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.
θ is given by $tan θ = | \frac {m – M} {1 + mM} |$
From θ = 100 degrees, tan θ is a known quantity H, say.
M, the slope of the line passing through (0, 10) and (10, 0) = … = –1
Therefore, $H = \frac {m + 1} {1 – m}$
:
$m = … = \frac {H – 1} {H + 1}$
Note: In case θ = 100 degrees, it is necessary to use the supplementary angle (80 degrees) instead to make sure that both sides are positive.
Best Answer
You can simply express this relative change as a fraction, namely
$$\frac ma=0.080=8.0\%\qquad\text{resp.}\qquad\frac nb\approx0.114=11.4\%$$