Calculating Height and Distance of a Mountain at Sea

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Discussion Overview

The discussion revolves around calculating the height of a mountain and the distances from two points at sea to the base of the mountain using trigonometric principles. Participants explore the geometry of the situation, involving angles of elevation and the distance between the two points.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes using two triangles to find the height of the mountain, defining variables for the height and distances involved.
  • Another participant provides a mathematical formulation for the distances from points A and B to the base of the mountain, suggesting a specific equation to calculate the distance from point B.
  • There are clarifications about the definitions of variables, with some participants emphasizing the importance of correctly identifying distances versus points.
  • Some participants express uncertainty about the correctness of earlier statements and suggest alternative approaches to the problem.
  • One participant acknowledges the learning process through mistakes and the value of respectful corrections in discussions.

Areas of Agreement / Disagreement

Participants generally express differing views on the best approach to solving the problem, with no consensus on a single method or solution. Some participants agree on the need for clarification in definitions, while others challenge earlier claims without reaching a definitive conclusion.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the relationships between the defined variables and the angles of elevation. The discussion reflects varying interpretations of the problem setup and the application of trigonometric principles.

Who May Find This Useful

This discussion may be useful for individuals interested in trigonometry, geometry, and problem-solving in mathematical contexts, particularly in applications involving angles of elevation and distances.

xyz_1965
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From points A to B at sea, the angles of elevation to the top of the mountain T are 37 degrees and 41 degrees, respectively. The distance between points A to B is 80 meters.

1. Find height of the mountain.

2. Find the distance from point A to the bottom of the mountain.

I think making two triangles makes sense here.

Triangle 1

tan(41°) = h/x

Let h = height of mountain.

Let x = the distance from point B out at sea to the bottom of the mountain.

Solving for x, I get h/tan(41°).

Triangle 2

tan(37°) = h/[80 + (h/tan(41°)]

I need to solve triangle 2 for h, the height of the mountain. After finding h, I can then find the distance between point B out at sea and the bottom of the mountain.

Lastly, the distance from point A out at sea to the bottom of the mountain is found by adding A + B.

1. Is any of this right?

2. Is there an easy way to solve
tan(37°) = h/[80 + (h/tan(41°)] for h?
 
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assuming A, B, and the mountain top are located in the same vertical plane ...

$h = x_B \tan(41)$

$h = x_A \tan(37)$

$x_A - x_B = 80 \implies x_A = x_B +80$

$x_B \tan(41) = (x_B+80)\tan(37)$

$x_B = \dfrac{80\tan(37)}{\tan(41)-\tan(37)}$

from here, calculate $x_B$, add 80 to get $x_A$, and use either value in one of the original equations to find $h$.
 
skeeter said:
assuming A, B, and the mountain top are located in the same vertical plane ...

$h = x_B \tan(41)$

$h = x_A \tan(37)$

$x_A - x_B = 80 \implies x_A = x_B +80$

$x_B \tan(41) = (x_B+80)\tan(37)$

$x_B = \dfrac{80\tan(37)}{\tan(41)-\tan(37)}$

from here, calculate $x_B$, add 80 to get $x_A$, and use either value in one of the original equations to find $h$.

Explain $x_B$ and $x_A$.
 
$x_B$ is the horizontal distance from point B to the point directly below the mountain top.

$x_A$ is the horizontal distance from point A to the point directly below the mountain top.
 
skeeter said:
$x_B$ is the horizontal distance from point B to the point directly below the mountain top.

$x_A$ is the horizontal distance from point A to the point directly below the mountain top.

Thank you for clearing this up for me.
 
A few comments
xyz_1965 said:
From points A to B at sea, the angles of elevation to the top of the mountain T are 37 degrees and 41 degrees, respectively. The distance between points A to B is 80 meters.

1. Find height of the mountain.

2. Find the distance from point A to the bottom of the mountain.

I think making two triangles makes sense here.

Triangle 1

tan(41°) = h/x

Let h = height of mountain.

Let x = the distance from point B out at sea to the bottom of the mountain.
It would make more sense to say these before you say "tan(41)= h/x!

Solving for x, I get h/tan(41°).
Specifically, x= h/tan(41).

Triangle 2

tan(37°) = h/[80 + (h/tan(41°)]
I think I would have first said "Let y be the distance from B to the base of the mountain" to get tan(37)= h/y and then use y- x= 80 so that y= 80+ x but that becomes the same thing.

I need to solve triangle 2 for h, the height of the mountain. After finding h, I can then find the distance between point B out at sea and the bottom of the mountain.
Yes.

Lastly, the distance from point A out at sea to the bottom of the mountain is found by adding A + B.
No! You were doing so well and then you forgot how you had defined "A" and "B'! "A" and "B" are points, not distances so you can't add them. "x" was the distance from point A to the base of the mountain.
1. Is any of this right?

2. Is there an easy way to solve
tan(37°) = h/[80 + (h/tan(41°)] for h?
A= h/(B+h/C)

Get rid of the fraction by multiplying both sides by B+ h/C:
AB+ (A/C)h= h

AB= h- (A/C)h= Ch/C- Ah/C= (C- A)h/C

h= ABC/(C- A)

Of course A= tan(37°)= 0.7536, approximately, B= 80, and C= tan(41°)= 0.8693 approximately.
 
Country Boy said:
A few comments

It would make more sense to say these before you say "tan(41)= h/x!Specifically, x= h/tan(41).I think I would have first said "Let y be the distance from B to the base of the mountain" to get tan(37)= h/y and then use y- x= 80 so that y= 80+ x but that becomes the same thing. Yes. No! You were doing so well and then you forgot how you had defined "A" and "B'! "A" and "B" are points, not distances so you can't add them. "x" was the distance from point A to the base of the mountain.

A= h/(B+h/C)

Get rid of the fraction by multiplying both sides by B+ h/C:
AB+ (A/C)h= h

AB= h- (A/C)h= Ch/C- Ah/C= (C- A)h/C

h= ABC/(C- A)

Of course A= tan(37°)= 0.7536, approximately, B= 80, and C= tan(41°)= 0.8693 approximately.

I thank you for the break down. It's ok to make mistakes here. Learning is not possible without errors.
 
That is true. And you learn by having those errors pointed out!
 
Country Boy said:
That is true. And you learn by having those errors pointed out!

I agree. This is why I don't mind being corrected as long as it is done respectfully.
 

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