Higher January 2023 Paper 2 Q14

EdexcelHigherCurrent spec5 marksCircle Theorems

14 \(A\), \(B\), \(C\) and \(D\) are points on a circle, centre \(O\)

\(EBF\) is the tangent to the circle at \(B\)

Circle centre O with cyclic quadrilateral ABCD; angle DAB = 40 degrees; OD and OB are joined; EBF is the tangent at B and angle ABF = 66 degrees

Diagram NOT accurately drawn

(a)

(i) Work out the size of angle \(DCB\) (1)

(ii) Give a reason for your answer to (a)(i) (1)

(b) Work out the size of angle \(ADO\) (3)

Mark scheme (a)

Scheme	Marks
(i) 140	B1
(ii) opposite angles of a cyclic quadrilateral (add to 180°) oe	B1
	(2)

Notes

B1: dep on B1 in (a)(i) or seeing 180 – 40 with no contradiction
oe eg angle at centre is double (2 ×) angle at circumference oe AND angles around a point (or point 360)

Mark scheme (b)

Scheme	Marks
\(ADB = 66\) or \(ABO = 90 - 66\;(= 24)\) or \(BAO = 90 - 66\;(= 24)\) or \(ODB = \dfrac{180 - 80}{2}\;(= 50)\) or \(DOB\) reflex = 280	M1
For 2 of: \(ADB = 66\) or \(ABO = 90 - 66\;(= 24)\) or \(BAO = 90 - 66\;(= 24)\) or \(ODB = \dfrac{180 - 80}{2}\;(= 50)\) \(DOB\) reflex = 280	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 16	A1
	(3)
	(5 marks)

Notes

M1: Clearly labelled in working or shown on diagram

M1: (award M2 for \(360 - (280 + 40 + 24)\) oe)