Question

13th and and 14th questions please

Answer 1

14) We know that angle subtended by an arc of a circle at the centre is double the angle subtended by it on remaining part of the circle. Arc CD subtends ∠COD at centre and subtends ∠BCD at B on the circle Hence ∠COD = 2∠BCD That is ∠COD = 2y [Since ∠BCD = y] Also ∠COD = ∠OCB = 50° (alternate interior angles) I.e, 2y = 50° y= 25° From the figure ∠AOD = 90° since ∠AEB = ∠AEC = 90° Therefore, ∠AOD = 2∠ABD That is 90° = 2∠ABD Hence ∠ABD = 45° In ΔAEB, ∠AEB + x + y + 45° = 180° 90° + x + 25° + 45° = 180° 160° + x = 180° x = 180° - 160° x = 20° ∴x=20° , y=25°