# triangle problem

You can find $\angle OAB$ by extending O to AB, thereby bisecting it. Now, recall that a circle's tangent forms a right angle with its radius at their point of intersection - i.e. $\angle OAT = 90^{\circ}$. You can then find $\angle BAT = \angle OAT - \angle OAB$.
Now extend a horizontal line from the middle of AB (call this point S) to point T (again bisecting AB to form AS and SB). Now we have a right triangle, $\triangle SAT$. You can then use trigonometry to find TA since: $\cos \left(\angle BAT\right) = \frac{AS}{TA} \: \: \iff \: \: TA = \frac{AS}{\cos \left(\angle BAT \right)}$