A different angle chasing route: In △BCD: ∠BCD = 180° - (15° + 40°) = 125° As BC is tangent to circle Q at C then ∠BCQ = 90° and so ∠QCD = 125° - 90° = 35° △CQD is isosceles so ∠QCD = ∠QDC = 35° and so ∠QDK = 40° - 35° = 5° △DQK is isosceles so ∠QKD = ∠QDK = 5° Circle P and circle Q are tangent at K, so P, K and Q are colinear. ∠PKB and ∠QKD are opposite angles so ∠PKB = ∠QKD = 5° △BPK is isosceles so ∠PBK = ∠PKB = 5° and so ∠PBK = 180° - (5° + 5°) = 170° ∠BAK (= x) is drawn from the same chord (BK) as ∠BPK so by the inscribed angle theorem ∠BAK = (∠BPK)/2 = 170°/2 = 85°
A different angle chasing route:
In △BCD: ∠BCD = 180° - (15° + 40°) = 125°
As BC is tangent to circle Q at C then ∠BCQ = 90° and so ∠QCD = 125° - 90° = 35°
△CQD is isosceles so ∠QCD = ∠QDC = 35° and so ∠QDK = 40° - 35° = 5°
△DQK is isosceles so ∠QKD = ∠QDK = 5°
Circle P and circle Q are tangent at K, so P, K and Q are colinear.
∠PKB and ∠QKD are opposite angles so ∠PKB = ∠QKD = 5°
△BPK is isosceles so ∠PBK = ∠PKB = 5° and so ∠PBK = 180° - (5° + 5°) = 170°
∠BAK (= x) is drawn from the same chord (BK) as ∠BPK so by the inscribed angle theorem ∠BAK = (∠BPK)/2 = 170°/2 = 85°
Thank you for the detailed explanation! I appreciate how you've broken down each step. Have a great rest of your day!