# Finding Height of Mountain Using Direction Cosines

• Uninspired
In summary, 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) We're trying to find the height of the mountain, the top of the mountain is point P. Direction cosines: Rap:cos theta x= .5179cos theta y= .6906cos theta z= .5048Rbp:cos theta x=-.3743cos theta y=.7486cos theta z

#### Uninspired

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:
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

• untitled.bmp
95.1 KB · Views: 511
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.

Last edited: