In the diagram, O is the centre of the circle - NSC Mathematics - Question 8 - 2022 - Paper 2

For angle M̂₁, we apply the inscribed angle theorem which states that the angle subtended at the center is twice the angle subtended at the circumference. Therefore:

$\angle M̂₁ + 64° = 90°$

Thus, M̂₁ = 90° - 64° = 26°.

To calculate Ôᵢ, we use the rule that the angle at the center is twice the angle at the circumference. Since M₁ is the angle at circumference:

$\angle Ôᵢ = 2 \times 26° = 52°$.

DE is parallel to BH because D and E are midpoints of AB and AG respectively, thus by the midpoint theorem, we have that DE || BH.

From the given information, we apply the intercept theorem:

1. We know that FC = \frac{1}{4} BF. Since BF = 6x - 2, we can denote: $FC = \frac{1}{4}(6x - 2)$

2. Given that GH = x + 1, and since we have established that DE || BH, we can set up the equation: $GH = FC$ Therefore, $x + 1 = \frac{1}{4}(6x - 2)$

3. Cross-multiplying yields: $4(x + 1) = 6x - 2$

4. Simplifying gives: $4x + 4 = 6x - 2$

5. Rearranging leads to: $6x - 4x = 4 + 2$ $2x = 6$ Thus, $x = 3$.