AB and CD intersect at the center O of the circle given in the above diagram. If ∠EBA = 33° and ∠EAC = 82°, find
(a) ∠BAE
(b) ∠BOC
(c) ∠ODB
Answer
(a) In △ BAE,
⇒ ∠AEB = 90° (Angle in a semicircle is a right angle)
By angle sum property of triangle,
⇒ ∠AEB + ∠EBA + ∠BAE = 180°
⇒ 90° + 33° + ∠BAE = 180°
⇒ ∠BAE + 123° = 180°
⇒ ∠BAE = 180° - 123° = 57°.
Hence, ∠BAE = 57°.
(b) From figure,
⇒ ∠EAC = ∠EAB + ∠BAC
⇒ 82° = 57° + ∠BAC
⇒ ∠BAC = 82° - 57° = 25°.
We know that,
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
⇒ ∠BOC = 2∠BAC = 2 × 25° = 50°.
Hence, ∠BOC = 50°.
(c) From figure,
∠BOC and ∠BOD form a linear pair.
∴ ∠BOC + ∠BOD = 180°
⇒ 50° + ∠BOD = 180°
⇒ ∠BOD = 180° - 50° = 130°.
In △ BOD,
⇒ ∠BOD + ∠ODB + ∠DBO = 180°
⇒ 130° + ∠ODB + 33° = 180°
⇒ ∠ODB + 163° = 180°
⇒ ∠ODB = 180° - 163° = 17°.
Hence, ∠ODB = 17°.