Question

O is the centre of the circle and ABC and EDC are tangents to the circle.
Find the angle of BCD.

Can someone please tell me the answer to this and also explain how you got to it. Please.

Answer 1

Answer:

28°

Step-by-step explanation:

Angle at centre is twice the angle at Circumference

hence ∠BOD = 76 * 2= 152°.

∠CDO = ∠CBO = 90° (Tangent and radius meet each at 90°).

Considering quadrilateral CBOD,

∠BCD = 360° - (90° + 90° + 152°) = 28°

Answer 2

The angle of BCD is 28°.

O is the centre of the circle and ABC and EDC are tangents to the circle.

We have to find the angle of BCD.

Concept :

1. angle at the centre of a circle is twice the angle at the circumference of the circle.
2. Tangent is perpendicular to radius of circle.

here, angle BFD = 76° , it is angle at the circumference.

∴ from (1), angle at the centre = angle BOD = 2 × angle BFD = 2 × 76° = 152°

⇒∠BOD = 152°

ABC and EDC are tangents, OB and OD are radii.

∴∠OBC = ∠ODC = 90° [ from (2), ]

now from quadrilateral OBCD,

∠BOD + ∠ODC + ∠BCD + ∠OBC = 360°

[∵ sum of angles of a quadrilateral is 360° ]

⇒152° + 90° + ∠BCD + 90° = 360°

⇒∠BCD = 360° - (180° + 152°)

⇒∠BCD = 360° - 332° = 28°

hence ∠BCD = 28°