Lau 發問時間: 科學及數學天文學及太空 · 2 個月前

Please answer the following questions?

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  • 2 個月前
    最佳解答

    h/AK = tanα, h/BK = tanβ, h/CK = tanγ

    ∴ AK = h cotα, BK = h cotβ, CK = h cotγ

    因 AC 是圓 ABC 的直徑,

    故 ∠CBA 為直角.

    cosθ = AB/AC

          = a/(AK+CK)

          = a/(h cotα+ h cotγ)

          = a/[h(cotα+ cotγ)]

    由三角形餘弦定律,

    BK^2 = AB^2 +AK^2 - 2(AB)(AK)cosθ

    所以

    cosθ = (AB^2+AK^2-BK^2)/[2(AB)(AK)]

      = (a^2 + h^2 cot^2 α- h^2 cot^2 β)/(2ah cotα)

    a/[h(cotα+ cotγ)]

      = (a^2 + h^2 cot^2 α- h^2 cot^2 β)/(2ah cotα)

    √3/[h(√3+1/√3)]

      = (3+3h^2-h^2)/(2√3h √3)

    h > 0

    ∴ 1/(1+1/3) = (3+2h^2)/6

    ∴ 3 + 2h^2 = 18/4

       h = √[(18/4-3)/2] = √3/2

    cosθ = √3/[h(√3+1/√3)] = (3/4)/h = √3/2

    ∵ 0 < θ < 90°

    ∴ θ = acos(√3/2) = π/6

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