In the figure given above TP and TQ are tangent to the circle with center O at B and C, respectively.
If angle PBA=60 degrees and angle ACQ =70 degrees, what is angle BTC in degrees?
+128473 arc BA = 2* anglePBA = 120°
arc CA = 2* angle ACQ = 140°
So minor arc BC = 360 - (120 + 140) = 100°
If we draw radii OB and OC then angles TBO and TCO = 90°
Then in quadrilateral TBOC the sum of the interior angles = 360°
And cental angle BOC = minor arc BC =100°
So
( 90 + 90 + 100 + angle BTC) = 360
280 + angleTBC =360
angle TBC = 360 - 280 = 80°
