# Why does using different units of angle affect the rate of change?

The question is

A triangle has sides of length 4cm and 9cm. The angle between them is increasing at a rate of 1$^\circ$ per minute. Find the rate in cm$^2$ per minute at which the area of the triangle is increasing when the angle is 45$^\circ$.

My solution was (and it is supposedly correct):

$A = \frac{1}{2}\cdot 4 \cdot 9\cdot \sin(\theta) = 18 \sin(\theta)$

$\frac{\Delta A}{\Delta \theta} = 18\cos(\theta)$

$\frac{\Delta \theta}{\Delta t} = 1^\circ = \frac{\pi}{180}^c$

$\frac{\Delta A}{\Delta t} = \frac{\Delta A}{\Delta \theta} \cdot \frac{\Delta \theta}{\Delta t} = \frac{\pi}{180} \cdot 18\cos(\theta) = \frac{\pi}{10} \cos(\theta)$

Let $\theta = \frac{\pi}{4}$

$\frac{\Delta A}{\Delta t} = \frac{\pi \sqrt{20}}{2}cm^2/min$

But I was wondering, why did we have to convert the angle from degrees to radians? I only converted to radians simply because I happened to prefer working with radians over degrees.

Obviously, by chain rule, if we stayed in degrees, we would have a different expression for $\frac{\Delta A}{\Delta t}$ and therefore a different answer for the rate of change. However, the rate of change shouldn't change simply because we used a different unit of measurement right? After all, $1^\circ = \frac{\pi}{180}^c$ and always will be.

Is there a fundamental reason why the angle must be measured in radians? My hypothesis is that the units have to match when working with related quantities but I can't seem to make the connection between units of angle and units of area.

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May be you know this, but I say it anyway. If you measure $\theta$ in degrees, then $\dfrac{\Delta A}{\Delta\theta}$ will be different. The derivative of $\sin\theta$ is $\cos\theta$ only when you measure $\theta$ in radians. If you use degrees, then another application of chain rule will cancel the change you noticed: $$\frac{\Delta\theta(\text{radians})}{\Delta\theta(\text{degrees})}=\frac{\pi}{1‌​80}.$$ – Jyrki Lahtonen Sep 28 '13 at 5:58

How do you know $\frac{d}{d\theta}\sin(\theta) = \cos(\theta)$? Is it still true if you measure $\theta$ in degrees instead of radians?
I'll leave out the details of proving the derivative formula for now, but the crux of the matter is that you need to know that $\sin(\theta)$ is very close to $\theta$ when $\theta$ is small, and this is only true if you measure $\theta$ in radians. The best explanation I've seen for that is actually here on math exchange: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
An important implicit point there is that the area of a wedge of a circle $\frac{\theta}{2} r^2$, which only holds if $\theta$ is measured in radians.