1999 - JAMB Mathematics Past Questions and Answers - page 5

< PST = < RQT = x(corresponding angle)

< PST = < PTS = < PTS x (PS = PT0

< SRQ = < SPT = 180^o (sum of < on straight line)

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

in < SPT, < PST = PTS = < PSt = 180^o

2x + 70 = 180^o

2x = 180^o - 70^o = 110^o

x = \(\frac{110^o}{2}\) = 55^o

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 =

xz² = xy² + yx² - 2(xy) (yz) cos 120^o

= 2² + 1² - 2(2) (1) cos 120²

= 4 + 1 - 4x - cos 60 = 5 - 4x - \(\frac{1}{2}\)

5 + 2 = 7

xz = \(\sqrt{7}\)m

degree of excellent in the pie-chart = 360^o - (90^o + 120^o + 80^o) = 360^o - 290^o = 70^o

= 360^o = 36 students

70 = \(\frac{70^o}{360^o}\) x 36 = 7