angles problem help please?

Question

The teacher leaves the school a little after six p.m. and notices the hands on her watch form a one-hundred ten degree angle. Returning before seven p.m. she again notices that the hands form a one-hundred ten degree angle. How many minutes have elapsed while the teacher was away?

i tried this problem but i dont know how to do it, esp when i remebered that the hour hand moves too! plz show work.

Answer 1

At 6:00 the hands are 180˚ apart. Let t be the number of minutes past 6:00. The minute hand moves 360˚ per hour, or 360/60 = 6˚/min. Likewise, the hour hand moves 360/12 degrees per hour, or 0.5˚/min.

Now, as time progresses, the movement of the hour hand increases the angle, and the movement of the minute hand decreases it. So, the current angle is represented by the expression 180 + 0.5t - 6t = 180 - 5.5t.

We want 110˚, so:
110 = 180 - 5.5t
5.5t = 70
t = 140/11

This seems reasonable, as it corresponds to a time of about 6:13. For the second time, note that the hands eventually come together, meaning 0˚, and then apart. At this point our expression will be negative, so find:

180-5.5t = -110
5.5t = 290
t = 580/11

This corresponds to a time of approximately 6:53. Good, that looks like a sensible time according to the problem. Now just subtract to find the elapsed time:

580/11 - 140/11 = 440/11 = 40 minutes

Answer 2

There are 30 degrees to each hour marking, so at x minutes after 6 pm the hour hand will be (x/2) degrees past the 6.

For the first time, "a little after 6 pm", we can assume the minute hand is 110° anticlockwise from the hour hand. The minute hand will be 6x degrees past the 12, so the angle between the two in degrees will be (180 + x/2) - (6x) = 110.
So 70 = 11x/2, i.e. x = 140/11 = 12 8/11 minutes. (Remember, this is the number of minutes past 6 pm.)

For the second time, "before 7 pm", the minute hand will be 110° clockwise from (i.e. past) the hour hand. So this time we have
(6x) - (180 + x/2) = 110
so 11x/2 = 290
so x = 580/11 = 52 8/11 minutes.

So the elapsed time is (580/11 - 140/11) = 440/11 = 40 minutes.

Answer 3

things to remember:
1) minute hand moves 360/60 degrees per minute
= 6 degress / minute

2) hour hand moves 360/12/60 degress per minute
= 0.5 degress / minute

now calculate how many minute will it take to form 110 degrees

At 6pm, the hands form a straight line at 180 degrees

so

let x be the number of minutes

then

180 - x(6degrees/min) + x(0.5 degrees/min) = 110

here you have to substract the angle from the minute hand and add to the hour hand

70 degrees = x(5.5 degrees/min)
x = 70/5.5
x = 12.7 minutes

now do the same thing for the return trip
At 7pm, the angle between the hands is 150 degrees

150 - x(6) + x(0.5) = 110
150 - 5.5x = 110
5.5x = 40
x = 7.3 minutes

therefore, the teacher left at 6:13pm and returned at 6:53pm making the time of trip at 40 minutes

Answer 4

in a minute hh (hour hand) moves 360/12*60 deg = 0.5 deg
and mh (minute hand) moves 360/60 deg = 6 deg.
if angle changed from 110 deg to 110 deg that means mh overtook hh and time lapsed is t minutes given by
6t = 2(110) + 0.5t
=> 5.5t = 220
=> t = 220/5.5 = 40
teacher remained out 40 minutes.