# Problems based on s r theta

In this section we will discuss problems based on s r theta formula. Here we will discuss the step-by-step solution of such problems.**$ \Theta = \frac{arc}{radius}$**

**$ \Theta = \frac{S}{r}$**

**S = r $\times \Theta$**

## Problems based on s r theta formula

1) The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? ($\pi$ =3.24)**Solution :**In 60 minutes, the minute hand of a watch completes one rotation which is $360^{'}$

∴ Angle traced by the minute hand in 1 minute = $\left ( \frac{360}{60} \right )^{0}= 6^{0}$

∴ Angle traced by the minute hand in 40 minutes

= $ \left ( 40 \times 6 \right)^{0}= 240^{0}$

= $\left ( 240 \times \frac{\pi}{180} \right )^{c}= \left ( \frac{4\pi}{3} \right)^{c}$

Now, $ \Theta = \frac{S}{r}$

⇒ $\frac{4\pi}{3} = \frac{S}{1.5}$

∴ S = $\left ( \frac{4\pi}{3} \times 1.5 \right )$

= 2$\pi$

= 2 $\times$ 3.14

= 6.28 cm

2) 2) If the angular diameter of the moon be $30^{'}$, how far from the eye a coin of diameter 2.2 cm be kept to hide the moon?

**Solution :**Suppose the coin is kept at a distance 'r' from the eye to hide the moon completely. Let E be the eye of the observer and let AB be the diameter of the coin. Then, arc AB = diameter AB = 2.2 cm.

$ \Theta = 30^{'} = \left ( \frac{30}{60} \right )^{0}= \left (\frac{1}{2} \times \frac{\pi}{180} \right)^{c} = \left ( \frac{\pi}{360} \right )^{c}$

Now, $ \Theta = \frac{S}{r}$

⇒ $\frac{\pi}{360} = \frac{2.2}{r}$

∴ r = $\left ( \frac{2.2 \times 360}{\pi} \right )$

r = $\frac{2.2 \times 360}{3.14}$

r = 252 cm

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