Finding Height of Mountain Using Direction Cosines

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Homework Help Overview

The discussion revolves around determining the height of a mountain using direction cosines and the coordinates of two points situated on the x-axis. The original poster provides specific altitude measurements and direction cosines for two reference points, A and B, which are 10,000 m apart.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the geometric relationships between points A, B, and the mountain peak P, considering the construction of right triangles and the role of direction cosines. Questions arise regarding the visualization of the problem and the significance of the y-axis in the context.

Discussion Status

There is ongoing exploration of the problem with various suggestions for visualizing the scenario, including drawing 3D representations and considering vector relationships. Some participants express uncertainty about specific terms and the overall setup, indicating a lack of consensus on the approach.

Contextual Notes

Participants note the absence of information regarding the y-axis and question how the provided direction cosines relate to the height calculation. The original poster expresses frustration with the progress made thus far.

Uninspired
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We're trying to find the height of a mountain.
First we have the measured altitude of 2 points at 3000 m above sea level.
Then, these two points are 10,000 m apart on the same axis(x axis, z axis is pointed up)

point A--> (0,0,0)
point B-->(10000,0,0)

We're trying to find the height of the mountain, the top of the mountain is point P.

Direction cosines:

Rap:
cos theta x= .5179
cos theta y= .6906
cos theta z= .5048

Rbp:
cos theta x=-.3743
cos theta y=.7486
cos theta z=.5472

Anybody? I've been stuck on this for a LONG time.

edit: well let me show what I've done so far (isn't much)

all I've written out is:

rapx/rap=.5179
rapy/rap=.6906
etc

but where do i go from here? I've drawn out triangles for each APx, APy, APz and same for BP, but none of this tells me anything or let's me continue. Nor do i know how the 10,000 plays a role in the equations other than |Rap| + |Rbp| =10,000m
 
Last edited:
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I'm not entirely sure I'm visualizing this correctly, or where the y-axis come into this because you didn't mention the y axis. Furthermore I'm not sure what you mean by Rap and Rbp...but

If you have two points on opposite sides of the mountain that are 10,000m apart, can you construct two right triangles?

Let A equal the point at (0, 0, 0)
Let B equal the point at (10,000, 0, 0)
Let P equal the point at the top of the mountain
Let Q equal a point on the line AB
Let the line PQ intersect the line AB at a right angle (aka, drops straight down from P)

You have two right triangles,
PQA
PQB

They both have the side PQ in common, so this side is equal for both right triangles...
 
i drew it out
 

Attachments

Are points A, B, and P all in the xz plane?
 
Severian596 said:
Are points A, B, and P all in the xz plane?
A and B are on the x axis, and P is xyz
 
anyone else? pleeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeease?
 
bizzump ttt
 
You could draw it out in 3D.

Draw out the x-, y-, and z-axes and plot the point P using the direction cosines.
Since you have direction cosines for two separate points, A and B, the vectors from these points, AP and BP, will intersect at P.

Once you have drawn out the 3D sketch, use trig to work out the height of P from the x-y plane.

Alternatively, use the direction cosines to write out the point P as a vector, AP, and as a vector AP=AB + BP.
Solve using vector algebra.
 
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