Question 1001663
What does vector AC have to do with points A, B, T, and the angles measured?
The situation described in the problem is not clear to me.
I am assuming that the angles are angles of elevation, and A, B, And T are on the same vertical plane,
If that is so, then a sketch of the situation, and the solution look like this:
{{{drawing(400,300,-2.8,1.2,-2.7,0.3,
green(arrow(2,0,-4,0)),green(arrow(0,0,1,0)),
green(arrow(2,-2.36,-4,-2.36)),green(arrow(0,-2.36,1,-2.36)),
triangle(0,0,0.42,-2.36,-2.4,0),locate(0.4,-2.36,T),
locate(-.02,0.15,B),locate(-2.42,0.15,A),
locate(0.1,-1.86,80^o),arc(0.42,-2.36,1,1,180,260),
green(arc(0.42,-2.36,1.4,1.4,180,220)),locate(-0.5,-2.06,green(40^o)),
green(arc(-2.4,0,1.4,1.4,0,40)),red(arc(0.42,-2.36,1.8,1.8,220,260)),
locate(-1.7,-0.1,green(x)),locate(-0.1,-1.4,red(y)),
locate(0.2,-0.05,green(horizontal)),locate(-2.4,-2.4,green(horizontal))
)}}}
The two horizontal green lines are parallel,
and the two angles marked with green arcs are alternate interior angles,
so they are congruent, meaning that {{{green(x=40^o)}}} .
Also, for the angles at T, {{{green(40^o)+red(y)=80^o}}} ---> {{{red(y)=80^o-green(40^o)}}} ---> {{{red(y=40^o)}}} .
Triangle TAB has congruent angles measuring {{{40^o}}} at T and A,
so the sides opposite T and A are congruent,
meaning {{{TB=AB=highlight(2.4miles)}}}