in the given figure O is the centre and ab is the diameter. if ∠aoe is 150° and ∠dao is 51° calculate ∠ceb and ∠cbe
Answers
Answer:
In Triangle OEB:
OB = OE [radii of same circle]
Thus, /_OBE = /_OEB [opposite angles of equal sides]
/_EOB = 180 - /_EOA [Linear pair]
/_EOB = 180 - 150
/_EOB = 30°
In Triangle OEB:
/_EOB + /_OBE + /_OEB = 180° [Angle Sum Property]
30° + 2/_OEB = 180
2/_OEB = 150
/_OEB = 75° = /_OBE
/_CBE = 180° - /_OBE
= 180 - 75
/_CBE = 105°
2/_ADE = /_AOE [Angke on centre is double of the angle on the circle]
/_ADE = 150/2
/_ADE = 75°
/_AOE + /_OED + /_EDA + /_DAO = 360° [Sum of all angles of a quadrilateral]
/_OED = 360 - 150 - 75 - 51
/_OED = 94°
/_CEB + /_OEB + /_OED = 180° [Linear Pair]
/_CEB = 180 - 94 - 75
/_CEB = 11°
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Answer:
In Triangle OEB:
OB = OE [radii of same circle]
Thus, /_OBE = /_OEB [opposite angles of equal sides]
/_EOB = 180 - /_EOA [Linear pair]
/_EOB = 180 - 150
/_EOB = 30°
In Triangle OEB:
/_EOB + /_OBE + /_OEB = 180° [Angle Sum Property]
30° + 2/_OEB = 180
2/_OEB = 150
/_OEB = 75° = /_OBE
/_CBE = 180° - /_OBE
= 180 - 75
/_CBE = 105°
2/_ADE = /_AOE [Angke on centre is double of the angle on the circle]
/_ADE = 150/2
/_ADE = 75°
/_AOE + /_OED + /_EDA + /_DAO = 360° [Sum of all angles of a quadrilateral]
/_OED = 360 - 150 - 75 - 51
/_OED = 94°
/_CEB + /_OEB + /_OED = 180° [Linear Pair]
/_CEB = 180 - 94 - 75
/_CEB = 11°
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