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
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=360∘1 h
hour hand: ωh=360∘12 h
2. angle between the two hands = α
α=(ωm−ωh)⋅tα=(360∘1 h−360∘12 h)⋅tα=360∘⋅(11 h−112 h)⋅tα=360∘⋅1112 h⋅t
3. Solution for 10:20, 6:55, 4:35, 2:27 and 11:09
t=10:20=10.ˉ3 hα=360∘⋅1112 h⋅10.ˉ3 hα=3410∘α=3410∘(mod360∘)α=170∘t=6:55=6.91ˉ6 hα=360∘⋅1112 h⋅6.91ˉ6 hα=2282.5∘α=2282.5∘(mod360∘)α=122.5∘t=4:35=4.58ˉ3 hα=360∘⋅1112 h⋅4.58ˉ3 hα=1512.5∘α=1512.5∘(mod360∘)α=72.5∘t=2:27=2.45 hα=360∘⋅1112 h⋅2.45 hα=808.5∘α=808.5∘(mod360∘)α=88.5∘t=11:09=11.15 hα=360∘⋅1112 h⋅11.15 hα=3679.5∘α=3679.5∘(mod360∘)α=79.5∘