General Mathematics Paper 2,May/June. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
Weakness/Remedies
Strength

Question 3

(a) In the diagram, L PQR = 125°, LQRS = r, LRST = 800 and LSTU = 44°. Calculate the value of r

b) .

In the diagram, TS is a tangent to the circle at A. ABI ICE, LAEC = sx", LADB = 60° and LTAE = xo. Find the value of x",

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Observation

Part (a) of this question required that candidates drew straight lines, one each passing through points Sand R and parallel to PQ and UT as shown in the diagram below but the report stated that majority of the candidates did not.From the diagram, LMQR = LQRN = 180 -125 = 55° ( alternate angles), LVST = 44° (alternate angle to LSTU). Therefore LNRS = LVSR = 80 - 44° = 360• Hence, r = 55 + 36 = 91°

In part (b), candidates' performance was reported to be worse than part (a). Candidates were reported to have exhibited poor understanding of circle theorems. Teachers were encouraged to do a lot of work in this area.

From the diagram, LBDA = LBAS = 600 (angles in the alternate segment). LBAE = 180° - 5x (adjacent angles on a transversal). Therefore LBAS + LBAE + LEAT = 60 + 180 - 5x + x = 180 (angles on a straight line). Solving this simple equation gave x = 150. Candidates' performance this question was described as poor.