As we know the relation between the angles and the arc length of any circle is given by the following relation:

$$
begin{align*}
s &= r ~ theta \
end{align*}
$$

As we know one revolution of a rotating object complete one cycle is $360^{circ}$. Also, the arc length of one cycle rotation is the circumference of a circle which is given by $2 ~ pi ~ r$. Substituting in the previous calculation, then we get

$$
begin{align*}
2 pi mathrm{~rad} &= 360^{circ} \
end{align*}
$$

So, if we want to convert an angle in degrees to the corresponding angle in rads, then we use the following relation:

$$
begin{align*}
theta_{rad} &= dfrac{ 2 pi }{ 360^{circ} } ~ theta^{circ} \
end{align*}
$$

Solve for $theta^{circ} = 30^{circ}$:

Substituting in the previous calculations, then we get

$$
begin{align*}
theta_{rad} &= dfrac{ 2 pi }{ 360^{circ} } ~ theta^{circ} \
&= dfrac{ 2 pi }{ 360^{circ} }times 30^{circ} \
&= dfrac{ pi }{ 6} ~ mathrm{~rad}
end{align*}
$$

So, $30^{circ}$ converts to $dfrac{ pi }{ 6} ~ mathrm{~rad}$.

Solve for $theta^{circ} = 45^{circ}$:

Substituting in the previous calculations, then we get

$$
begin{align*}
theta_{rad} &= dfrac{ 2 pi }{ 360^{circ} } ~ theta^{circ} \
&= dfrac{ 2 pi }{ 360^{circ} }times 45^{circ} \
&= dfrac{ pi }{ 4} ~ mathrm{~rad}
end{align*}
$$

So, $45^{circ}$ converts to $dfrac{ pi }{ 4} ~ mathrm{~rad}$.

Solve for $theta^{circ} = 90^{circ}$:

Substituting in the previous calculations, then we get

$$
begin{align*}
theta_{rad} &= dfrac{ 2 pi }{ 360^{circ} } ~ theta^{circ} \
&= dfrac{ 2 pi }{ 360^{circ} }times 90^{circ} \
&= dfrac{ pi }{ 2} ~ mathrm{~rad}
end{align*}
$$

So, $90^{circ}$ converts to $dfrac{ pi }{ 2} ~ mathrm{~rad}$.

Solve for $theta^{circ} = 180^{circ}$:

Substituting in the previous calculations, then we get

$$
begin{align*}
theta_{rad} &= dfrac{ 2 pi }{ 360^{circ} } ~ theta^{circ} \
&= dfrac{ 2 pi }{ 360^{circ} }times 180^{circ} \
&= pi ~ mathrm{~rad}
end{align*}
$$

So, $180^{circ}$ converts to $pi ~ mathrm{~rad}$.

$45^{circ}$ converts to $dfrac{ pi }{ 4} ~ mathrm{~rad}$.

$90^{circ}$ converts to $dfrac{ pi }{ 2} ~ mathrm{~rad}$.

$180^{circ}$ converts to $pi ~ mathrm{~rad}$.

As we know the relation between the angles and the arc length of any circle is given by the following relation:

$$
begin{align*}
s &= r ~ theta \
end{align*}
$$

As we know one revolution of a rotating object complete one cycle is $360^{circ}$. Also, the arc length of one cycle rotation is the circumference of a circle which is given by $2 ~ pi ~ r$. Substituting in the previous calculation, then we get

$$
begin{align*}
2 pi mathrm{~rad} &= 360^{circ} \
end{align*}
$$

So, if we want to convert an angle in rads to the corresponding angle in degrees, then we use the following relation:

$$
begin{align*}
theta^{circ} &= dfrac{ 360^{circ} }{ 2 pi } ~ theta_{rad} \
end{align*}
$$

Solve for $theta_{rad} = dfrac{ pi }{ 6} ~ mathrm{~rad}$:

Substituting in the previous calculations, then we get

$$
begin{align*}
theta^{circ} &= dfrac{ 360^{circ} }{ 2 pi } ~ theta_{rad} \
&= dfrac{ 360^{circ} }{ 2 pi } times dfrac{ pi }{6} \
&= 30^{circ}
end{align*}
$$

So, $dfrac{ pi }{ 6} ~ mathrm{~rad}$ converts to $30^{circ}$.

Solve for $theta_{rad} = 0.7 ~ mathrm{~rad}$:

Substituting in the previous calculations, then we get

$$
begin{align*}
theta^{circ} &= dfrac{ 360^{circ} }{ 2 pi } ~ theta_{rad} \
&= dfrac{ 360^{circ} }{ 2 pi } times 0.7 mathrm{~rad} \
&= 40.11^{circ}
end{align*}
$$

So, $0.7 mathrm{~rad}$ converts to $40.11^{circ}$.

Solve for $theta_{rad} = 1.5 pi ~ mathrm{~rad}$:

Substituting in the previous calculations, then we get

$$
begin{align*}
theta^{circ} &= dfrac{ 360^{circ} }{ 2 pi } ~ theta_{rad} \
&= dfrac{ 360^{circ} }{ 2 pi } times 1.5 pi \
&= 30^{circ}
end{align*}
$$