Answer:
last option
Step-by-step explanation:
To prove that ΔEFG is also a right triangle, you must prove that KL = EF so that in ΔKLM c² = a² + b² which would make ΔEFG a right triangle.
1. Angle PAB is 90 degrees, as it is formed from the tanget to the circle at A, and the radius drawn to A.
2. AB=BC, because tangents drawn to a circle from the same point are equal.
3. PB is Common, so by the side-side-side congruence postulate, triangles ABP and CBP are congruent.
4. So measure of m(BPA)=x/2 and m(ABP)=73/2.
5.



, x= 107 degrees.
5/12 - 4/5 = -23/60. You can't simplify -23/60. Hope this is right