Recently, I was discussing the relative virtues of four-door and two-door cars with a friend. I prefer four-door cars, because they make it much easier for back-seat passengers to get in and out (they make it easier to access the back seat, in general). My friend prefers two-door cars, because he seldom has back-seat passengers, and the larger doors of two-door cars make it easier for the front-seat occupants to get in and out.

“But,” I say, “on a two-door car, the doors are larger and heavier. Also, if you’re in a limited space, the doors can’t open as far, and the smaller opening actually makes it *harder* for the people in the front seat.”

I also recently read Kate Nowak’s post “Introducing Right Triangle Trig”, from her blog *f(t) [Function of Time]*.^{[1]}

Putting the two together, I thought it would be fun^{[2]} to turn the car-door discussion into a trig problem.

Suppose I park my car in a garage. The side of the car is some fixed distance from the garage wall, limiting the amount that the door can open. *Figure 1* illustrates that: the car is parked at a distance of *a* from the wall (because mathematicians like to confuse things by representing everything with incomprehensible letters), the door has a length of *d*, and when the edge of the door hits the wall, the angle the door makes with the car is *t*. (Click on the figures to enlarge them.)

What we want to know is how big the opening (*o*, in figure 1) is, so we know how easy it will be for a person to get in when the door is as far open as it can be. Noting that the closed-door position, the open door, and the opening form a triangle, we bet we can sort this out with trigonometry.

The first step is to drop a line from the tip of the door to the car, perpendicular to the car. The dashed line labelled *a* in *figure 2* forms a right triangle.

With respect to angle *t*, the opposite side has length *a* and the hypotenuse has length *d*. The sine of the angle, *sin(t)*, is the opposite side divided by the hypotenuse, or *a/d*. So we have our first formula, which we can solve for *t*:

sin(t) = a / dt = arcsin(a / d)

Now we’ll note that the two sides of length *d* form an isosceles triangle with base *o*. That means that if we bisect angle *t* we’ll get two equivalent right triangles (*figure 3*).

We now have an angle *t/2*, a hypotenuse *d*, and an opposite side *o/2*, giving us our second formula, which we can solve for *o*:

sin(t / 2) = (o / 2) / do = 2 * d * sin(t / 2)

It’s time to plug in the numbers.

Let’s assume that a four-door car’s front door is 4 feet long, and a two-door car’s door is 6 feet long. And let’s say we park the car 3 feet from the wall. So, for a 4-door car:

a = 36 in, d = 48 inAnd for a 2-door car:sin(t) = 36 in / 48 in

t = arcsin(36 / 48) = arcsin(.75) = 48.59 degrees

sin(48.59 / 2) = (o / 2) / 48 in

o =2 * 48 in * sin(24.295) = 2 * 48 in * .41143 =39.5 in

a = 36 in, d = 72 insin(t) = 36 in / 72 in

t = arcsin(36 / 72) = arcsin(.5) = 30.00 degrees

sin(30 / 2) = (o / 2) / 72 in

o =2 * 72 in * sin(15) = 2 * 72 in * .25882 =37.27 in

So, though the door on the two-door car is longer, with the wall restricting it the door will create an opening that’s about two and a quarter inches (5.6%) smaller. That’s actually less of a difference than I’d expected. It was good to do the exercise, which showed that I was right... but not by enough to matter.

^{[1]} I love that blog name!

^{[2]} Demonstrating, as it does, why we mathematicians are in such demand at parties.

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