Finding height using the direction cosines

[janahan]

Sorry for the re-post

Homework Statement

http://books.google.ca/books?id=nYR...g direction cosines to measure height&f=false

The obove link has a coopy of the question as it is hard to describe.

If you find the link is too hard to read here is the data
Direction cosines:

of Rap(from point a to top of mountain):
cos theta x= .5179
cos theta y= .6906
cos theta z= .5048

of Rbp(from point b to top of mountain):
cos theta x=-.3743
cos theta y=.7486
cos theta z=.5472

b and a are 10000m apart. Find how high point p is.
It is on page 57 #2.82 (The mount everest question)

Homework Equations

The Attempt at a Solution

I attemted the question in a number of ways but can't seem to come to an answer. I tried Looking at it as two right angled triangles and using trig. I am not really sure what i can do with the direction cosines either.

 

[rl.bhat]

Draw a perpendicular PO on the base of the mountain. Let PD, PE and PF on x, y and z axis passing through A.
AD = AP*cos(α)
AE = AP*cos(β)
[cos(α) = 0.5179, cos(β) = 0.6906]
Now tan(OAD) = AE/AD. Find angle OAD. Similarly find angle OBD. from these two angles find the third angle AOB. Using sine rule in the triangle AOB, find Ao and BO.
cos(γ) = 0.5048 is given. From that find tan(γ) which is equal to PO/AO. AO is known. Find PO.