For the Geometrically Proficient...?

Question

Two segments tangent to a circle form a 60 angle where they meet at pt. P. Each segment has a length of 8 in. How far is P from the center of the circle???

Answer 1

A line from P to the center divides the 60 into two 30 degree angles. Since a radius is perpendicular to the tangent, you have a 30-60-90 triangle. Opposite the 60 is the tangent 8 inches,

x sq rt 3 = 8

x = 8/ sq rt 3 = 8 sq rt 3/3

Distance to center = 2x = 16 sq rt 3/3

Answer 2

Another way you could do it is extend both segments 8 inches away from P, then connect their endpoints, so that your circle is inscribed in an equilateral triangle. The center of the circle is now also the intersection of the 3 altitudes of the triangle. And conveniently, with equilateral triangles, that intersection is always 2/3 of the altitude's length away from the vertices. So, split the triangle in half and use a 30-60-90 to determine the altitude length to be 8sqrt3, and then the distnce of the center from each vertex is 16/3sqrt3. Since P _is_ one of the vertices, the distance from P to the center is about 9.238.

So that turned out to be way more trouble than it was worth, but still. That's what you could do.

Answer 3

We can calculate the cord determinated by the points of contact
T and T´ which is 8 as the triangle PTT´is equilateral
If M is the midpoint of the cord OTM is half of a equilateral triangle and the radius is
TM = r sqrt3/2 so r = 8/sqrt3 and MO = sqrt(64/3-16 )=4/sqrt3 =4sqrt3 /3
PM = 8sqrt3/2
so
PO = 8sqrt3/2 +4sqrt3 /3 = 16/3 sqrt3
I did not use trigonometry