see the picture attaced and many thanks to people who will help me solve this problem and the other problems I have posted before. I really appreaciate the contributors very much.I give thanks to everyone that answer to my question
see the picture attaced and many thanks to people who will help me solve this problem and the other problems I have posted before. I really appreaciate the contributors very much.I give thanks to everyone that answer to my question
You can find $\displaystyle \angle OAB$ 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. $\displaystyle \angle OAT = 90^{\circ}$. You can then find $\displaystyle \angle BAT = \angle OAT - \angle OAB$.
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, $\displaystyle \triangle SAT$. You can then use trigonometry to find TA since: $\displaystyle \cos \left(\angle BAT\right) = \frac{AS}{TA} \: \: \iff \: \: TA = \frac{AS}{\cos \left(\angle BAT \right)}$