Find the height of the tree - understanding the task

AI Thread Summary
The discussion focuses on understanding a geometry problem involving a tree and a road with a 7% grade. The key points include the need to visualize the scenario correctly, particularly that the tree is vertical, not tilted. The participants clarify the relationships between the angles and sides of the triangles involved, specifically using the Law of Sines to find the tree's height. It is emphasized that the 500 feet measurement is likely along the road rather than as a horizontal leg of a triangle. The conversation concludes with a reminder that the angle formed by the road's grade cannot be 90 degrees.
Vital
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Homework Statement


Hello!
Please, help me to understand the task - I seem to fail to understand what goes where, and hence cannot proceed to solving the exercise. Please, take a look at the task, and then my questions. The task is on using the law of sines. Before trying to solve it I need to understand the exercise itself.

Homework Equations


Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing

perfectly plumb alongside the road. From a point 500 feet downhill from the tree, the angle

of inclination from the road to the top of the tree is 6. Use the Law of Sines to find the

height of the tree. (Hint: First show that the tree makes a 94 angle with the road.)

The Attempt at a Solution


I would like to draw what's going on with this road. I am attaching a horrific picture (horrific because I draw it on the trackpad, and as it is a pure square it was not easy to draw anything), but at least it gives some idea.

Do I assume correctly that:
(1) 500 feet is the length of the horizontal leg of a right triangle, CB on the picture;
(2) then the trunk of a tree forms a vertical leg of a right triangle, AB on the picture;
(3) A is the top of the tree, so angle of inclination 6 is angle formed by ACQ (C in the middle);
(4) I need to find the angle QCB (with C in the middle).

Are these correct assumptions and a correct understanding of the task?
Thank you!
 

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Vital said:

Homework Statement


Hello!
Please, help me to understand the task - I seem to fail to understand what goes where, and hence cannot proceed to solving the exercise. Please, take a look at the task, and then my questions. The task is on using the law of sines. Before trying to solve it I need to understand the exercise itself.

Homework Equations


Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing

perfectly plumb alongside the road. From a point 500 feet downhill from the tree, the angle

of inclination from the road to the top of the tree is 6. Use the Law of Sines to find the

height of the tree. (Hint: First show that the tree makes a 94 angle with the road.)

The Attempt at a Solution


I would like to draw what's going on with this road. I am attaching a horrific picture (horrific because I draw it on the trackpad, and as it is a pure square it was not easy to draw anything), but at least it gives some idea.

Do I assume correctly that:
(1) 500 feet is the length of the horizontal leg of a right triangle, CB on the picture;
(2) then the trunk of a tree forms a vertical leg of a right triangle, AB on the picture;
(3) A is the top of the tree, so angle of inclination 6 is angle formed by ACQ (C in the middle);
(4) I need to find the angle QCB (with C in the middle).

Are these correct assumptions and a correct understanding of the task?
Thank you!
You have drawn the tree tilted over so it forms a 90 degree angle with the road. The problem says that the tree is "plumb", which means that it is vertical. Can you make a more detailed drawing showing the tree being vertical?
 
berkeman said:
You have drawn the tree tilted over so it forms a 90 degree angle with the road. The problem says that the tree is "plumb", which means that it is vertical. Can you make a more detailed drawing showing the tree being vertical?
Actually, no, the tree on my picture doesn't form a 90 degree with the road. The tree is AB. Road is CQ. And CA is the view point from the road to the top of the tree.
Is this incorrect?
 
Vital said:
...
Do I assume correctly that:
(1) 500 feet is the length of the horizontal leg of a right triangle, CB on the picture;
(2) then the trunk of a tree forms a vertical leg of a right triangle, AB on the picture;
(3) A is the top of the tree, so angle of inclination 6 is angle formed by ACQ (C in the middle);
(4) I need to find the angle QCB (with C in the middle).

Are these correct assumptions and a correct understanding of the task?
Thank you!
For (1) :
It seems more likely that the 500 feet is along the road. That's CQ in your sketch.​

For (2) :
The tree trunk is AQ .​

(3) is fine.

For (4) :
If you are going to use the Law of Sines to get the tree height directly, you should consider using ∠CAQ (It's opposite the 500 foot side) and use the 6° angle, ∠ACQ.
.
 
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SammyS said:
For (1) :
It seems more likely that the 500 feet is along the road. That's CQ in your sketch.​

For (2) :
The tree trunk is AQ .​

(3) is fine.

For (4) :
If you are going to use the Law of Sines to get the tree height directly, you should consider using ∠CAQ (It's opposite the 500 foot side) and use the 6° angle, ∠ACQ.
.
I see. Thank you. I will try it.
 
I am stuck. Don't see a picture.
 
Vital said:
I am stuck. Don't see a picture.
It's the picture you provided.
screen-shot-2017-05-05-at-19-13-45-png.198781.png


AQ is the tree.

CQ is the road (length: 500 ft.).
 
SammyS said:
It's the picture you provided.
View attachment 199293

AQ is the tree.

CQ is the road (length: 500 ft.).
Yes, yes. By saying that I don't see a picture, I was talking about a big picture, metaphorically :) Not about the one I draw :)
Do I understand correctly that if the road CQ has 7% grade, then ∠CQA cannot be 90°?
 
Vital said:
Yes, yes. By saying that I don't see a picture, I was talking about a big picture, metaphorically :) Not about the one I draw :)
Do I understand correctly that if the road CQ has 7% grade, then ∠CQA cannot be 90°?

Well, you can see that for yourself: ∠CBA = 90°, so what does that tell you?
 
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