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Find the size of the smallest angle between the two hands of a clock displaying these times. 10:20, 6:55, 4:35, 2:27 and 11:09

 Jun 15, 2017
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Find the size of the smallest angle between the two hands of a clock displaying these times.

10:20, 6:55, 4:35, 2:27 and 11:09

 

1. angular speed = ω

minutes hand: ωm=3601 h

hour hand: ωh=36012 h

 

2. angle between the two hands = α

α=(ωmωh)tα=(3601 h36012 h)tα=360(11 h112 h)tα=3601112 ht

 

3. Solution for 10:20, 6:55, 4:35, 2:27 and 11:09

t=10:20=10.ˉ3 hα=3601112 h10.ˉ3 hα=3410α=3410(mod360)α=170t=6:55=6.91ˉ6 hα=3601112 h6.91ˉ6 hα=2282.5α=2282.5(mod360)α=122.5t=4:35=4.58ˉ3 hα=3601112 h4.58ˉ3 hα=1512.5α=1512.5(mod360)α=72.5t=2:27=2.45 hα=3601112 h2.45 hα=808.5α=808.5(mod360)α=88.5t=11:09=11.15 hα=3601112 h11.15 hα=3679.5α=3679.5(mod360)α=79.5

 

 

laugh

 Jun 15, 2017
edited by heureka  Jun 15, 2017

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