Given: The sides of triangle ACD are tangent to circle O at points B, X, and Y. Also, AB=BC= 10 and measure of arch XY=100.

Question

Given: The sides of triangle ACD are tangent to circle O at points B, X, and Y. Also, AB=BC= 10 and measure of arch XY=100. show that triangle ACD is isosceles. find the measures of the angles of triangle ACD. find the measure of the sides of triangle ACD. find the radius of circle O.

Answer 1

draw a diagram

draw radii to points B, X and Y

these radii are perpendicular to the sides of the triangle because they touch the sides at the points of tangency

AB=10, BC=10

also, CX=10 and AY=10 because two tangent segments from the same point to a circle are congruent

therefore arc YB=130° and arc BX=130° (100+130+130=360)

this is sufficient to show the triangle is isosceles

angle A=50° and angle C=50° because 90+90+130+50=360

therefore angle D=80°

use the law of cosines to find AD and CD

AC^2=AD^2+DC^2-2(AD)(DC)cos80°

to simplify things call AD and DC "n"

20^2=n^2+n^2-2*n*n*0.173648

400=2n^2-0.347296n^2

400=1.6527n^2

242.0282=n^2

n=15.557

therefore AD=DC=15.557

find the area of the triangle using A=absinC/2

A=(15.6)(15.6)(sin80°)/2

A=15.6*15.6*0.9848

A=119.83

now use A=(1/2)(r)(perimeter)

119.83=(1/2)(r)(51.114)

119.83=25.557*r

r=119.83/25.557

r=4.6887