The sides of a triangle are 15, 20 and 28. How long are the segments into which the bisector of the largest angle separates the opposite side

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  1. they are in the ratio 3:4 as provided by the angle bisector theorem.

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  2. The largest angle will be opposite the side 28
    let it be 2Ø, so each bisected angle is Ø
    let the bisector form angles A and B along the 28 side, so that A + B = 180° --> B = 180-A
    and we know sinA = sin(180-A) = sinB

    let the 28 side be split into x and 28-x, where x is adjacent the side 20

    Now use the sine law in each of the smaller triangles
    sinØ/x = sinA/20
    sinØ = x sinA/20

    sinØ/(28-x) = sinB/15
    sinØ = (28-x)sinB/15

    x sinA/20 = (28-x)sinB/15 , but remember sinA = sinB, so dividing them out
    x/20 = (28-x)/15
    15x = 560 - 20x
    35x = 560
    x = 16

    so the side 28 is cut into parts 16 and 12

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  3. good call Steve!
    (at least I got the right answer, lol)

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