   SEARCH HOME Math Central Quandaries & Queries  Question from amber, a student: i am working with an arched entry. i know that the radius is 25' and the height is 20'-11". i need to know the length of the arch and degree of bend of the arch. how do i find these? Amber,

I am assuming it is to be a circular arch, since you haven't indicated otherwise. 20'11" is 20.917 ' (I want to use decimals).

Take a look at http://mathcentral.uregina.ca/QQ/database/QQ.09.07/s/bruce1.html
which has similar geometry to yours.

h = r - r cos theta.

20.917 = 25 - 25 cos theta

20.917 / 25 = 1 - cos theta
cos theta = 1 - 20.917 / 25
theta = arccos (1 - 20.917 / 25)
theta = arccos ( 0.16332 )
theta = 80.6 degrees.

Now the angle you are interested in is actually 2 times theta, so that is 161.2 degrees.

The circumference of a circle is twice the radius times Pi. So in your case the full circle would be 2 x 25 x 3.14159 feet = 157.0795 feet.

Since you want 161.2 / 360 degrees of a full circle, your arc is ( 161.2 / 360 ) x 157.0795 feet = 70.337 feet, which is about 70'4".

Cheers,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.