Posted by Reiny on Thursday, May 16, 2013 at 10:09pm.
Yes please. I still need it. Thanks!
I made a sketch of triangle ABC
drew in the bisector of angle A and
drew the altitude AM
In triangel ANC
NC = 21, angle NAC = 45°
21/sin45 = AC/sinØ
sinØ = (sin45)(AC)/21
sinØ = √2 AC/42
in triangle ABN
BN = 1, angle BAN = 45°
AB/sin(180-Ø) = 1/sin45
sin(180-Ø) = √2AB/2
but sin(180-Ø) = sinØ
√2AC/42 = √2AB/2
2AC = 42AB
AC = 21 AB
let AB = k , then AC = 21k
in triangle ABC
k^2 + (21k)^2 = 22^2
442k^2 = 484
k = 22/√442
so AB = 22/√442 = appr 1.046
AC = 21(22)/√442 = 462/√442 = appr 21.975
let's find angle Ø
sinØ/AC = sin45/21
sinØ = (√2/2)AC/21
sinØ = (√2/2)(462/√442)/21
= 11/√221
Ø = 47.726 or Ø = 132.274°
from my diagram in triangle ACN, Ø = 132.274 ,
making angle C = 2.726°
and angle ANM = 47.726°
almost there!!!
in right-triangle AMC
sinC = AM/AC
AM = ACsinC = 1.0452
In right - triangle AMN
tan 47.726 = AM/MN
MN = 1.0452/tan47.726 = .950226
so CM = 21.950226
BM = 1-.950226 = .049774
CM : BM = 21.950226 : .049774
= 441 : 1
( I was going to keep all the square roots to have "exact" answers , but it got a bit tedious, but I did store all intermediate answers in my calculator's multi-memory locations.
Check my arithmetic, in all that mess it is easy to make errors.