Question

O is the centre of the circle find x if angle aob 80 degree​

Answer 1

Answer:

∠AOC + 80° = 180° [LINEAR PAIR PROPERTY]

∠AOC = 100°

NOW,

OA = OC

→ ∠OAC = ∠OCA = x° [say]

now.

∠OAC + ∠OCA +∠AOC = 180°

x + x + 100 = 180°

2x = 80°

x = 40

Step-by-step explanation:

Answer 2

∠AOB = 80°

∠ACB = ?

According to the theorem of circle,

Angel at the circumference is twice the angle at centre

∠ACB = ½ × ∠AOB

➡ ∠ACB = 40°

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$\huge\mathfrak{Theorems}$

• There is only one tangent at a point of the circle.
• The tangent to a circle is perpendicular to the radius through the point of contact.
• The lengths of the two tangents from an external point to a circle are equal.

• The angle at the centre is twice the angle at the circumference.
• The angle in a semicircle is a right angle.
• Angles in the same segment are equal.
• Opposite angles in a cyclic quadrilateral sum to 180°
• The angle between the chord and the tangent is equal to the angle in the alternate segment.