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

Answer:

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

Now, we know that in a triangle,

$cosx=\frac{adjacent}{hypotenuse}$

i.e. $cosx=\frac{w}{d}$

i.e. $secx=\frac{d}{w}$.

i.e. $wsecx=d$.

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

Further, we have that,

$tanx=\frac{opposite}{adjacent}$

i.e. $tanx=\frac{l}{w}$

i.e. $tanx=\frac{2w}{w}$ ( 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°

Answer 2

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°