Why is $tan^{-1}\left(\frac{100}{100}\right)\neq90^\circ$ when it's said to have a slope of $100\%$?

Question

I don't see why, but I "know" $tan^{-1}\left(\frac{100}{100}\right)=45^\circ$. It's not immediately intuitive for me if we think of it as driving a car on a road with 45 degree angle uphill, and then it's said to be a 100% slope?

Answer 1

Say we have a line $y = ax + b$. Then, if we look at a point $(x_0,y_0)$ on that line, and at another point $(x_0+\Delta x, y_0+\Delta y)$ on that line, we have that $$y_0 + \Delta y = a(x_0 + \Delta x) = ax_0 + a\Delta x = y_0 + a\Delta x \implies a = \frac{\Delta y}{\Delta x}.$$

Graphically, we are considering right triangle

so, $a = \frac{\Delta y}{\Delta x} = \tan\alpha$.

Now, if $a = 100\% = 1$, then $\Delta x = \Delta y$. So, our right triangle is isosceles. It follows that $\alpha = 45^\circ$.

Answer 2

It could just as well have been. But whoever decided on the standard we use today decided that it shouldn't be. Slope is defined as the ratio of vertical distance to horizontal distance, not the ratio of vertical distance to distance along the slope (i.e. the $\tan$ of the angle, not the $\sin$), and that's just the way it is. This means it's entirely possible to have slopes of $200\%$, and a completely vertical wall has infinite slope.

Note that this way it actually coincides with the notion of slope from calculus, where a $45^\circ$ line (given, for instance, by $y = x$) is said to have slope $1 = 100\%$.