(In reply to
Simple Trigonometry (spoiler) by Steve Herman)
Wow. I didn't see it as quite so simple.
I instead solved by finding AO, then the area of triangle BAO, then working backwards from this triangle to find its altitude.
It turns out to be AB*sin(45º).
In general the distance is AB*sin(∠BAC)
∠ABD and the length of AD are not independent, nor are they so simple.
|
Posted by Jer
on 2013-05-14 16:01:20 |