For Anubhav

posted by .

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 -

    Yes please. I still need it. Thanks!

  • For Anubhav -

    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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