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

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- Aug 15th 2008, 06:59 PMhelloyingtriangle problem
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

- Aug 15th 2008, 08:16 PMo_O
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: $\displaystyle \cos \left(\angle BAT\right) = \frac{AS}{TA} \: \: \iff \: \: TA = \frac{AS}{\cos \left(\angle BAT \right)}$