03 mc ce maths Q.49

圖片參考：http://hk.geocities.com/stevieg_1023/ms2003.gif

From the above figure ,
CD=PQ=x cm
Let PC=QD=z cm
SC=z/tanα and DR=z/tanβ
SC+DR+CD=y
(z/tanα)+(z/tanβ)+x=y
z[(1/tanα)+(1/tanβ)]=y-x
z=(y-x)/[(1/tanα)+(1/tanβ)]
z=[(sinαsinβ)(y-x)]/(cosαsinβ+cosβsinα)
PS=z/sinα
PS=[(sinαsinβ)(y-x)]/[(sinα)(cosαsinβ+cosβsinα)]
PS=[(sinβ)(y-x)]/sin(α+β)
Answer : D
[Note : sin(α+β)=cosαsinβ+cosβsinα]