Determine, with reasons, the size of \(\hat{B}_2\) in terms of \(x\).
About this question
Question 10.2.2 of Mathematics Grade 12 Paper 2, NSC Examination (May/June) (DBE, May/June 2026). It tests the Term 3 ATP topic Euclidean Geometry.
- TermTerm 3
- ATP topicEuclidean Geometry
- PaperPaper 2
- Marks3 marks
O is the centre of the circle. Chord AS is perpendicular to diameter BOD at K. OR is drawn with R on BS. AB, AD and OS are drawn. \(\hat{B}_1 = x\).
Angle labels: \(\hat{B}_1 = \angle ABD\), \(\hat{B}_2 = \angle DBS\). \(\hat{A}_1 = \angle BAS\), \(\hat{A}_2 = \angle SAD\). \(\hat{K}_3 = \angle OKS\) (\(= 90^\circ\)). \(\hat{O}_1 = \angle BOR\), \(\hat{O}_2 = \angle ROS\), \(\hat{O}_3 = \angle SOK\). \(\hat{S}_1 = \angle OSK\), \(\hat{S}_2 = \angle OSR\). \(\hat{R}_1 = \angle ORB\), \(\hat{R}_2 = \angle ORS\).
- \(\hat{S}_1 + \hat{S}_2 = \hat{D} = 90^\circ - x\) (\(\angle\)s in the same segment)
- In \(\triangle BKS\): \(\hat{K}_3 = 90^\circ\) (given \(AS \perp BD\))
- \(\hat{B}_2 = 180^\circ - 90^\circ - (90^\circ - x)\) (sum of \(\angle\)s in \(\triangle\))
\(\hat{B}_2 = x\)
Other points that earn marks
- Alternative: \(\hat{A}_2 = x\) (in \(\triangle AKD\), or since \(\hat{A}_2 = 90^\circ - \hat{D}\)), and \(\hat{B}_2 = \hat{A}_2\) (\(\angle\)s in the same segment).
More Questions On Euclidean Geometry
\(\hat{D}_1 = E\hat{A}G = x\) (tan chord theorem, small circle)
\(\hat{C}_1 = \hat{D}_1 = x\) (tan chord theorem, small circle)
\(\hat{C}_4 = \hat{C}_1 = x\) (vert opp \(\angle\)s \(=\))
\(A\hat{F}G = E\hat{A}G = x\) (tan chord theorem, large circle)
\(\hat{D}_1\), \(\hat{C}_1\), \(\hat{C}_4\) and \(A\hat{F}G\) all equal \(x\)
- \(\hat{D}_1 = 180^\circ - 27^\circ - 92^\circ = 61^\circ\) (sum of \(\angle\)s in \(\triangle BED\))
- \(\hat{A}_2 = \hat{D}_1\) (ext \(\angle\) of cyclic quad ABDE)
\(\hat{A}_2 = 61^\circ\)
Construction: draw diameter \(CF\) and join \(FB\).
- \(F\hat{C}E = 90^\circ\) (tan \(\perp\) radius)
- So \(B\hat{C}E = 90^\circ - F\hat{C}B\)
- \(F\hat{B}C = 90^\circ\) (\(\angle\) in semi-circle)
- So \(\hat{F} = 90^\circ - F\hat{C}B\) (sum of \(\angle\)s in \(\triangle\))
- \(\therefore B\hat{C}E = \hat{F}\)
- But \(\hat{F} = \hat{A}\) (\(\angle\)s in the same segment)
- \(\therefore B\hat{C}E = \hat{A}\)
- \(\hat{L}_2 = 90^\circ\) (\(\angle\) in semi-circle; TR is a diameter of the smaller circle)
- \(\hat{R}_2 = 90^\circ\) (\(\angle\) in semi-circle; PL is a diameter of the larger circle)
- \(\therefore \hat{L}_2 = \hat{R}_2\)
- \(\therefore LT \parallel PR\) (alt \(\angle\)s =)
- \(\hat{S}_2 = \hat{B}_2 = x\) (\(\angle\)s opp equal sides; \(OB = OS\))
- \(\hat{S}_1 = (\hat{S}_1 + \hat{S}_2) - \hat{S}_2 = 90^\circ - x - x\)
\(\hat{S}_1 = 90^\circ - 2x\)
In \(\triangle PQR\): \(\dfrac{QE}{QR} = \dfrac{PW}{PR}\) (line \(\parallel\) one side of \(\triangle\); \(PQ \parallel WE\))
\(\therefore PR = \dfrac{PW \cdot QR}{QE}\)