8.1 Complete the following theorem: The angle subtended by the diameter at the circumference of the circle is … 8.2 In the diagram below, O is the centre of circle BCDE with diameter BE - NSC Technical Mathematics - Question 8 - 2019 - Paper 2

(a) ∠CBO:

Using the theorem that the angle subtended by the diameter at the circumference is 90°, we note that:

$$∠CBO = 90° - 30° = 60°.$$

Therefore, ∠CBO = 60°.

(b) ∠D̂2:

Since ∠D̂2 is subtended by the chord DE, we can apply the same reasoning as for ∠CBO:

$$∠D̂2 = 30° ext{ (as it is equal to that of the opposite angle)}.$$

Thus, ∠D̂2 = 30°.

(c) ∠Ô1:

The angle at the center (O) is twice the angle at the circumference (B):

$$∠Ô1 = 2 imes ∠CBO = 2 imes 60° = 120°.$$

Hence, ∠Ô1 = 120°.

(d) ∠Ô2:

Using the same logic, since ∠Ô2 subtends the same chord DE:

$$∠Ô2 = 2 imes ∠D̂2 = 2 imes 30° = 60°.$$

Thus, ∠Ô2 = 60°.

From the diagram, we observe that:

• Triangles FCB and FDE are equal in design because:
  • FC and FD are opposite sides.
  • BC = DE (equal chords of the circle).
  • They share the angle ∠C and ∠D, respectively.

Thus, by the properties of congruent triangles:

$$FC = FD.$$