Please answer the following questions?

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