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Rain is falling vertically downwards wit...

Rain is falling vertically downwards with a speed of `4 km h^-1`. A girl moves on a straight road with a velocity of `3 km h^-1`. The apparent velocity of rain with respect to the girl is.

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The situation is shown in Fig.
OA represents velocity `vecv_(c) ` of cyclist , which is `3km hr^(-1)`
OB represents velocity `vecv_(R)` of vertically falling rain , which is `4km hr^(-1)`.
OC represents `-vecv_(c)` opposite of velocity of the cyclist .
OD represents `vecv_(R)+(-vecv_(c))=vecv_(R)-vecv_(C)` velocity of rain relative to cyclist .
In parallelogram OBDC.
`OC=3km hr^(-1), OB=4km hr^(-1)`
`/_BOC=90^(@)`
Hence `OD=sqrt((OC)^(2)+(OB)^(2))=sqrt((3)^(2)+(4)^(2))=sqrt(9+16)=sqrt(25)=5 km hr^(-1)`
`tan beta =(BD)/(OB)=(OC)/(OB)=3/4=0.75=tan 36^(@)52.`,i.e., `beta=36^(@)-52.`east of vertical .
the cyclist must hold his umbrella at `36^(@)-52` with vertical in the direction of his motion(i.e. east).
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