$\sin\left(\frac{p}{q}\pi\right)=\sin\left(\frac{p}{q}180^\circ\right)$
is always algebraic for $\frac{p}{q}\in\mathbb{Q}$:
Let
$$
\alpha=e^{\frac{i\pi}{q}}=\cos\frac{\pi}{q}+i\sin\frac{\pi}{q}.
$$
Then $\alpha^q+1=0$, i.e. $\alpha$ is an (algebraic)
$2q^\text{th}$ root of unity, i.e. it is a root of $x^{2q}-1$.
Hence, so is its power $\alpha^p$ and reciprocal/conjugate power,
which for $p$ an $q$ in lowest terms are roots of $x^q-(-1)^p=0$.
Therefore, so too are
$$
\cos\frac{p\pi}{q}=\frac{\alpha^p+\alpha^{-p}}{2}
\qquad\text{and}\qquad
\sin\frac{p\pi}{q}=\frac{\alpha^p-\alpha^{-p}}{2i},
$$
by the closure of the algebraic numbers as a field.
Ivan Niven gives a nice proof at least that $\sin x$
is irrational for (nonzero) rational $x$.
As @Aryabhata points out, the
Lindemann-Weierstrass theorem
gives us that these values of $\sin$ and $\cos$
are transcendental (non-algebraic),
by using the fact that the field extension $L/K$
of $L=\mathbb{Q}(\alpha)$
over $K=\mathbb{Q}$
has transcendence degree 1.
Why is it that when we convert radians to degrees we multiply radians $\times \frac{180}π$ , but when we convert slope per radians to slope per degrees we have to multiply the inverse conversion formula slope per radians $\times \frac{π}{180}$
If we want to know an hour in terms of minutes, we multiply 1 hour $\times \frac{60}{1}$, given the result in minutes. If we want to know how convert 180 minutes, we divide $180$ minutes by $60$, i.e., multiply $180 \times \frac 1{60}$.
You'll find this phenomenon in any conversion: To convert temperature in degrees Celsius to temp in Fahrenheit, we have $F = \frac 95 C + 32$. To convert to from F to C, we need to invert this: $C = \frac 59(F-32)$
$R \text{ radians}\;\times \dfrac{180^\circ}{\pi \;\text{radians}} = \dfrac{R\times 180^\circ}{\pi}$.
"Radians" cancels as the unit, leaving a numeric value expressed in degrees.
$D \text{ degrees}\; \times \dfrac{\pi \;\text{radians}}{180\; \text{degrees}} = \dfrac{D\pi\;}{180}\;\text{ radians}$.
"Degrees" cancels as the unit, leaving the value expressed in radians.
Moved from comments:
Note that slope per radian is a ratio: $\;\dfrac{\text{slope}}{\text{radians}}.\;$ So to obtain "slope per degree", you need to have $π$ radians in the numerator, to cancel the unit "radians" from the denominator, and $180$ degrees in the denominator, to end with slope/degrees.
Slope itself is not a "unit" per se, meaning it isn't a degree, or radian, a mm, or foot. It is unit-free: even if we assign distance units (meters, say) to displacement: e.g. $Δ\,y\text{m}=y_2\,\text{m}−y_1\,\text{m}$ and $Δ\,x\,\text{m}=x_2\text{m}−x_1\,\text{m},$ then we have $$\text{slope}\,= \dfrac{Δy\,\text{ m}}{Δx\,\text{m}}=\frac{Δy}{Δx},$$ you see that the units "m: meter" attached to "change in y" and "change in x" cancel in the ratio defining slope, leaving us with a unit-free scalar which slope really is.
Best Answer
Although your calculator is evaluating $\sin x$ assuming $x$ is in degrees, it is integrating $x$ from $0$ to $30$. What it is doing would usually be written $$\int_0^{30} \sin \frac {\pi x}{180}\;dx\approx 7.676$$ where the argument of $\sin$ is now in radians. Note that the result of the integral is a pure number, not degrees or radians.