General Mathematics Paper 2, May/June. 2011  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
General Comments

Question 4






In the diagram, PQRST is a quadrilateral.  PT//QS, ÐPTO = 42o, ÐTSQ = 38o and ÐQSR = 30o.
If ÐQTS = x and ÐPQT = y, find:

  • X;
  • y.





In the diagram, PQRS is a circle centre O.  If PÔQ = 150o, ÐQSR = 40o and ÐSQP = 45o, calculate ÐRQS.


This question was reported to be quite unpopular among the candidates. Majority of the candidates who attempted it performed poorly. They were unable to recall the appropriate theorems correctly as expected. Teachers were encouraged to emphasize this topic as candidates’ performance appears to be consistently low over the years.
In part (a), candidates were reported not to recall the properties of parallel lines and triangles. They were expected to show that ∠PTQ = ∠TQS (alternate angles). Hence ∠TQS = 42o. Also, x + 42o + 38o = 180o (sum of the angles of triangle QTS). This gave x = 100o. ∠RQS = 60o(sum of angles of a triangle RQS). Therefore y + 60o + 42o = 180o (sum of angles on a straight line). This gave y = 78o.
In part (b), candidates were expected to recall and apply circle theorems correctly. They were to show that ∠PSQ = 75o (angle subtended by a chord at the centre of a circle = twice the angle it subtends at circumference). Also, ∠PQS + ∠QSR + ∠RQS + ∠SQP = 180o (opposite angles of a cyclic quadrilateral) i.e. 75o + 40o + ∠RQS + 45o = 180o. This implied that ∠RQS = 20o.

Powered by Sidmach Technologies(Nigeria) Limited .
Copyright © 2015 The West African Examinations Council. All rights reserved.