Answer:
1. Arc angle
2. Angle at circumference
3. Base angle of an isosceles triangle
4. Sum interior of angles in a triangle
5. Sum interior of angles in a triangle
6. The sum of angles on a straight line
7. Sum interior of angles in a triangle
8. Sum interior of angles in a triangle
9. Angle from a tangent to the radius of a circle
10. Alternate angles of line crossing two parallel lines
Step-by-step explanation:
Here, we note that, the angles where derived as follows;
1. mBC = 90°, therefore, angle subtended by the arc BC at the center of the circle = 1 = 90°
2. 90° The angle at the center which is the straight line diameter with angle 180° = 2 × angle at the circumference
∴ The angle at the circumference, angle 2 = 180°/2 = 90°
3. Angle subtended by the sum of arcs CD + DE at the center = 50°, therefore, ∡EOC = 50° and ∡EOA = 180° - 50° = 130° (the sum of angles on a straight line), since ΔAOE is an isosceles triangle, therefore, ∡OAE = ∡OEA = (180° - 130°)/2 = 25°
4. ∡A + ∡E + ∡C = 180° (Sum interior of angles in a triangle)
∴ 25° + 90° + ∡C = 180°
Hence ∡C = 180° - (90° + 25°) = 65°
5. From arc CD = 30°, ∡COD = 30°, ∴ ∡OBD = 30° = base angle of isosceles triangle with sides equal to the radius
Hence ∡8 = 30° from ∡8 + ∡1 + ∡OBD = 180° which gives
∡8 + 90° + 30° = 180°, hence;
∡8 = 180° - (90° + 30°) = 60°
∡8 = opposite ∡8 = 60°
Opposite ∡8 + ∡C + ∡5 = 180° (Sum interior of angles in a triangle)
Therefore ∡5 = 180° - (opposite ∡8 + ∡C) = 180° - (60° + 65°) = 55°
6. Angles 6 plus angle 5 = 180° (The sum of angles on a straight line)
Therefore, ∡6 = 180° - ∡5 = 180° - 55° = 125°
7. ∡7 + opposite ∡5 + ∡E = 180° (Sum interior of angles in a triangle)
∡E + ∡2 = 180° (The sum of angles on a straight line)
∴ ∡E = 180° - ∡2 = 180° - 90° = 90°
Hence, ∡7 + opposite ∡5 + ∡E = 180° = ∡7 + 55° + 90° = 180°
∴ ∡7 = 35°
8. ∡8 = 60° as found above
9. ∡9 = 90° (Angle from a tangent to the radius of a circle)
10. ∡10 = ∡8 = 60° (Alternate angles).
