Hello, aeroflix!
The wording of the problem is confusing.
I'll rewrite it to my interpretation.
A pole leans $\displaystyle 15^o$ from the vertical directly towards the sun.
The angle of elevation of the sun is $\displaystyle 61^o.$
The pole casts a 15-meter shadow on the ground.
Determine the length of the pole. Code:
A o
\ *
\ *
\ *
L \ | *
\15| *
\ | *
\| 61 *
B o - - - - - - - - - - - o C
15
The pole is: $\displaystyle L = AB.$
The shadow is: $\displaystyle BC = 15$ m.
$\displaystyle \angle C = 61^o,\;\;\angle ABC = 105^o$
Hence: .$\displaystyle \angle A \;=\;180^o - 61^o - 105^o \;=\;14^o$
Law of Sines: .$\displaystyle \dfrac{L}{\sin61^o} \:=\:\dfrac{15}{\sin14^o} \quad\Rightarrow\quad L \:=\:\dfrac{15\sin61^o}{\sin14^o}$
Therefore: .$\displaystyle L \;=\;54.22946763 \;\approx\;54.23$ m