Question 726536
{{{drawing(375,300,-2,13,-6,6,
circle(0,0,1), circle(5,0,4),
circle(0,0,0.1),circle(5,0,0.1),
arrow(-1.667,0,12,0),arrow(-1.667,0,6.333,6),arrow(-1.667,0,6.333,-6),
line(2.6,-3.2,5,0),line(-0.6,-0.8,0,0),
locate(-2,0.4,A),locate(-0.2,0.6,B),locate(4.8,0.6,D),
locate(-0.9,-0.8,C),locate(2.3,-3.2,E)
)}}} {{{BC=1}}} and {{{DE=4}}} are radii to the points of tangency.
They are perpendicular to tangent AE, so ABC and ADE are similar right triangles.
If {{{AB=y}}}, {{{AD=AB+1+4=y+5}}}.
In triangles ABC and ADE,
{{{sin(DAE)=1/y=4/(5+y)}}} --> {{{4y=5+y}}} --> {{{3y=5}}} --> {{{y=5/3}}}
So {{{sin(DAE)=1/(5/3)=3/5}}} ---> {{{cos(DAE)=4/5}}}
{{{x=2(DAE)}}} so {{{sin(x)=sin(2DAE)=2sin(DAE)cos(DAE)=2(3/5)(4/5)=24/25=0.96}}}