(i)

(ii)

SOLUTION:

(i) According to the given question,

\[\angle AOB\text{ }=\text{ }2\angle AOB\text{ }=\text{ }2\text{ }x\text{ }{{50}^{o}}~=\text{ }{{100}^{o}}\]

[Angle at the center is double the angle at the circumference subtend by the same chord]

Also, \[OA\text{ }=\text{ }OB\]

\[\angle OBA\text{ }=\angle OAB\text{ }=\text{ }c\]

\[c\text{ }=\text{ }({{180}^{o}}-\text{ }{{100}^{o}})/\text{ }2\text{ }=\text{ }{{40}^{o}}\]

(ii) According to the given question,

We have, \[\angle APB\text{ }=\text{ }{{90}^{o}}~\][Angle in a semicircle]

\[\angle BAP\text{ }=\text{ }{{90}^{o}}-\text{ }{{45}^{o}}~=\text{ }{{45}^{o}}\]

Now, \[d\text{ }=\angle BCP\text{ }=\angle BAP\text{ }=\text{ }{{45}^{o}}\]

[Angles subtended by the same chord on the circle are equal]