Question

15. In the given figure, TBP and TCQ are tangents, to the
circle whose centre is 0. Also ZPBA
60° and
ZACQ = 70° Determine BAC and ZBTC​

Answer 1

130

Step-by-step explanation:

join OA , OB AND OC in triangle OAB angle OBA = 90 - 60 = 30 OB = OA (radii) angle OAB = 30 angle AOB = 180 - ( 30 + 30) = 120 in triangle OAC angle OCA = 90 - 70 = 20 OC = OA (RADII) angle OCA = OAC = 20 angle AOC = 180 - (20 +20) 140 thus angle BAC = 50 angle BTC = 180 - 50 = 130

I hope it is help you ❤