# Thread: trig word problem help

1. ## trig word problem help

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

2. 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:

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

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

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

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

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

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 = $\displaystyle w\sqrt{5}$

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

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