Problem: The points A,BA, BA,B, and CCC are on a circle 000. The tangent line at AAA and the secant BCB CBC intersect at PPP. If BC=20B C=20BC=20 and PA=103P A=10 \sqrt{3}PA=103, then PBP BPB equals
Answer Choices:
A. 555
B. 101010
C. 10310 \sqrt{3}103
D. 202020
E. 303030 Solution:
PB‾⋅PC‾=PA2‾.PB‾(PB‾+20)=300;∴PB‾=10\overline{P B} \cdot \overline{P C}=\overline{P A^{2}} . \quad \overline{P B}(\overline{P B}+20)=300 ; \quad \therefore \overline{P B}=10PB⋅PC=PA2.PB(PB+20)=300;∴PB=10.