Solution:
Given:
To solve the angle, we use the theorem of angles in a cyclic quadrilateral.
The theorem states that the opposite angles in a cyclic quadrilateral are supplementary (i.e. they add up to 180°).
Hence,
[tex]\begin{gathered} \angle A+\angle C=180^0 \\ \angle B+\angle D=180^0 \end{gathered}[/tex]
Thus,
[tex]\begin{gathered} \angle A+\angle C=180^0 \\ (x+2)+(x-2)=180^0 \\ x+x+2-2=180 \\ 2x=180 \\ x=\frac{180}{2} \\ x=90^0 \end{gathered}[/tex]
Hence,
[tex]\begin{gathered} \angle A=x+2=90+2 \\ \angle A=92^0 \\ \\ \angle C=x-2=90-2 \\ \angle C=88^0 \\ \\ \angle D=x-10=90-10 \\ \angle D=80^0 \\ \\ \text{Also, } \\ \angle B+\angle D=180^0 \\ \angle B+80=180 \\ \angle B=180-80 \\ \angle B=100^0 \end{gathered}[/tex]
Therefore, the measure of each angle of the quadrilateral is;
[tex]\begin{gathered} \angle A=92^0 \\ \angle B=100^0 \\ \angle C=88^0 \\ \angle D=80^0 \end{gathered}[/tex]
