trig word problem help

**triggy22** · November 17th 2008, 02:17 PM

The problem is attached to the post
I really dont know how to turn that into a graph

**earboth** · November 18th 2008, 05:31 AM

Originally Posted by triggy22

The problem is attached to the post
I really dont know how to turn that into a graph

to a):
Use the definition of Cosine:

$\dfrac wd = \cos(x)~\implies~d=\dfrac w{\cos(x)} = w\cdot \sec(x)$

to b):
$\min(x) = 0~\implies~d = w$

If P = L then $\max(x) = \arctan\left(\dfrac{2w}w\right)=\arctan(2) \approx 1.1071...$

to c)
If x = 0 then d = w
if $x = \arctan(2)$ then $d = \sqrt{(2w)^2+w^2}=w\sqrt{5}$

Therefore $w\leq d \leq w\sqrt{5}$

to d)
I'll leave this part for you.

to e):
That depends on how fast the life-guard is able to run or to swim.
Maxiumum way to run = 3w
Maximum way to swim = $w\sqrt{5}$

Let $v_r$ denote the running speed and $v_s$ the swimming speed then the lifeguard will swim if

$\dfrac{w\sqrt{5}}{v_s} < \dfrac{3w}{v_r}$

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