1999 - JAMB Mathematics Past Questions and Answers - page 5

< PST = < RQT = x(corresponding angle)
< PST = < PTS = < PTS x (PS = PT0
< SRQ = < SPT = 180o (sum of < on straight line)
< SPT = 180o - 110o = 70o
in < SPT, < PST = PTS = < PSt = 180o
2x + 70 = 180o
2x = 180o - 70o = 110o
x = (\frac{110^o}{2}) = 55o
Users' Answers & Comments
In (\bigtriangleup)WOP, b2 = r2 - (4.502...........(i)
In (\bigtriangleup)WOT, b(\bigtriangleup) + (4.5 + x)2 = a2...........(ii)
In (\bigtriangleup)ZOT, a2 = 62 + r2 ..........(iii)
subst. equation (iii) and (i) in eqn (ii)
r2 - (4.5)2 + (4.5 + x)2 = 36 = 20.25
= (4.5 + x)2 = (\sqrt{56.25}) = (4.5 + x)
= 56.25
= 4.5 + x = 7.5
x = 3.0m
x =
Users' Answers & Comments
xz2 = xy2 + yx2 - 2(xy) (yz) cos 120o
= 22 + 12 - 2(2) (1) cos 1202
= 4 + 1 - 4x - cos 60 = 5 - 4x - (\frac{1}{2})
5 + 2 = 7
xz = (\sqrt{7})m
Users' Answers & Comments
degree of excellent in the pie-chart = 360o - (90o + 120o + 80o) = 360o - 290o = 70o
= 360o = 36 students
70 = (\frac{70^o}{360^o}) x 36 = 7
Users' Answers & Comments