To solve this problem we will use the Pythagorean theorem twice:
[tex]c^2=a^2+b^2,[/tex]
where c is the length of the hypotenuse and, a, and b are the lengths of the legs.
Now, from the triangle OPD, we get ( we will omit the units to simplify the calculations):
[tex]OP^2=1.2^2+OD^2.[/tex]
Notice that OD is a radius of the circle, therefore OD=1, and OP=OC+CP=1+CP, then we can rewrite the above equation as:
[tex]\begin{gathered} 1+CP=\sqrt{1.2^2+1^2}, \\ CP=\sqrt{\frac{11}{5}}-1\approx0.56. \end{gathered}[/tex]
Analogously, we get that:
[tex]\begin{gathered} AP^2=AD^2+1.2^2=2^2+1.2^2, \\ AB+BP=1.7+BP=\sqrt{4+1.2^2}, \\ BP=\sqrt{4+1.2^2}-1.7\approx0.63. \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} Distance(P,C)\approx0.56\text{ }in, \\ Distance(P,B)\approx0.63\text{ in.} \end{gathered}[/tex]
Answer:
[tex]Point\text{ C.}[/tex]
