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Question:

greenturtle
greenturtle
Student
High School - 10th Grade

We construct fi angle between the median and the bisectrix traced from the B vertex of right triangle ABC (A=90degrees). Calculate tg^3 (B/2).

In this calculus, we'll depend on fi angle.

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Posted by greenturtle on Wednesday June 24, 2009 at 12:21 AM and tagged with alfa, angle, bisectrix, math, median, triangle, vertex.

Answers:

  1. giorgiana1976
    giorgiana1976 Teacher
    Doctorate

    eNotes Editor

    Let's have ABM=alfa, CBM=gama, angles fomed by the median MB.

    With the help of sine theorem in triangles ABM, CMB we have:

    sin alfa/sin gama= sinA/sinC

    Using the proportions properties:

    (sin alfa-sin gama)/(sin alfa+sin gama)=(sinA-sinC)/(sinA+sinC)

    By transforming the sum in product, we'll have:

    sin alfa-sin gama=2cos((alfa+gama)/2)sin((alfa-gama)/2)

    sin alfa+sin gama=2sin((alfa+gama)/2)cos((alfa-gama)/2)

    sinA-sinC=2cos((A+C)/2)sin((A-C)/2)

    sinA+sinC=2sin((A+C)/2)cos((A-C)/2)

    But ((alfa-gama)/2)=fi, (alfa+gama)/2=B/2

    Knowing that A=90, then (A+C)/2=pi/2-B/2

    (sin alfa-sin gama)/(sin alfa+sin gama)=

    =2cos(B/2)sin fi/2sin(B/2)cos fi=ctg(B/2)tg fi (1)

    2sin(pi/2-B/2)cos((A-C)/2)=2cos(B/2)cos((A-C)/2

    2cos(pi/2-B/2)sin((A-C)/2)=2sin(B/2)sin((A-C)/2

    {2sin(B/2)sin((A-C)/2}/{2cos(B/2)cos((A-C)/2}=

    =tg(B/2)tg(A-C)/2 (2)

    From (1) and (2), ctg(B/2)tg fi=tg(B/2)tg(A-C)/2

    (1/tg(B/2))tg fi=tg(B/2)tg(A-C)/2

    tg^2 (B/2)=tg fi/tg(A-C)/2

    tg^3(B/2)=tg fi

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    Posted by giorgiana1976 on Wednesday June 24, 2009 at 4:26 AM