In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate : ∠BDC
Exercise 17 (C) | Q 6.1 | Page 266
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Given that BD is a diameter of the circle.
The angle in a semicircle is a right angle.
∴ ∠ BCD = 90°
Also given that ∠ DBC = 58°
In Δ BDC ,
∠ BCD + ∠ BCD + ∠ BDC = 180°
⇒ 58° + 90° + ∠ BDC = 180°
⇒ 148° + ∠ BDC = 180°
⇒ ∠ BDC = 180° - 148°
⇒ ∠ BDC = 32°
The angle in a semicircle is a right angle.
∴ ∠ BCD = 90°
Also given that ∠ DBC = 58°
In Δ BDC ,
∠ BCD + ∠ BCD + ∠ BDC = 180°
⇒ 58° + 90° + ∠ BDC = 180°
⇒ 148° + ∠ BDC = 180°
⇒ ∠ BDC = 180° - 148°
⇒ ∠ BDC = 32°
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