ANSWER:
[tex] \large \boxed{\boldsymbol{x = 15^{\circ}}} [/tex]
SOLUTION:
[tex] \angle CPD [/tex] is an exterior angle of [tex] \odot\ O [/tex]. Note that the measure of an exterior angle of a circle is equal to half the difference of its intercepted arcs. In the figure shown, we have
[tex] \bold{Exterior\ Angle:}\ m\angle CPD = x [/tex]
[tex] \bold{Intercepted\ Arcs:}\ \begin{cases} m\overset{\frown}{BC} = (x + 10)^{\circ} \\ m\overset{\frown}{CD} = (5x - 20)^{\circ}\end{cases} [/tex]
Now, we can solve for [tex] x. [/tex]
[tex] \begin{array}{l} m\angle CPD = \dfrac{m\overset{\frown}{CD} - m\overset{\frown}{BC}}{2} \\ \\ x = \dfrac{5x - 20 - (x + 10)}{2} \\ \\ 2x = 4x - 30 \\ \\ 2x = 30 \\ \\ \boxed{\boldsymbol{x = 15^{\circ}}} \end{array} [/tex]
