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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

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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

Guest 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 = $$\omega$$

minutes hand: $$\omega_{\text{m}} = \frac{360^{\circ}}{1\ h}$$

hour hand: $$\omega_{\text{h}} = \frac{360^{\circ}}{12\ h}$$

2. angle between the two hands = $$\alpha$$

$$\begin{array}{rcll} \alpha &=& ( \omega_{\text{m}} - \omega_{\text{h}} ) \cdot t \\ \alpha &=& \Big( \frac{360^{\circ}}{1\ h} - \frac{360^{\circ}}{12\ h} \Big)\cdot t \\ \alpha &=& 360^{\circ}\cdot \Big( \frac{1}{1\ h} - \frac{1}{12\ h} \Big)\cdot t \\ \mathbf{\alpha} & \mathbf{=} & \mathbf{360^{\circ}\cdot \frac{11}{12\ h} \cdot t} \\ \end{array}$$

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

$$\begin{array}{|rcll|} \hline t=10:20=10.\bar{3}\ h & \alpha & = & 360^{\circ}\cdot \frac{11}{12\ h} \cdot 10.\bar{3}\ h \\ & \alpha & = & 3410^{\circ} \\ & \alpha & = & 3410^{\circ} \pmod {360^{\circ} } \\ & \mathbf{\alpha} & \mathbf{=} & \mathbf{170^{\circ}} \\\\ t=6:55=6.91\bar{6}\ h & \alpha & = & 360^{\circ}\cdot \frac{11}{12\ h} \cdot 6.91\bar{6}\ h \\ & \alpha & = & 2282.5^{\circ} \\ & \alpha & = & 2282.5^{\circ} \pmod {360^{\circ} } \\ & \mathbf{\alpha} & \mathbf{=} & \mathbf{122.5^{\circ}} \\\\ t=4:35=4.58\bar{3}\ h & \alpha & = & 360^{\circ}\cdot \frac{11}{12\ h} \cdot 4.58\bar{3}\ h \\ & \alpha & = & 1512.5^{\circ} \\ & \alpha & = & 1512.5^{\circ} \pmod {360^{\circ} } \\ & \mathbf{\alpha} & \mathbf{=} & \mathbf{72.5^{\circ}} \\\\ t=2:27=2.45\ h & \alpha & = & 360^{\circ}\cdot \frac{11}{12\ h} \cdot 2.45\ h \\ & \alpha & = & 808.5^{\circ} \\ & \alpha & = & 808.5^{\circ} \pmod {360^{\circ} } \\ & \mathbf{\alpha} & \mathbf{=} & \mathbf{88.5^{\circ}} \\\\ t=11:09=11.15\ h & \alpha & = & 360^{\circ}\cdot \frac{11}{12\ h} \cdot 11.15\ h \\ & \alpha & = & 3679.5^{\circ} \\ & \alpha & = & 3679.5^{\circ} \pmod {360^{\circ} } \\ & \mathbf{\alpha} & \mathbf{=} & \mathbf{79.5^{\circ}} \\\\ \hline \end{array}$$

heureka  Jun 15, 2017
edited by heureka  Jun 15, 2017