Can someone please solve this, and show the process? Thanks in advance.
Answers
Given : Two circles intersecting at A and B
P , A , B and X lies on one circle
Q , A, B and Y lies on other circle
PAQ and XAY are straight lines
∠BAX = 58° , ∠PBX=26° , ∠ABY = 23°
To Find : ∠AQB , ∠AYQ
Solution:
∠PAX = PBX angle by same arc PX
=> ∠PAX = 26°
∠YAQ = ∠PAX ( vertically opposite angles)
=> ∠YAQ = 26°
∠YBQ = ∠YAQ angle by same arc YQ
=> ∠YBQ = 26°
∠QBA = ∠YBQ + ∠ABY
=> ∠QBA = 26° + 23°
=> ∠QBA = 49°
∠QBA + ∠AYQ = 180° ( sum of opposite angles of cyclic Quadrilateral)
=> 49° + ∠AYQ = 180° '
=> ∠AYQ = 131°
∠BAX + ∠BAY = 180° Linear pair
∠BQY + ∠BAY = 180° ( sum of opposite angles of cyclic Quadrilateral)
=> ∠BQY = ∠BAX
=> ∠BQY = 58°
∠AQY = ∠ABY = 23° angle by same arc AQ
∠AQB = ∠BQY - ∠AQY
=> ∠AQB = 58° - 23°
=> ∠AQB = 35°
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