Answer:
(a)x=128 degrees
(b)[tex]\angle APD\neq 116^\circ[/tex]
Step-by-step explanation:
[tex]\angle ABD+\angle CBD =180^\circ$ (Linear Postulate)\\\angle ABD+116^\circ =180^\circ\\\angle ABD=180^\circ-116^\circ\\\angle ABD=64^\circ[/tex]
(a) Now:
[tex]\angle AOD=2 \times \angle ABD$ (Angle at the centre is twice the angle at the circumference)\\x=2 \times 64^\circ\\x=128^\circ[/tex]
(b)
[tex]\angle APD=2 \times \angle ABD$ (Inscribed Angle Theorem)\\\angle APD=2 \times 64^\circ\\ \angle APD=128^\circ[/tex]
[tex]\angle APD\neq 116^\circ[/tex]
