Finding height using the direction cosines

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SUMMARY

The discussion focuses on calculating the height of a point (P) on a mountain using direction cosines from two points (A and B) that are 10,000 meters apart. The direction cosines provided are cos θx, cos θy, and cos θz for both points A and B. The solution involves constructing right-angled triangles and applying trigonometric relationships, specifically using the sine rule to find angles and subsequently the height of point P. The key equations utilized include AD = AP*cos(α) and AE = AP*cos(β), with the final goal of determining PO using the known values of AO and tan(γ).

PREREQUISITES
  • Understanding of direction cosines in three-dimensional geometry
  • Knowledge of trigonometric functions and their applications in right-angled triangles
  • Familiarity with the sine rule for triangle calculations
  • Ability to interpret and manipulate geometric relationships in 3D space
NEXT STEPS
  • Study the application of direction cosines in 3D geometry problems
  • Learn advanced trigonometric identities and their use in solving height-related problems
  • Explore the sine rule and its derivation for triangle calculations
  • Practice solving similar problems involving height measurements using trigonometry
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Students in mathematics or engineering fields, particularly those studying geometry and trigonometry, as well as professionals involved in surveying or geographical measurements.

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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.
 
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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.
 

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