SOLUTION
The angles at points L and O make up a straight line.
These angles are
![65^o,35^o,51^o\text{ and angle LOQ}](https://img.qammunity.org/2023/formulas/mathematics/high-school/m5cz5dm4i1aljmoafdh0pwasiudqtlbltq.png)
Angles on a straight line = 180 degrees. So
![\begin{gathered} 65+35+51+angleLOQ=180^o \\ 151+\text{ angle LOQ = 180} \\ \text{LOQ = 180 - 151 = 29}^o \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/6whfn963tt75eynvmt0dwiuq96l3wwazom.png)
Therefore, angle LOQ = 29 degrees
Angle OPQ at point P is opposite to the angle at point L.
The angle at point L = 65 + 35 = 100 degree
Opposite angle of a parallelogram are equal.
Therefore, angle OPQ = 100 degrees
Angle OPL is alternate to angle PLQ. And angle PLQ = 65 degrees
Alternate angles are always equal.
Therefore, angle OPL = 65 degrees
Before we find LQP, let's find LOP.
Recall that LOQ = 29 degrees. So, LOP = 29 + 51 = 80 degrees
LQP is opposite to LOP. Since opposite angles of a parallelogram are equal,
Therefore, LQP = 80 dgrees.
Angle LPQ is alternate to angle OLP. OLP = 35 degrees
Since alternate angles are equal,
Angle LPQ = 35 degrees