# Converting angle in inches to degrees

#### paulm

Hi forum, my window air conditioner instructions indicate that the unit must be tilted back 1/4" for proper drainage. I'd like to figure this out in degrees, since I took a college trig a few years ago and feel like I should know how to solve something so basic. My initial attempt has me drawing a unit circle with with a right angle measuring .25 on the y axis.

How to approach this problem? Thanks in advance.

#### skipjack

Forum Staff
The instructions are ambiguous, as they do not state which part of the air conditioner has to move 1/4â€ when the air conditioner is tilted.

#### paulm

Hi, a window air conditioner is one, self-contained unit that tilts back so that excess condensate drains outside. The tilt of the unit is specified as 1/4" angle downward (to the outside).

If someone might help me get started converting that 1/4" tilt to a degree angle I'd be most appreciative.

#### hanslanda

Hi, a window air conditioner is one, self-contained unit that tilts back so that excess condensate drains outside. The tilt of the unit is specified as 1/4" angle downward (to the outside).

If someone might help me get started converting that 1/4" tilt to a degree angle I'd be most appreciative.

Assuming the air conditioner is hinging on the bottom mating corner, we'd need to know the length of the unit to calculate the angle. However measuring total inch drop at the back of the unit is more accurate in all practicality.

• 1 person

#### skeeter

Math Team
What is the distance from the front to the rear of the unit along the bottom?

• 1 person

#### paulm

What is the distance from the front to the rear of the unit along the bottom?
Thanks, that got me started. The back-half of unit is already angled, and instructions indicate measuring from where window meets the angled part. So the angled back-half of unit is 7.8". I found the last (adjacent) side of triangle through Pythag. Ther. (1/4)^2 + x^2 = (7.8 )^2, thus adjacent side = 7.79, and then solving cos = .998 and acos = 3.62Â°, which is exactly the angle of back-half of unit, using bubble level.

Basically the unit could be level (but not at all tilting forward) and drain properly, given the already-angled back.

I wanted to know how to approach this using trig so thanks for reminding me how to get started.

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#### Country Boy

Math Team
I would have done this a little differently. If the height of the air conditioner (measured from top to bottom) is "h" inches then the angle is given by $$\displaystyle \sin(\theta)= \frac{\frac{1}{4}}{h}= \frac{1}{4h}$$ so $$\displaystyle \theta=\arcsin\left(\frac{1}{4h}\right)$$.

But I must say that I think that, as a practical matter, you will find it easier to measure the "1/4 inch" than the angle.

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#### Burt

using 7.8" as the hypotenuse and .25" as the opposite leg figure the angle using square root.
square 7.8" = 60.84 or (7.8" times itself). Hypotenuse side
square .25" = .0625 Opposite side
subtract .25 from 60.84 = 60.7775
get square root of 60.7775 = 7.79599256" Adjacent side
divide .25 by 7.8 = .032051282 this is SINE of the smallest angle of this right angle
divide .25 by 7.79599256 = .032067757 this is TANGENT of this angle
inverse Sine (.032051282) = 1.836717753 or decimal degree of angle
inverse TANGENT (.032067757) = 1.836717753 or decimal degree of angle
using calculator put in 1.836717753 (D.D.) then degree, minutes, seconds key =
1 degree 50 minutes and 12.18 seconds this will be the tilt angle for your air cond.