+1459 Lines $XQ$ and $XR$ are tangent to a circle, as shown below. If $\angle QXR = 60^\circ$, then find $\angle TAU$.
+129885 Let O be the center of the circle
Angles OTX and OUX = 90°
Angle TXU = angle QXR = 60°
So TOUX is a quadrilateral whose interior angles sum to 360°
So angle TOU = 360 - 90 - 90 -60 = 120° = minor arc TU
And angle TAU is an inscribed angle intercepting the same minor arc TU
So TAU measures (1/2) (120) = 60°
