+0

# 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

0
206
1

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
Sort:

### 1+0 Answers

#1
+19089
+1

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

### 32 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details