Application of a right triangle: Sand Dune

Jakotheshadows

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Quote:
"While visiting the Sand Dunes National Park in Colorado, Cole approximated the angle of elevation to the top of a sand dune to be 20°. After walking 800 ft closer, he guessed that the angle of elevation had increased by 15°. Approximately how tall is the dune he was observing?"

If I could give you the picture I would, but I've tried everything I can think of to figure this out and the closest I came to my own solution was completely wrong.

I first noted that the angle between a hypothetical rod inserted into the highest point in the dune all the way to the dune's base and Cole's original position would have been 70° as the complement to the angle of elevation. I then assumed that the angle of elevation at 800 ft closer to the dune was 35°, and of course its complement is 55°. I tried approaching the problem by unsuccessfully figuring the original hypotenuse (or distance from Cole's original position to the top of the dune). Then I tried to figure out how much distance was left between Cole and the center of the dune (in order to make it a right triangle) and failed.

My next attack was to make a right triangle with the existing angle of elevation and using 800ft as one of the legs. With a 20/70/90 right triangle, I figured that the triangles other leg would be approx 291.2 ft and the hypotenuse would be approx 851.3 ft. My next train of thought brought me to the connection that when Cole got 800 ft closer, the angle of elevation increased by 15°. Assuming that the angle of elevation would increase by 15° for every 800 ft closer that Cole moved towards the dune, I mathematically came to the conclusion that from the center of the dune at ground level, Cole would have to dig a tunnel approximately 3,508.9(Cos 55°) ft straight up to get to the top of the dune.

Anyone care to shed some light upon my silly ignorance?
 
We can construct two equations in two variables and solve for x and y. y being the height of the dune.

y800+x=tan(20)\displaystyle \frac{y}{800+x}=tan(20)

yx=tan(35)\displaystyle \frac{y}{x}=tan(35)

Now, can you solve for x and y?.
 

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Jakotheshadows said:
Quote:
"While visiting the Sand Dunes National Park in Colorado, Cole approximated the angle of elevation to the top of a sand dune to be 20°. After walking 800 ft closer, he guessed that the angle of elevation had increased by 15°. Approximately how tall is the dune he was observing?"

I first noted that the angle between a hypothetical rod inserted into the highest point in the dune all the way to the dune's base and Cole's original position would have been 70° as the complement to the angle of elevation.

This is true, but I would make my first step: "Let h represent the height of the dune."

I then assumed that the angle of elevation at 800 ft closer to the dune was 35°,

This should not be an assumption! 20 + 15 = 35

and of course its complement is 55°.

This is correct, and you can use this angle to find the length of the hypotenuse of the larger right triangle.

I tried approaching the problem by unsuccessfully figuring the original hypotenuse (or distance from Cole's original position to the top of the dune).

This is a good approach; what went wrong?

The oblique triangle has base length 800 with angles of 20° and 55° at each end. This represents the Angle-Side-Angle (ASA) case.

The ASA case first uses the fact that all angles in a triangle add to 180°; second, the Law of Sines is used to find remaining sides.


Then I tried to figure out how much distance was left between Cole and the center of the dune (in order to make it a right triangle) and failed.

Your first approach is easier.

My next attack was to make a right triangle with the existing angle of elevation ...

The phrase "existing angle" is ambiguous because both angles exist.

... and using 800ft as one of the legs. With a 20/70/90 right triangle, I figured that the triangles other leg would be approx 291.2 ft and the hypotenuse would be approx 851.3 ft.

This approach will get messy. It would be better to create a system of two equations if you want to use the smaller right triangle as a strategy.

My next train of thought brought me to the connection that when Cole got 800 ft closer, the angle of elevation increased by 15°. Assuming that the angle of elevation would increase by 15° for every 800 ft closer that Cole moved towards the dune, ...

This is an incorrect assumption.

.. I mathematically came to the conclusion that from the center of the dune at ground level, Cole would have to dig a tunnel approximately 3,508.9(Cos 55°) ft straight up to get to the top of the dune.

This result is way off.

Hi Jako:

Use the Law of Sines to find the hypotenuse of the large right triangle (ASA Case).

Then, multiply it by sin(20°) to find h.

Cheers,

~ Mark :)
 
galactus said:
We can construct two equations in two variables and solve for x and y. y being the height of the dune.

y800+x=tan(20)\displaystyle \frac{y}{800+x}=tan(20)

yx=tan(35)\displaystyle \frac{y}{x}=tan(35)

Now, can you solve for x and y?.

Tan 20°= y / (800 + x) y = (800 + x) * Tan 20° and x = y / Tan 35°
I've found equations for x and y, but I'm unsure of how to proceed with them. Substitution of one into the other seems to lead only to frustration.

(After much frustration and a little meditation)
substituting my x into my y equation:
y = [800 + (y / Tan 35°)](Tan 20°)
y = 800Tan 20° + (yTan 20° / Tan 35°)
y = 291.2 + .52y
.48y = 291.2
y = 606.7
.... Thank you. That is the approximate solution. Case closed.
 
mmm4444bot said:


Hi Jako:

Use the Law of Sines to find the hypotenuse of the large right triangle (ASA Case).

Then, multiply it by sin(20°) to find h.

Cheers,

~ Mark :)

I don't believe the Law of Sines is introduced until the next chapter in my book. Of course, they never tell you the stuff that makes it all easy until you've ripped all your hair out figuring it out the long way. Your advice is appreciated all the same :)
 
Jakotheshadows said:
galactus said:
We can construct two equations in two variables and solve for x and y. y being the height of the dune.

y800+x=tan(20)\displaystyle \frac{y}{800+x}=tan(20)

yx=tan(35)\displaystyle \frac{y}{x}=tan(35)

Now, can you solve for x and y?.

Tan 20°= y / (800 + x) y = (800 + x) * Tan 20° and x = y / Tan 35°
I've found equations for x and y, but I'm unsure of how to proceed with them. Substitution of one into the other seems to lead only to frustration.

(After much frustration and a little meditation)
substituting my x into my y equation:
y = [800 + (y / Tan 35°)](Tan 20°)
y = 800Tan 20° + (yTan 20° / Tan 35°)
y = 291.2 + .52y
.48y = 291.2
y = 606.7
.... Thank you. That is the approximate solution. Case closed.
Jak, you can really cut that down by cross multiplication:
y = (800 + x)TAN(20)
y = xTAN(35)

So xTAN(35) = (800 + x)TAN(20)
This'll give you x right away, from which you calculate y
 
Denis said:
Jak, you can really cut that down by cross multiplication:
y = (800 + x)TAN(20)
y = xTAN(35)

So xTAN(35) = (800 + x)TAN(20)
This'll give you x right away, from which you calculate y

That way looks much sexier than the way I did it.
 
There's a very good reason for learning the long way first .

Jakotheshadows said:
I don't believe the Law of Sines is introduced until the next chapter in my book. Of course, they never tell you the stuff that makes it all easy until you've ripped all your hair out figuring it out the long way.

Hi Jako:

Two pilots went to flight schools. One pilot learned the long way first; they other pilot did not.

One day, each pilot is flying his own plane, and the same problem arises for both pilots: the controls for the tail rudder stop working.

The pilot who learned the long way first, removes a plate from the floor and manually pulls cables to control the rudder; the other pilot crashes and burns.

So, buy a wig. :lol:

Cheers,

~ Mark
 
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