Respuesta :

Answer:

[tex]p=\frac{4\sqrt{10}}{5}[/tex]  

Step-by-step explanation:

step 1

Find the radius of the circle

Remember that AB is a tangent to the circle at point B

so

AB is perpendicular to OA

The radius of the circle is equal to the segment OA

In the right triangle OAB

[tex]sin(30\°)=\frac{OA}{OB}[/tex]

[tex]OA=(OB)sin(30\°)[/tex]

substitute the values

[tex]OA=(16)0.5=8\ units[/tex]

step 2

Find the value of p

In the right triangle OPQ

see the attached figure to better understand the problem

Applying the Pythagoras Theorem

[tex]OP^2=OQ^2+PQ^2[/tex]

substitute the values

Remember that OP is the radius

[tex]8^2=p^2+(3p)^2[/tex]

[tex]64=p^2+9p^2[/tex]

[tex]64=10p^2[/tex]

[tex]p^2=\frac{64}{10}[/tex]

[tex]p=\frac{8}{\sqrt{10}}[/tex]

simplify

[tex]p=\frac{4\sqrt{10}}{5}[/tex]  

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