Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

Convert \[{{40}^{\circ }}\,{{20}^{'}}\]into radian measure. Select the correct option
A . \[\dfrac{121}{540}\pi \] radians
B. \[\dfrac{121}{570}\pi \]radians
C. \[\dfrac{120}{513}\pi \]radians
D. None

Answer
VerifiedVerified
583.5k+ views
Hint: To convert the given angle from degrees to radians, we will use the conversion factor that are as follows: \[{{60}^{'}}={{1}^{\circ }}\] and \[{{1}^{\circ }}=\dfrac{\pi }{180}\] radians.

Complete step by step solution:
Now, in the given question, we have to convert \[{{40}^{\circ }}\,{{20}^{'}}\]into radians.
So, here we will first convert \[{{20}^{'}}\] minutes in degrees, using the conversion factor
\[\begin{align}
  & \Rightarrow 60'={{1}^{\circ }} \\
 & \Rightarrow 20'={{\left( \dfrac{1}{60}\times 20 \right)}^{\circ }} \\
 & \Rightarrow 20'={{\left( \dfrac{1}{3} \right)}^{\circ }} \\
\end{align}\]
Now, \[{{40}^{\circ }}\,{{20}^{'}}\] can be written in the degree as \[\,{{\left( 40\dfrac{1}{3} \right)}^{\circ }}\] Next this is the mixed fraction, that can be converted in the improper fraction as \[\,{{\left( \dfrac{121}{3} \right)}^{\circ }}\].
Next, we will use the conversion \[{{1}^{\circ }}=\dfrac{\pi }{180}\] radians in order to convert \[\,{{\left( \dfrac{121}{3} \right)}^{\circ }}\] in radians.
So we will get:
\[\begin{align}
  & \Rightarrow {{1}^{\circ }}=\dfrac{\pi }{180}\,\,\,\,radian \\
 & \Rightarrow \,{{\left( \dfrac{121}{3} \right)}^{\circ }}=\dfrac{\pi }{180}\times \dfrac{121}{3}\,\,\,radian \\
 & \Rightarrow \,{{\left( \dfrac{121}{3} \right)}^{\circ }}=\dfrac{121}{540}\,\pi \,\,radian \\
\end{align}\]
So this is the required value in radian and hence option A is the correct answer.

Note: It is important here that we don’t get confused with the conversion factors for conversions from radian to degree and from degree to radian. It is important that we first convert the minutes & seconds in degrees and then convert the degree in radians. Also, \[{{60}^{'}}={{1}^{\circ }}\] and not \[60'\ne {{0.6}^{\circ }}\]