Posted by **Reiny** on Thursday, May 16, 2013 at 10:09pm.

re

http://www.jiskha.com/display.cgi?id=1368707176
I have a solution, but what a mess.

Before I type it all out, let me know if you still need the solution. I certainly don't feel like typing it all out unless you will actually look at it.

- For Anubhav -
**Anubhav**, Thursday, May 16, 2013 at 11:03pm
Yes please. I still need it. Thanks!

- For Anubhav -
**Reiny**, Friday, May 17, 2013 at 12:27am
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.

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