Answer:
a) ∠OAB = 54°
b) ∠OCB = 36°
c) ∠ABC = 90°
Explanation:
a)
If O is the center of the circle, then OA and OB are both the radius. Therefore, OA = OB
This means that ΔAOB is an isosceles triangle, as is has two sides of equal length (OA and OB).
The angle between the equal sides (∠AOB) is called the vertex angle. The side opposite the vertex angle (AB) is called the base and base angles are equal. Therefore, ∠OAB = ∠OBA
Given:
- The sum of the interior angles of a triangle is 180°
- ∠AOB = 72°
- ∠OAB = ∠OBA
⇒ ∠OAB + ∠OBA + ∠AOB = 180
⇒ ∠OAB + ∠OBA + 72 = 180
⇒ ∠OAB + ∠OBA = 108
⇒ ∠OAB = 108 ÷ 2 = 54°
b)
Again, we can see that OB and OC are both the radius, so OB = OC. This means that ΔBOC is also an isosceles triangle, with ∠BOC as the vertex and BC as the base.
Calculate ∠BOC:
Given:
- Angles on a straight line add up to 180°
- ∠AOB = 72°
⇒ ∠AOB + ∠BOC = 180
⇒ 72 + ∠BOC = 180
⇒ ∠BOC = 108°
Calculate ∠OCB:
Given:
- The sum of the interior angles of a triangle is 180°
- ∠BOC = 108°
⇒ ∠OBC + ∠OCB + ∠BOC = 180
⇒ ∠OBC + ∠OCB + 108 = 180
⇒ ∠OBC + ∠OCB = 72
⇒ ∠OCB = 72 ÷ 2 = 36°
c)
Thales's theorem is a special case of the inscribed angle theorem and states that:
- If A, B, and C are distinct points on a circle where the line AC is a diameter, then ∠ABC is a right angle.
Given that AC is the diameter of this circle, and points A, B and C are on the circle, then ∠ABC = 90°