Hotel corridor arc

Solution: (1) As shown in the figure, let the center of the arc FG be Q, the point passing through Q be the vertical line of CD, the vertical foot be T, the point passing through MN or its extension line be S, and connect PQ, then point N is the vertical line of TQ, and the vertical foot be W in Rt△NWS, because NW=2, ∠SNW=θ, so

Because MN and arc FG are tangent to point P, PQ⊥MN is in Rt△QPS, because PQ= 1, ∠PQS=θ, QS= 1cosθ, QT-QS=2- 1cosθ.

①S is on the TG line, then TS=QT-QS,

In Rt△STM, MS=TSsinθ=QT? QSsinθ，

So MN=NS+MS=NS+QT? QSsinθ。

② If S is on the extension line of line segment GT, then TS=QS-QT. In Rt△STM,

MS=TSsinθ=QT？ QSsinθ, so MN=NS-MS=NS-QS? QTsinθ=NS+QT？ QSsinθ，

f(θ)=MN=NS+QT？ QS sinθ= 2 cosθ+(2 sinθ- 1 sinθcosθ)= 2 sinθ+2 cosθ？ 1sinθcosθ(0