Mathematics
In the given figure, ∠DBC = 58° and BD is a diameter of the circle. Calculate :
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC
Answer
(i) Given that BD is a diameter of the circle.
We know that,
Angle in a semicircle is a right angle.
So, ∠BCD = 90°
Also given that,
∠DBC = 58°
In ∆BDC,
⇒ ∠DBC + ∠BCD + ∠BDC = 180° [Angle sum property of triangle]
⇒ 58° + 90° + ∠BDC = 180°
⇒ 148° + ∠BDC = 180°
⇒ ∠BDC = 180° - 148° = 32°.
Hence, ∠BDC = 32°.
(ii) We know that, the opposite angles of a cyclic quadrilateral are supplementary.
So, in cyclic quadrilateral BECD
⇒ ∠BEC + ∠BDC = 180°
⇒ ∠BEC + 32° = 180°
⇒ ∠BEC = 180° - 32° = 148°
Hence, ∠BEC = 148°.
(iii) In cyclic quadrilateral ABEC,
⇒ ∠BAC + ∠BEC = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]
⇒ ∠BAC + 148° = 180°
⇒ ∠BAC = 180° - 148° = 32°.
Hence, ∠BAC = 32°.
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