(Geometry) The figure shows that (BP) is a line passing the center O and (PT) is a tangent line to the circle at point T

Solution:

Since PT is tangent to the circle, we have \(\angle\ OTP\ =\ 90^o\ \ and\ \ \angle\ AOT\ =\ 180^o\ -\ 20^o-\ 90^o=\ 70^o\) .

Since the triangle is an isosceles triangle with AO = OT, we have \(\angle\ OTA\ =\ \angle\ OAT\ \ =\ \angle\ x\ \)

Thus \(\ \angle\ x\ =\ \frac{180^o-70^o}{2}\ =55^o\) .

Therefore, D is the correct answer.

Answer: D