Home Technical Mathematics Languages Science Social Science Art Literature Arabic Islamic Studies C.R.K HistoryMusicVisual Art Clothing/Textile Home Management Shorthand
 General Mathematics Paper 2,Nov/Dec. 2013
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
Weakness/Remedies
Strength

Question 10

Question 10

Town Q is 20 km due north of P.  The bearing of town R from Q is 140o.  If R is 8 km from Q, calculate:

1. the bearing of R from P, to the nearest degree;
2. how far north of P, R is, correct to 2 significant figures.

_____________________________________________________________________________________________________
Observation

The Chief Examiner reported that majority of the candidates who attempted this question did not illustrate the given information in a diagram correctly.  This also affected their being able to solve the question correctly.  Candidates were expected to represent the information in a diagram as shown:

Q     140o

8 km
20km                                                     R
r
o
P

From the diagram, ∠PQR = 180o – 140o = 40o.  Using the cosine rule,
|PR|2 = 202 + 82 – 2(20)(8) cos 40o simplifying this equation gave
|PR| = 14.79411km.  Using the sine rule  =  where  is the bearing of R from P.  Solving this equation gave  = 20.34o i.e. bearing of R from P is 020o or N20oE.  Let r be the distance of R north of P.  Then r = |PR| cos  = 13 km, correct to 2 significant figures.