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Math Help - geometry

  1. #1
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    geometry

    Question
    In a triangle ABC, AC = 2AB and BC/BA = 3/2. H is the foot of the perpendicular from B to AC. Prove that CH/AH = 21/11.

    Solution
    In the triangle ABC, the sides are:
    1. AB,
    2. BC=1.5* AB
    3. AC=2* AB
    and CH+AH = AC.
    COS(angle A) = AH / AB
    COS(angle C) = CH / BC = CH / (1.5* AB)

    CH / AH = (1.5* AB)*COS(angle C) / (AB * COS(angle A) )
    CH / AH = (1.5)*COS(angle C) / (2 - COS(angle C) )
    Hence, CH / AH = 1.90909...= 21 / 11


    Is this the correct way to do it?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by xwrathbringerx View Post
    Question
    In a triangle ABC, AC = 2AB and BC/BA = 3/2. H is the foot of the perpendicular from B to AC. Prove that CH/AH = 21/11.

    Solution
    In the triangle ABC, the sides are:
    1. AB,
    2. BC=1.5* AB
    3. AC=2* AB
    and CH+AH = AC.
    COS(angle A) = AH / AB
    COS(angle C) = CH / BC = CH / (1.5* AB)

    CH / AH = (1.5* AB)*COS(angle C) / (AB * COS(angle A) )
    CH / AH = (1.5)*COS(angle C) / (2 - COS(angle C) )
    Hence, CH / AH = 1.90909...= 21 / 11


    Is this the correct way to do it?
    Maybe I am still rusty, but how did you get
    [(1.5)cos(angle C)] / [2 -1.5cos(angle C)] = 1.90909...?
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  3. #3
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    Exclamation

    Can anyone help me solve this problem?!?
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  4. #4
    MHF Contributor
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    Quote Originally Posted by xwrathbringerx View Post
    Can anyone help me solve this problem?!?
    Question
    In a triangle ABC, AC = 2AB and BC/BA = 3/2. H is the foot of the perpendicular from B to AC. Prove that CH/AH = 21/11.

    Let AB = u
    Then, AC = 2u
    And, BC = (3/2)u = 1.5u

    Let also AH = v
    So, HC = 2u -v

    In right triangle AHB, by Pythagorean theorem,
    (BH)^2 = (AB)^2 -(AH)^2 -------(i)

    In right triangle BHC,
    (BH)^2 = (BC)^2 -(HC)^2 ------(ii)

    (BH)^2 = (BH)^2, so,
    (AB)^2 -(AH)^2 = (BC)^2 -(HC)^2
    Substitutions,
    u^2 -v^2 = (1.5u)^2 -(2u -v)^2
    u^2 -v^2 = 2.25u^2 -4u^2 +4uv -v^2
    The -v^2 cancels out,
    u^2 -2.25u^2 +4u^2 = 4uv
    2.75u^2 = 4uv
    2.75u = 4v
    v = (2.75 /4)u = (5.5 /8)u = (11 /16)u -------**

    Now,
    (CH) / (AH)
    = (2u -v) / v
    = [2u -(11/16)u] / [(11/16)u]
    = [(32u -11u) /16] / [11u /16]
    = [21u /16] *[16 /(11u)]
    = 21/11 -------------------------proven.
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