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Points A & B Out At Sea

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xyz_1965

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Jul 26, 2020
81
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?
 

skeeter

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MHB Math Helper
Mar 1, 2012
658
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 $h$.
 
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xyz_1965

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Jul 26, 2020
81
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 $x_A$.
 

skeeter

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MHB Math Helper
Mar 1, 2012
658
$x_B$ 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.
 
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xyz_1965

Member
Jul 26, 2020
81
$x_B$ 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.
 

Country Boy

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MHB Math Helper
Jan 30, 2018
428
A few comments
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.
 
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xyz_1965

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Jul 26, 2020
81
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.
 

Country Boy

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MHB Math Helper
Jan 30, 2018
428
That is true. And you learn by having those errors pointed out!
 
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xyz_1965

Member
Jul 26, 2020
81
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.