Patrick仔 asked in 科學及數學數學 · 1 decade ago

a.maths trigonometry

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  • wy
    Lv 7
    1 decade ago
    Favorite Answer

    Note: I have used x instead of theta to represent the angle for easy typing.

    Q.1

    Angle EBD = angle BDC - angle BED = 2x - (90 - x ) = (3x - 90). (Exterior angle of triangle.)

    Q.2

    Let AB = m.

    For triangle BAE, applying sine rule, we get

    m/sin[180 -(90-x)] = a/sin(angle ABE). Angle ABE = 90 - x - x = (90 -2x) (Exterior angle of triangle).

    Therefore, m/sin(90+x) = a/sin(90 -2x). That is m/cosx = a/cos2x.........(1)

    For triangle ABD, applying sine rule, we get

    m/sin(180-2x) = (a+b)/sin(angleABD). Angle ABD = 2x -x = x (Exterior angle of triangle).

    That is m/sin(180 -2x) = (a+b)/sinx

    Or m/sin2x = (a+b)/sinx..................................(2)

    (1) /(2) we get sin2x/cosx = asinx/[(a+b)cos2x]

    2sinx = asinx/[(a+b)cos2x]

    Therefore, cos2x = a/[2(a+b)].

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  • 1 decade ago

    問題的中心係幾個基礎三角形定律

    上半題:

    (三角形內角和180度)

    角EBD = 180 - ( 90 -X) - (180 -2X) = 3X - 90

    下半題:

    (三角形內角和180度)

    (sin rule)

    同樣地,角ABE = 180 -X -[180 -(90-X)] = 90 + 2X

    SO, 角ABD = 3X -90 +90 -2X = X

    SO, 三角形ABD 為等腰三角形

    AD = DB = a+b

    sin rule:

    sin角ABE/a = sin角BAE/BE

    sin(90-2X)/a = sinX/BE

    cos2x = a*sinX /BE

    sin角BED/BD = sin角BDE/BE

    sin(90 - X)/(a+b) = sin (180 -2X)/BE

    BE = sin2X*(a+b)/cosX

    BE = 2sinXcosX*(a+b)/cosX

    BE = 2sinX*(a+b)

    so, cos2X =[a*sinX]/[2sinX*(a+b)]

    cos2X = a/2(a+b)

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