# Relation between Degrees and Radians

In this section we will discuss about the relation between degrees and radians.
1 radian = $\frac{180^{0}}{\pi}$

OR
1 degree = $\frac{\pi}{180^{0}}$

Relation between degrees and radian :
Consider a circle with center 'O' and radius 'r'. Let A and P be any point on the circle such that arc AP = r. Let the $\angle AOP$= 1 radian. From the diagram we can see that , $\angle AOB$ = 2 right angles = $180^{0}$
As we know that angle subtending at the center of the circle is proportional to the arc subtended in it .
∴ $\frac{\angle AOP}{\angle AOB} = \frac{arc AP}{arc APB}$

⇒ $\frac{\angle AOP}{180} = \frac{r}{\pi r}$

⇒ $\angle AOP = \frac{180}{\pi}$

⇒ $1^{c} = \frac{180^{0}}{\pi}$

Remark 1 : When an angle is expressed in radians, the word radian is generally omitted.
Remark 2 : $180^{0} = \pi$ radians ⇒ $1^{0} = \frac{\pi}{180}$
Remark 3 : $180^{0} = \pi$ radians
∴ $1^{0} = \frac{\pi}{180}$ radian = $\left ( \frac{22}{7 \times 180} \right )$ radian = 0.01746 radians
Remark 4 : $\pi radians =180^{0}$
∴ 1 radian = $\frac{180}{\pi}$=$\left (\frac{180}{22}\times 7\right )^{0}$ = $57^{0}16^{'}22^{"}$

## Examples on relation between degrees and radians

Find the radian measures corresponding to the following degree measures:
(i) $340^{0}$
Solution : As we know that $1^{0} = \left ( \frac{\pi}{180}\right )^{c}$
∴ $340^{0} = \left (340\times \frac{\pi}{180}\right )^{c}$
= $\left ( \frac{17\pi}{9}\right )^{c}$

(ii) $75^{0}$
Solution : As we know that $1^{0} = \left ( \frac{\pi}{180}\right )^{c}$
∴ $75^{0} = \left (75\times \frac{\pi}{180}\right )^{c}$
= $\left ( \frac{5\pi}{12}\right )^{c}$

(iii) $520^{0}$
Solution : As we know that $1^{0} = \left ( \frac{\pi}{180}\right )^{c}$
∴ $520^{0} = \left (520\times \frac{\pi}{180}\right )^{c}$
= $\left ( \frac{26\pi}{9}\right )^{c}$

(iv) $40^{0}20^{'}$
Solution : As we know that $1^{0} = \left ( \frac{\pi}{180}\right )^{c}$
∴ $40^{0}20^{'} =40^{0} + \frac{20}{60} = 40^{0} + \frac{1}{3} = \frac{121}{3}$

=$\left (\frac{121}{3}\times \frac{\pi}{180}\right )^{c}$= $\left ( \frac{121\pi}{540}\right )^{c}$

Find the degree measures corresponding to the following radian measures:
(i)$\left ( \frac{2\Pi }{15} \right )^{c}$
Solution : As we know that $1^{c}= \frac{180^{0}}{\pi}$
$\left ( \frac{2\Pi }{15} \right )^{c} = \left (\frac{2\pi}{15}\times \frac{180 }{\pi} \right )^{c} = 24^{0}$

(ii) $\left ( \frac{\Pi }{8} \right )^{c}$
Solution : As we know that $1^{c}= \frac{180^{0}}{\pi}$
$\left ( \frac{\Pi }{8} \right )^{c} = \left (\frac{\pi}{8}\times \frac{180 }{\pi} \right )^{c} = \left ( 22\frac{1}{2} \right )^{0} = 22^{0}\left (\frac{1}{2}\times 60 \right )^{'}= 22^{0}30^{'}$

(iii) $\left ( \frac{1 }{4} \right )^{c}$
Solution : As we know that $1^{c}= \frac{180^{0}}{\pi}$
$\left ( \frac{1 }{4} \right )^{c} = \left (\frac{1}{4}\times \frac{180 }{22} \times 7\right )^{c} = \left ( \frac{315}{22} \right )^{0} = 14^{0}\left (\frac{7}{22} \right )^{0}$

= $14^{0}\left (\frac{7}{22} \times 60\right )^{'}= 14^{0}\left (19\frac{1}{11} \right )^{'}$

= $14^{0}19^{'}\left ( \frac{1}{11} \times 60\right )^{"}$

= $14^{0}19^{'}5^{"}$

11th grade math

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