29%2A360+=+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
-------------------
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.
-------------------
Let t= the elapsed time in hours after 2 o'clock when the hands
are opposite eachother
360%2F1= the rate of travel of the minute hand in degrees/ hr
30%2F1 = 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%2F12+=+1%2F6 of the 360 degrees
%281%2F6%29%2A360+=+60 degrees
So, when the hands are opposite the hour hand is at
60+%2B+%2830%2F1%29%2At
I want the minute hand to be this position + 180 degrees.
I also know that 360%2F1{ is the minute's hand rate of travel)
so, I can write
60+%2B+%2830%2F1%29%2At+%2B+180+=+%28360%2F1%29%2At
330t+=+240
t+=+24%2F33
t+=+.7273hrs
t+=+.7273%2A60min
t+=+43.64min
and
.64%2A60+=+38sec
The hands are opposite eachother at 43 min and 38 sec past 2 o'clock
---------------------
check answer:
In degrees the hour hand has moved
.7273%2A%2830%2F1%29+=+21.82 degrees
In degrees, the minute hand has moved
.7273%2A%28360%2F1%29+=+261.82 degrees
261.82+-+21.82+=+240 degrees
If I subtract the 60 degree headstart the hour hand had,
240+-+60+=+180