This helps...
http://www.mrpk.org/Portals/11/usersdat ... 0TABLE.pdf
JDC wrote:tan(45)=1
150*1=150
JDC wrote:tan(10) = .176327
150*.176327 = 26.44905
150-26.44905 = 123.551
am I even close?
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rocket path:
l *. v
a / | ` . i
u / | ` . e
n / |h ` . w
c /45* | 10* ` . e
h *---------*------------------* r
|<===========150============>|
Since the triangle on the launch side is 45 degrees then the base is h too. Then the base of the viewer's triangle is 150-h.
triangle for A:
l A*. v
a / | ` . i
u / | ` . e
n / |h ` . w
c /45* | 10* ` . e
h *---------*------------------* r
| h | 150-h |
|<===========150============>|
So do the tangent of the 10 degrees & make it equal to h/(150-h).
JDC wrote:Is there a quicker way of doing this?
JDC wrote:Sorry, but, I could not understand the solution you tried to walk me through, would you mind posting something similar?
triangle for A:
l A*. v
a / | ` . i
u / | ` . e
n / |h ` . w
c /45* | 10* ` . e
h *---------*------------------* r
| h | 150-h |
|<===========150============>|
triangle for A:
l A*. v
a / | ` . i
u / | ` . e
n / |h ` . w
c / L* | V* ` . e
h *---------*------------------* r
| x | y |
|<===========d==============>|