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Exercise 2.2 · Q10

Q.At what time between 3'o clock and 4'o clock are the hands of a clock three degrees apart?

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Solving ∣30(3)−5.5M∣=3∘|30(3)-5.5M|=3^\circ gives two very close moments just after 3:15 — 3 ⁣: ⁣159113\!:\!15\frac{9}{11} and 3 ⁣: ⁣1610113\!:\!16\frac{10}{11}.

θ=∣30H−5.5M∣\theta = |30H-5.5M| degrees.

Given: H=3H=3, θ=3∘\theta=3^\circ.

  1. Set up the equation:

∣30(3)−5.5M∣=3 ⇒ ∣90−5.5M∣=3|30(3) - 5.5M| = 3 \ \Rightarrow\ |90-5.5M|=3

  1. Case 1 (90−5.5M=390-5.5M=3, minute hand slightly behind the 3∘3^\circ-before point):

5.5M=87⇒M=875.5=17411=15911 min5.5M = 87 \Rightarrow M = \dfrac{87}{5.5} = \dfrac{174}{11} = 15\tfrac{9}{11}\ \text{min}

  1. Case 2 (90−5.5M=−390-5.5M=-3, minute hand slightly past): 5.5M=93⇒M=935.5=18611=161011 min5.5M = 93 \Rightarrow M = \dfrac{93}{5.5} = \dfrac{186}{11} = 16\tfrac{10}{11}\ \text{min} …

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