SOLUTION: A 120º arc of a circle has end points (√6,0) and (0, -√6). How long is radius of this circle (in simplest form).

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: A 120º arc of a circle has end points (√6,0) and (0, -√6). How long is radius of this circle (in simplest form).       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1004217: A 120º arc of a circle has end points (√6,0) and (0, -√6). How long is radius of this circle (in simplest form).
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!

Points A%28sqrt%286%29%2C0%29 and B%280%2C-sqrt%286%29%29 both are at distance sqrt%286%29 from O%280%2C0%29 , the origin.
So, triangle ABO is a right isosceles triangle, with 45%5Eo angles at a and B .
Points A and B are on the same arc of a circle of radius r ,
so both are at distance r from the center of the circle, C .
Since the center is at the same distance from A and B ,
the center is on the perpendicular bisector of AB ,
which contains M , the midpoint of AB .
The perpendicular bisector of the base of an isosceles triangle passes through the vertex,
so the perpendicular bisector of AB is OM .
The distance from A to B is
AB=sqrt%286%2B6%29=sqrt%2812%29=2sqrt%283%29 ,
so AM=MB=2sqrt%283%29%2F2=sqrt%283%29 .
Adding point C to the sketch and drawing isosceles triangle ABC we get

AngleACB=120%5Eo , and as ABC is an isosceles triangle,
altitude AC splits it into two congruent 30-60-90 triangles: ACM and BCM .
AngleBAC=30%5Eo , and AC%2Acos%28BAC%29=AM .
Substituting the measures we know
r%2Acos%2830%5Eo%29=sqrt%283%29 , and since cos%2830%5Eo%29=sqrt%283%29%2F2 ,
r%2A%28sqrt%283%29%2F2%29=sqrt%283%29-->highlight%28r=2%29 .