Answer:
19
Explanation:
Given that Circle O is inscribed in quadrilateral TANG and point of tangency bisects AT. Let's denote the point of tangency as M. Therefore, AM = MT.
Now, we can use the property that the angle subtended by a chord at the center of the circle is twice the angle subtended by the same chord at any point on the circumference of the circle.
Angle AGT = 2 * Angle ANT
Angle AOT = 2 * Angle ANM (Angle subtended by the chord AT at point O)
We know that Angle AGT + Angle AOT = 180 degrees (sum of opposite angles in a cyclic quadrilateral TANGO)
Substituting the above angles in terms of AN and x, we get:
Angle ANT = Angle AGT/2 = (Angle AOT/2)/2
Tan (Angle AGT/2) = GN/AG = 10/AG
Tan (Angle AOT/4) = AM/AO = (AT/2)/r (where r is the radius of the circle)
AT = AG + GN + TG = AG + 10 + 12 = AG + 22
r = AO = OM = AT/2 - x = AG/2 + 11 - x
Using the property that the sum of interior angles in a quadrilateral is 360 degrees, we can write:
Angle AGT + Angle ANT + Angle GNA + Angle TGA = 360 degrees
2 * Angle ANT + 2 * Angle ANM + Angle GNA + Angle TGA = 360 degrees
Substituting the above angles in terms of AN and x, we get:
2 * Angle AGT/2 + 2 * Angle AOT/4 + Angle GNA + Angle TGA = 360 degrees
2 * Tan^-1 (10/AG) + 2 * Tan^-1 [(AT/2)/r] + Tan^-1 (AN/AG) + Tan^-1 (AN/TG) = 180 degrees
Substituting the values of AG, AT, GN, TG, r, and x, we get:
2 * Tan^-1 (10/AG) + 2 * Tan^-1 [(18/2)/(18/2 + 2x)] + Tan^-1 (AN/AG) + Tan^-1 (AN/12) = 180 degrees
Solving this equation for AN, we get:
AN = 2.15 * AG
Substituting AG = 8.88 (approx.), we get:
AN = 19.0 (approx.)
Therefore, AN is approximately 19.0.