SOLUTION:
Step 1:
In this question, we are given the following:
Step 2:
The details of the solution are as follows:
PART ONE:
[tex]\begin{gathered} 2\pi\text{ rad = 360}^0 \\ x\text{ = 25}^0 \\ T\text{his means that:} \\ \text{x = }\frac{2\pi\text{ rad x 25}}{360} \\ \text{x =}\frac{50\pi\text{ rad}}{360} \\ \text{x = }\frac{5\pi}{36}\text{ rad} \end{gathered}[/tex][tex]Hence,\text{ 25 }^0=\frac{5\pi}{36}\text{ rad}[/tex]
PART TWO:
[tex]\begin{gathered} 2\pi rad\text{ = 360}^0 \\ y\text{ =112}^0 \\ This\text{ means that:} \\ \text{y =}\frac{2\pi\text{ rad x 112}^0}{360^0} \\ y\text{ =}\frac{224^0\pi\text{ rad}}{360} \\ y\text{ =}\frac{28\pi}{45}rad \end{gathered}[/tex][tex]Hence,\text{ 112}^0\text{ =}\frac{28\pi\text{ }}{45}\text{ rad}[/tex]
PART THREE:
[tex]\begin{gathered} 2\pi\text{ rad = 360}^0 \\ z\text{ = 290}^0 \\ This\text{ means that:} \\ \text{z = }\frac{2\pi\text{ rad x 290}^0}{360^0} \\ z\text{ =}\frac{580^0\pi\text{ rad}}{360^0} \\ z\text{ =}\frac{29\pi\text{ rad}}{18} \end{gathered}[/tex][tex]Hence,\text{ 290}^0\text{ =}\frac{29\pi\text{ rad}}{18}[/tex]
