Question 1201107
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Notation like 40°30' is read out as "40 degrees 30 minutes".
"minutes" refers specifically to <a href = "https://en.wikipedia.org/wiki/Minute_and_second_of_arc">arc minutes</a>


There are 60 arc minutes in a full degree.
30 arc minutes is 30/60 = 1/2 = 0.5 of a degree.
15 arc minutes is 15/60 = 1/4 = 0.25 of a degree.


Therefore,
40 degrees 30 minutes = 40.5 degrees 
37 degrees 15 minutes = 37.15 degrees 
which represent the angles of elevation.


This is what it looks like from a birds eye view
{{{
drawing(400,400,-6,7,-5,5,
blue(
line(-5,0,-4.75,0),line(-4.5,0,-4.25,0),line(-4,0,-3.75,0),line(-3.5,0,-3.25,0),line(-3,0,-2.75,0),line(-2.5,0,-2.25,0),line(-2,0,-1.75,0),line(-1.5,0,-1.25,0),line(-1,0,-0.75,0),line(-0.5,0,-0.25,0),line(0,0,0.25,0),line(0.5,0,0.75,0),line(1,0,1.25,0),line(1.5,0,1.75,0),line(2,0,2.25,0),line(2.5,0,2.75,0),line(3,0,3.25,0),line(3.5,0,3.75,0),line(4,0,4.25,0),line(4.5,0,4.75,0)
),

line(0,0,-3,-3),
line(0,0,4,-4),
line(-3,-3,4,-4),
line(5,4,5,2),
line(4,3,6,3),

locate(4.88,4.52,"N"),
locate(6.28,3.2,"E"),
locate(4.94,1.76,"S"),
locate(3.54,3.22,"W"),
locate(-4,4,matrix(1,3,"Birds","eye","view")),
locate(-4,3.5,matrix(1,3,"Not","to","scale")),

locate(-3.78,-2.8,"A"),
locate(-0.16,0.5,"C"),
locate(4.26,-3.68,"B"),
locate(-0.62,-3.68,matrix(1,2,"75","m")),
locate(-1.62,-0.34,25^o),
locate(1.06,-0.34,23^o)
)
}}}
A = location of the surveyor
B = second observation location
C = location of the tower
Notation like W 25° S means "aim directly west, then turn 25° toward the south"
The blue dashed line in the diagram above helps us set up the angles of 25° and 23°


Refer to this page for a few other examples
<a href = "http://academic.brooklyn.cuny.edu/geology/leveson/core/linksa/comp.html">http://academic.brooklyn.cuny.edu/geology/leveson/core/linksa/comp.html</a>


Also, refer to this problem involving similar compass bearings notation.
<a href = "https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1200155.html">https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1200155.html</a>


Angle ACB is sandwiched between the 25 and 23 degree angles.
Those three angles add to 180.
25+(angleACB)+23 = 180
angleACB + 48 = 180
angle ACB = 180-48
angle ACB = 132


Let's update the drawing with that angle.
{{{
drawing(400,400,-6,7,-5,5,
blue(
line(-5,0,-4.75,0),line(-4.5,0,-4.25,0),line(-4,0,-3.75,0),line(-3.5,0,-3.25,0),line(-3,0,-2.75,0),line(-2.5,0,-2.25,0),line(-2,0,-1.75,0),line(-1.5,0,-1.25,0),line(-1,0,-0.75,0),line(-0.5,0,-0.25,0),line(0,0,0.25,0),line(0.5,0,0.75,0),line(1,0,1.25,0),line(1.5,0,1.75,0),line(2,0,2.25,0),line(2.5,0,2.75,0),line(3,0,3.25,0),line(3.5,0,3.75,0),line(4,0,4.25,0),line(4.5,0,4.75,0)
),

line(0,0,-3,-3),
line(0,0,4,-4),
line(-3,-3,4,-4),
line(5,4,5,2),
line(4,3,6,3),

locate(4.88,4.52,"N"),
locate(6.28,3.2,"E"),
locate(4.94,1.76,"S"),
locate(3.54,3.22,"W"),
locate(-4,4,matrix(1,3,"Birds","eye","view")),
locate(-4,3.5,matrix(1,3,"Not","to","scale")),

locate(-3.78,-2.8,"A"),
locate(-0.16,0.5,"C"),
locate(-0.46,-0.5,132^o),
locate(4.26,-3.68,"B"),
locate(-0.62,-3.68,matrix(1,2,"75","m")),
locate(-1.62,-0.34,25^o),
locate(1.06,-0.34,23^o)
)
}}}


Here's a 3D look at what's going on.
Pay close attention to the points A,B,C how they are laid out. 
Compare their relative locations to the previous diagram shown above.
{{{
drawing(400,400,-6,7,-5,5,

blue(
line(2.2,-1.2,3,-2),line(3,-2,2.6722,-1.9122),line(3,-2,2.9122,-1.6722),
locate(-2,-0.5,40.5^o),
locate(1.2,-0.5,37.25^o),
line(-1.7,-1.1,-2.2,-1.6),line(-2.2,-1.6,-2.1451,-1.3951),line(-2.2,-1.6,-1.9951,-1.5451)
),

line(-3,-2,4,-3),
line(4,-3,1,-1),
line(1,-1,1,3),
line(4,-3,1,3),
line(-3,-2,1,3),
line(-3,-2,1,-1),
locate(0,-2.5,matrix(1,2,"75","m")),
locate(1.26,0.82,"x"),

locate(-3.32,-1.64,"A"),
locate(4.14,-2.68,"B"),
locate(0.5,-0.7,"C"),
locate(0.5,-1.2,132^o),
locate(0.9,3.34,"D"),

locate(-4,4,matrix(1,2,3D,"view")),
locate(-4,3.5,matrix(1,3,"Not","to","scale"))

)
}}}
triangle ABC is along the flat ground
A = location of the surveyor
B = second observation location
C = base of the tower
D = top of the tower
x = height of the tower = length of segment CD
angle ACB = 132 degrees
angle CAD = 40.5 degrees
angle CBD = 37.25 degrees


Here is the side view of right triangle ACD
{{{
drawing(400,400,-6,7,-5,5,

blue(locate(-2.2,-1.2,40.5^o)),

line(1,-2,1,3),
line(1,3,-3,-2),
line(-3,-2,1,-2),

locate(1.26,0.82,"x"),

locate(-3.32,-1.64,"A"),
locate(1.2,-1.7,"C"),
locate(0.9,3.34,"D"),

line(1,-1.7,0.7,-1.7),
line(0.7,-1.7,0.7,-2),

locate(-4,4,matrix(1,2,"Side","view")),
locate(-4,3.5,matrix(1,3,"Not","to","scale"))

)
}}}


Here is the side view of right triangle BCD
{{{
drawing(400,400,-6,7,-5,5,
blue(locate(1.8,-1.2,37.25^o)),

line(1,-2,1,3),
line(1,3,4,-2),
line(4,-2,1,-2),

locate(0.65,0.8,"x"),
locate(4,-2,"B"),
locate(0.65,-1.9,"C"),
locate(0.9,3.34,"D"),

line(1,-1.7,1.3,-1.7),
line(1.3,-1.7,1.3,-2),

locate(-4,4,matrix(1,2,"Side","view")),
locate(-4,3.5,matrix(1,3,"Not","to","scale"))

)
}}}


At this point I get stuck. 
I don't think there's enough information to determine x.
If we knew the length of segment AC or the length of segment BC, then we could use trigonometry to determine x.
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