26 Feb '08 07:19>
Originally posted by smaiaI got about 19.85 degrees, but i want to find the nice simple way
You will be amazed how beautiful and simple the solution is.
No need to use calculators or perform complex computations.
It can be resolved mentally if you discover a few hidden things.
(there are omitted degrees symbols....)
let x = AB = BC = AC
BD = DC = x/2
using simple trig, BED = 50 deg, EBD = 60 deg, BDE = 70 deg
ED/sin(60) = (x/2)/sin(50) (sine rule)
ED = x*sin(60)/sin(50) = sqroot(3)*x/(2sin(50))
sqr(EC) = sqr(ED) + sqr(DC) - 2*ED*DC*cos(110) (cosine rule, 110 deg from basic trig)
EC = sqroot(3sqr(x)/(4sqr(sin(50))) + sqr(x)/4 - 2 * sqroot(3)*sqr(x)/(4sin(50)) * cos(110)) = 1.39x
DC/sin(DEC) = EC/sin(110)
sin(DEC) = x/2 * sin(110)/EC
DEC = 19.849404914... degrees
very messy....
correct me if im wrong please