Hello, euclid2!
Did you make a sketch?
Find the horizontal and vertical components of a force of 75N
that acts in a direction forming an angle of 51 degrees with the vertical.
In the diagram, "d" represents "degrees".
Code:
| o A
| * :
| 75 * :
| * :
|51d :
| * 39d :
O * - - - - - + - -
| B
In right triangle $\displaystyle ABO\!:\;\;O\!A = 75,\;\angle AOB = 39^o$
The horizontal component is $\displaystyle OB.$
The vertical component is $\displaystyle AB.$
Therefore: .$\displaystyle \begin{Bmatrix} \cos39^o \:=\:\dfrac{OB}{75} & \Rightarrow & OB \;=\;75\cos39^o \;\approx\;58.3 \\ \\[-3mm]
\sin39^o \:=\:\dfrac{AB}{75} & \Rightarrow & AB \:=\:75\sin39^o \;\approx\;47.2 \end{Bmatrix} $