Question 1101149
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With angle AOT 60 degrees, triangle AOT is a 30-60-90 right triangle.<br>
OT is the radius of the circle, which is 8; so AT is 8*sqrt(3) and AO is 16.<br>
AX=XY=m; so TX = 8*sqrt(3)-m and XO = 8+m.<br>
Then in right triangle TOX we have
{{{8^2 + (8sqrt(3)-m)^2 = (m+8)^2}}}
{{{64 + 192 - 16sqrt(3)m + m^2 = m^2 + 16m + 64}}}
{{{192 - 16sqrt(3)m = 16m}}}
{{{12 - sqrt(3)m = m}}}
{{{12 = m(sqrt(3)+1)}}}
{{{m = 12/(sqrt(3)+1) = (12(sqrt(3)-1))/((sqrt(3)+1)(sqrt(3)-1)) = (12(sqrt(3)-1))/2 = 6(sqrt(3)-1)}}}<br>