How do you convert $\dfrac{{11\pi }}{2}$ radians to degree?

Hint: In this question we need to convert $\dfrac{{11\pi }}{2}$ radians to degree. Here, we know that to convert $\dfrac{{11\pi }}{2}$ radians to degree we need to multiply the radians with $\dfrac{{180}}{\pi }$ and by evaluating it we will get the required degree of the given radian.

Complete step by step answer:
Here, we need to convert $\dfrac{{11\pi }}{2}$ radians to degree.
We know that pi radians is equal to $180$ degrees, i.e., $\pi \,rad = 180^\circ $.
One radian is equal to $57.295779513$ degrees, i.e., $1\,rad = \dfrac{{180^\circ }}{\pi } = 57.295779513^\circ $
The angle $\alpha $ in degrees is equal to the angle $\alpha $ in radians times $180$ degrees divided by pi constant is,
$\deg = rad \times \dfrac{{180^\circ }}{\pi }$
To convert radians to degrees we need to multiply the radians with $\dfrac{{180}}{\pi }$ .
 $ = \dfrac{{11\pi }}{2} \times \dfrac{{180^\circ }}{\pi }$
$ = 11 \times 90^\circ $
$ = 990^\circ $
Hence, the degree of $\dfrac{{11\pi }}{2}$ radians is $990^\circ $.

Note: Radian is a unit of measure of angles. One radian is the angle made at the center of a circle by an arc whose length is equal to the radius of the circle. The radian is denoted by the symbol $rad$. The length of an arc of a unit circle is numerically equal to the measurement in radians of the angle that it subtends.
A degree is an measure of an angle, one degree is $\dfrac{1}{{{{360}^{th}}}}$ part of a full circle, since there are $360^\circ $ in a full rotation. A degree usually denoted by $^\circ $.
The measure of an angle is determined by the amount of rotation from the initial side to the terminal side. In radians, one complete counterclockwise revolution is $2\pi $ and in degrees, one complete counterclockwise revolution is $360^\circ $. So degree measure and radian measure are related by the equations, $360^\circ = 2\pi \,radians$ and $180^\circ = \pi \,radians$. Hence, to convert radians to degrees we need to multiply the radians with $\dfrac{{180}}{\pi }$ .