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 7

Question 7
Q

5 cm
R
o                        61o

P                     3 cm                   S

In the diagram, PQS is a right angled triangle, ∠PSQ = 90o, |PS| = 3 cm, |PQ| = 5 cm, ∠PRS = 61o and ∠QPR = xo.  Calculate, correct to one decimal place, the

1. value of x;
2. area of PQR.

_____________________________________________________________________________________________________
Observation

This question was reportedly attempted by majority of the candidates and they performed well in it.  However, it was also observed that a few of them applied the wrong trig. ratio.
In part(a) ∠QPS = cos-1 = 53.13o.  xo = 53.13o – (90o – 61o) = 24.1o
In part(b), QS =  = 4 cm.  Area of PQS =  x 3 x 4 = 6 cm2
|RS| = 3tan29o = 1.6629.  Area of PRS =  x 3 x 1.6629 = 2.49439
Area of PQR  = Area of  PQS – Area of PRS = 6 – 2.49439 = 3.5 cm2, correct to 1 decimal place.