Connect BC, because AB is the diameter of a circle, so the triangle ABC is a right triangle.
In the right triangle ABC, AB=2 and AC= root number 3, so the angle CAB=30 degrees.
Note that PC=AC, so the angle CPA=30 degrees and the angle ACP= 120 degrees.
OA and OC are both radii of a circle, so they are equal, so angle OCA= angle OAC=30 degrees.
Therefore, angle OCP= angle ACP- angle OCA= 120 -30 degrees =90 degrees,
So PC is tangent to circle O.
2. Solution:
Note that angle CAB= angle ABD=30 degrees, so angle BDP= angle D-30 degrees.
The other angle ABD= angle BDP+ angle APD, so angle APD=60 degrees-angle D.
In the right triangle CDP, the sine and cosine of the angle d can be found.
In this way, the sine value of the angle APD can be finally obtained as (root number 2 1)/ 14.