Circle Theorems: Eduqas GCSE Higher June 2025 Paper 2 Q16

Higher June 2025 Paper 2 Q16

16. The diagram below shows points A, B and C on the circumference of a circle with centre O.
The line DAE is a tangent to the circle.

Diagram not drawn to scale

Show that \(A\hat{B}O = \left(\frac{1}{2}x - 25\right)^\circ\).

Give a reason for each step of your answer. [4]

Mark scheme

Scheme	Marks	Comments
Method using angles \(B\hat{A}C\) and \(O\hat{A}B\)		Angles may be seen on the diagram
\(B\hat{A}C = \frac{1}{2}x\) (Angle at circumference = half angle at centre)	B1
\(O\hat{A}C\) (\(= 90 - 65\)) \(= 25(^\circ)\) OR \(O\hat{A}B = 90 - (115 - \frac{1}{2}x)\) (Tangent perpendicular to radius)	B1
\(B\hat{A}O = \frac{1}{2}x - 25\) AND a statement that \(A\hat{B}O = O\hat{A}B\) (Base angles of an isosceles triangle are equal)	B1
Reasons must include at least two of the three reasons Angle at circumference = half angle at centre oe Tangent perpendicular to radius Base angles of an isosceles triangle are equal	E1	Award E1 for the reasons provided at least B1 B1 awarded.

Scheme	Marks	Comments
		Angles may be seen on the diagram
\(A\hat{B}C = 65(^\circ)\) (Alternate segment theorem)	B1	Must be correctly placed on the diagram or appropriately labelled
\(O\hat{B}C\) or \(O\hat{C}B = \frac{180 - x}{2}\) or \(90 - \frac{1}{2}x\) (Base angles of an isosceles triangle are equal)	B1
\(A\hat{B}O = 65 - \left(\frac{180 - x}{2}\right)\) or \(65 - (90 - \frac{1}{2}x)\) convincingly leading to \(\frac{1}{2}x - 25\)	B1
Reasons must include alternate segment theorem AND isosceles triangle	E1	Award E1 for the reasons provided at least B1 B1 awarded.

Scheme	Marks	Comments
		Angles may be seen on the diagram
\(O\hat{A}C\) (\(= 90 - 65\)) \(= 25(^\circ)\) (Tangent perpendicular to radius)	B1
\(A\hat{O}B\) (\(= 360 - (180 - 2\times25) - x\)) \(= 230 - x\) (Isosceles triangle and angles at a point)	B1
\(A\hat{B}O = \frac{180 - (230 - x)}{2} = \frac{1}{2}x - 25\) (Base angles of an isosceles triangle are equal)	B1
Reasons must include at least Tangent perpendicular to radius Base angles of an isosceles triangle are equal	E1	Award E1 for the reasons provided at least B1 B1 awarded.

Scheme	Marks	Comments
\(A\hat{B}C = 65(^\circ)\) (Alternate segment theorem)	B1
\(O\hat{B}C = 65 - (\frac{1}{2}x - 25) = 90 - \frac{1}{2}x\)	B1
\(O\hat{B}C = O\hat{C}B\) so \(B\hat{O}C = 180 - 2\times(90 - \frac{1}{2}x) = x\) (Isosceles triangle and angles in a triangle)	B1
Reasons must include alternate segment theorem AND isosceles triangle	E1	Award E1 for the reasons provided at least B1 B1 awarded.

Scheme	Marks	Comments
\(B\hat{A}C = \frac{1}{2}x\) (Angle at circumference = half angle at centre)	B1
\(B\hat{A}O = A\hat{B}O = \frac{1}{2}x - 25\) AND \(O\hat{A}C = \frac{1}{2}x - (\frac{1}{2}x - 25) = 25(^\circ)\) (Isosceles triangle)	B1
\(E\hat{A}C = 90 - 25 = 65(^\circ)\) OR \(O\hat{A}C = 90 - 65 = 25(^\circ)\) (Tangent perpendicular to radius)	B1
Reasons must include at least two of the three reasons Angle at circumference = half angle at centre oe Tangent perpendicular to radius Base angles of an isosceles triangle are equal	E1	Award E1 for the reasons provided at least B1 B1 awarded.

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