# [Icehouse] Isosceles triangle angle

• FromDavid Artman <david@xxxxxxxxxxxxxxx>
• DateMon, 05 Mar 2007 12:06:39 -0700
```OK, I found the angle of an Icehouse peak:

The peak angle (A) is divided by two, to get two right triangles
using the base and half the height, with 1/2A = T (theta).

tanT = "opposite over adjacent" = (B/2)/H
T = arctan[(B/2)/H]

...so for a large:
T = arctan [ (1.00/2) / 1.75 ]
= arctan [0.28571428571428571428571428571429]
= ~15.95 degrees

FYI:
A = 2T = 2 * 15.945395900922854797657689523261 =
31.890791801845709595315379046522 degrees (31.90 degrees)
That's the peak angle... but we only need theta (T).

Knowing T, once can write the formula for Height as:

Height = [ (B/2) / tanT ]
= [ (Base/2) / tan(15.945395900922854797657689523261) ]
= Base / 0.57142857142857142857142857142857

THEREFORE, for any given X pip count, after using this to get Base:
Base = (Base of X-1) + 0.21875

...this will calculate the Height:
Height = Base / 0.57142857142857142857142857142857

And, yes, I took my degrees in Liberal Arts majors, too. But I
never forgot SOHCAHTOA. ;)

David

```