Circle Tangent Challenge

$\angle ACO$ and $\angle PBA$ are tangent chord angle so $\angle BOA=120°$ and $\angle COA=140°$ .

$\angle OCT=\angle OBT=90°$ because $TP$ and $TO$ are tangent lines.

Focus on quadrilateral $BOCT$ , where $\angle OCT+\angle OBT=180°.$

Then $\angle BTC+\angle COB=180°$ but $\angle COB=360°-140°-120°=100°$ $\therefore \boxed{\angle BTC=80°}$

An exterior angle created by two tangents is equal to 1/2(measure of the major intercepted arc - measure of the minor intercepted arc). So in this case, the measure of angle BTC is equal to 1/2 (major arc BC - minor arc BC)

The measure of an arc formed by a cord and a tangent is equal to twice the angle they form. So the measure of arc BA: 2x60 degrees = 120 degrees, and the measure of arc BC is 2x70 = 140 degrees.

The measure of major arc BC is the sum of those two arcs we found, so 120 degrees + 140 degrees = 260 degrees. Because a circle is 360 degrees, the minor arc BC is 360 degrees - 260 degrees = 100 degrees!

Therefore the measure of angle BTC = 1/2(260 degrees - 100 degrees) = 80 degrees. Great problem!

Let $\stackrel \frown {CAB} = \alpha$ , $\stackrel \frown {BC} = \beta$

It's known that $\angle BTC = \dfrac{\alpha-\beta}{2}$

But $\alpha = 2 \times 60^\circ + 2 \times 70^\circ = 260^\circ$ and $\beta = 360^\circ-\alpha = 360^\circ-260^\circ = 100^\circ$

So $\angle BTC = \dfrac{260-100}{2} = \boxed{80^\circ}$

1.Join OB and OC . 2.Angle OBP=90 , therefore angle OBA=30, Similarly angle OCA=20. 3.Angle OAB=30 and similarly angle OAC=20 (Isosceles Triangle prop.) 4.Angle BOC=100 by the property that angle subtended at the centre is twice the angle subtended on rest of the circle. 5.Angle BTC+ angle OBT +angle BOC +angle OCT=360. Therefore Angle BTC=80.

Let BTC= x, Now BTC+TCA+CAB+TBA= 360 X+120+110+BAC=360 ==> BAC= 130-X BOC = 2BAC ==> Hence BOC = 260-2X nOW, BTC+BOC= 180 ==> X+ 260- 2X = 180 ==> X= 80

draw line BC NOW ANGLE CBA=70 &ANGLE ACB= 60 HENCE ANGLE BAC=180-(70+60)=50 ALSO ANGLE ABT=120 ANGLE ACT=110 HENCE ANGLE BTC=360-(50+110+120)=80

Join $BC.$ By the alternate segment theorem, $\angle ACQ = \angle CBA \Rightarrow \angle CBP = 130^{\circ} \therefore \angle CBT = 50^{\circ}.$ Also, since $TB = TC \Rightarrow \angle BCT = 50^{\circ}.$ So we can conclude that $\angle BTC = 80^{\circ}$ .