Draw a triangle with A on the West end and B on the East. The third point (under the UFO) is C. Drop a perpendicular from C onto AB and label that point D. this divides the base into AD and DB.
1) CAD = 60 deg (given)
2) CBD = 45 deg (given)
3) ADC = CDB = 90 deg (by construction)
4) ACD = 30 deg (all triangles have 180 deg)
5) DCB = 45 deg (all triangles have 180 deg)
6) CD = AD/tan 30 (definition of tangent)
7) CD = DB = BC/(sqrt 2) (property of 45-45 right triangle)
8) AD + DB = 100 (by inspection)
9) AD + CD = 100 (By substitution, since CD = DB)
10) AD + AD/(tan 30) = 100 (Substitution from #6)
11) Solve #10 for AD
12) DB = CD = 100 - AD (# 7 and 8)
Now solve for the lengths of the bearing shots in the xy plane
BC = DB(sqrt 2) = 89.66' (# 7)
AC = sqrt (AD^2+CD^2) = 73.21' (Pythagoras)
The elevation of the "UFO" is
ELEV = 73.21 (tan 82)
OR
ELEV = 89.66 (tan 80)
I got a degree in math in 1971 but it took me 4 hours to figure this out.
Enzyme breakdown is not a pretty sight.
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high school math
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