圓形數學問題難解

Construct a point S on AB such that QS⊥AB, a point T on OP such that QT⊥OP.
as OP⊥AB (radius ⊥ tangent), so the distance from Q to AB is :
QS
= OP - OT
= r - r cos∠QOP
= r - r cos(360° - reflex angle QOP)
= r - r cos(360° - 2(α/2)) ..... (∠ at centre, twice ∠ at circumference)
= r - r cos(360° - α)
= r - r cos α
= r(1 - cos α)

ans : (D)

加一條由Q畫起的水平線,加上acute angle of QOP = 360-a

(r-x)/r=cos(360-a)
r-x=rcosa
x=r(1-cosa)

當α = 0 (即∠POQ = 0), Q_AB = 0

A.
Q_AB = r tan(0/2) = 0

B.
Q_AB = r (1 - sin0) = r

C.
Q_AB = r (1 + cos0) = 2r

D.
Q_AB = r (1 - cos0) = 0

所以B,C都不是答案。

當α = 360 (即∠POQ = 180), Q_AB = 2r

A.
Q_AB = r tan(360/2) = 0

D.
Q_AB = r (1 - cos360) = 2r

所以A不是答案, 答案是D。

2014-05-24 16:37:23 補充：
∠POQ(反角) = 2∠PRQ = 2 (α/2) = α (圓心角等於2倍圓周角)

∠POQ = 360 - ∠POQ(反角) = α

作一點X由Q垂直至OP(∠XOQ = ∠POQ = α),

OX = r cos∠XOQ = r cosα


Q_AB = r - OX = r - r cosα = r (1 - cosα)