# TrigonometryFinding the width of the gorge

#### daveyc3000

##### New member
"Greg and Kristine are on opposite ends of a zip line that crosses a gorge. Greg went across the gorge first, and he's now on a ledge that's 15 m above the bottom of the gorge. Kristen is at the top of a cliff that is 72 m above the bottom of the gorge. Jon is on the ground at the bottom of the gorge, below the zip line. He sees Kristen at a 65 degree angle of elevation and Greg at a 35 degree angle of elevation,. What is the width of the gorge to the nearest metre?"

#### topsquark

##### Well-known member
MHB Math Helper
Re: need help solving this problem..ims tuck

As Dr, Peterson asked you on FMH: "What have you tried so far?" (Aside from posting the problem on just about any Math forum.)

-Dan

#### Greg

##### Perseverance
Staff member
Re: need help solving this problem..ims tuck

... and if you're stuck doing that try making a diagram if you haven't done so already.

#### daveyc3000

##### New member
Re: need help solving this problem..ims tuck

Nothing but I have found the answer and now understand the problem

Thanks !

#### MarkFL

Staff member
Re: need help solving this problem..ims tuck

Nothing but I have found the answer and now understand the problem

Thanks !
I've given this thread a useful title, and now, let's make the content useful to others by actually showing the work.

We are not told where along the bottom of the gorge Jon is, so let's let his distance from the taller side be $$x$$. All measures are in meters.

And then we may state:

$$\displaystyle \tan\left(65^{\circ}\right)=\frac{72}{x}$$

$$\displaystyle \tan\left(35^{\circ}\right)=\frac{15}{w-x}$$

The second equation implies:

$$\displaystyle w=15\cot\left(35^{\circ}\right)+x$$

The first equation implies:

$$\displaystyle x=72\cot\left(65^{\circ}\right)$$

Hence:

$$\displaystyle w=15\cot\left(35^{\circ}\right)+72\cot\left(65^{\circ}\right)\approx54.99637148829162$$