Triangles in 3 directions?

Question

A perfectly vertical, level wall is 14 degrees high 10 degrees to the right of the closest point. How many degrees high is the highest part of the wall?

All references are as seen from X.

top view

A'________B'
|_|................./
|.................../
|................../
|................./
|................/
|.............../
|............../
|............./
|............/
|.........../
|........../
|........./
|......../
|......./
|10º/
|..../
|.../
|../
|./
|/
X observer



left view

B
|..\
|...\
|....\
|76º\
|.......\
|........\
|.........\
|..........\
|...........\
|_..........\
|_|____ \
B'...........X, observer



front view, as seen from X

___A______B________
........|_|.........|_|..................
........|...............|..................
........|_.........._|..................
........|_|____|_|..................
.......A'........... B'................
.........................................

What is angle A-X-A'?

.
A' is the closest point of the wall to X
.

Answer 1

Let the distance between A' and X be d. Then the distance between B' and X is d / cos 10°. Since the elevation of the wall in this direction is 14°, the height of the wall is (d / cos 10°) tan 14°.
So, the angle of elevation θ at the closest point is given by
tan θ = ((d / cos 10°) tan 14°) / d
So θ = arctan(tan 14° / cos 10°)
= 14.2° to the nearest 0.1°.