MHB Finding the width of the gorge

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The problem involves calculating the width of a gorge based on the heights of Greg and Kristine and their angles of elevation as viewed by Jon at the bottom. Using trigonometric functions, the equations tan(65°) = 72/x and tan(35°) = 15/(w-x) were established. By substituting and solving these equations, the width of the gorge was found to be approximately 55 meters. The discussion emphasizes the importance of diagramming the problem for clarity. Ultimately, the calculations confirm the width of the gorge to the nearest meter is 55 m.
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"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?"

Answer: 55 m.
 
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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
 
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. :)
 
Re: need help solving this problem..ims tuck

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

Thanks !
 
Re: need help solving this problem..ims tuck

daveyc3000 said:
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:

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

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

The second equation implies:

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

The first equation implies:

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

Hence:

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