2011 - WAEC Mathematics Past Questions and Answers - page 5

< NOP = 180 - 125 = 55^o(< s on a straight line)

But < NOP = < ONM (alternate < s)

< ONM = 55^o

< M + < N + < MQN = 180^o (sum of interior < s of a \(\bigtriangleup\)) i.e

65^o + 55^o + < MQN = 180

< MQN = 180 - 12o = 60

< MQR + < MQN = 180 (< s on straight line)

< MQR + 60 = 180

< MQR + 60 = 180

< MQR = 180 - 60

= 120^o

< TUW = 110^o = 180^o (< s on a straight line)

< TUW = 180^o - 110^o = 70^o

In \(\bigtriangleup\) XTU, < XUT + < TXU = 180^o

i.e. < YTS + 70^o = 180

< XTU = 180 - 110^o = 70^o

Also < YTS + < XTU = 180 (< s on a straight line)

i.e. < YTS + < XTU - 180(< s on straight line)

i.e. < YTS + 70^o = 180

< YTS = 180 - 70 = 110^o

in \(\bigtriangleup\) SYT + < YST + < YTS = 180^o(Sum of interior < s)

SYT + 40 + 110 = 180

< SYT = 180 - 150 = 30

< SYT = < XYW (vertically opposite < s)

Also < SYX = < TYW (vertically opposite < s)

but < SYT + < XYW + < SYX + < TYW = 360

i.e. 30 + 30 + < SYX + TYW = 360

but < SYX = < TYW

60 + 2(< TYW) = 360

2(< TYW) = 360^o - 60

2(< TYW) = 300^o

TYW = \(\frac{300}{2}\) = 150^o
< SYT

student that like swimming = x + 2

where 2 is the number of students who like reading, dancing and swimming. To find x from the venn diagram of swimming

6 +3 + 2 + x = 16

11 + x = 16

x = 16 - 11 = 5

no. of students that like dancing and swimming

x + 2 = 5 + 2 = 7

< QPR = < STU = 65^o (Corresponding angles)

245 + < QPR = x = 360^o (< s at a point)

i.e. 245 + 65 + x = 360

x = 360 - (245 + 65)

x = 360 - 310

x = 50^o

< QPR = < STU = 65^o (Corresponding angles)

245 + < QPR = x = 360^o (< s at a point)

i.e. 245 + 65 + x = 360

x = 360 - (245 + 65)

x = 50^o