Question

A lifeguard at position L spots a swimmer in trouble at one corner of the pool, S. She runs down the length of the pool to position P, and then dives in and swims a distance d from P and S.

Show that the swimming distance is given by the relation d=wsecx
if l=2w, determine the range of values that x may take on, given that P can be anywhere along the length of the pool.

Answer 1

1 ) cos x = w / d
d = w / cos x ( sec x = 1 / cos x )
d = w sec x
2 )
Min. value of x :
tan x = w / 2 w = 1/2 = 0.5
x = tan^(-1) 0.5 = 26.56°
The range of values that x can take on:
26.56° < x < 90°

Answer 2

Let us assume the angle which the lifeguard makes at the point S be x°.

Now, we know that in a triangle,

`cosx=(adjacent)/(hypotenuse)`

i.e.
`cosx=(w)/(d)`

i.e.
`secx=(d)/(w)`.

i.e.
`wsecx=d`.

Hence, the distance is given by i.e.
`wsecx=d`.

Further, we have that,

`tanx=(opposite)/(adjacent)`

i.e.
`tanx=(l)/(w)`

i.e.
`tanx=(2w)/(w)` ( As l=2w )

i.e.
`tanx=2`

i.e.
`x=\arctan2`

i.e.
`x=1.107`

So, the lifeguard can go from the path PS to MS i.e. the range of x is 1.107° to 90°