Question

in this figure lines pq and st inersect at o. if angle POR=90° and x:y = 3:2 then Z is equal to
a) 126° b) 144° c) 136° d) 154°

Answer 1

x:y=3:2
let M be the common multiple
x=3m
y=2m
<POR+<ROT+<TOQ=180
-LINEAR PAIR
<POR=90
90+3M+2M=180
90+5M=180
5M=180-90
5M=90
M=90/5
M=18
x=3m=3x18
54
y=2m=2x18
36

now
<QOS+<TOQ=180
-LINEAR PAIR
Y+Z=180
36+Z=180
Z=180-36
Z=144

Answer 2

Given:-POR=90°

POR+RXT+TYQ=180°(PQ is a straight line )

90°+3x+2x=180°

90°+5x=180°

5x=180°-90°

5x=90°

x=90°/5

x=18°

therefore RXT=18×3=54°

TYQ=18×2=36°

TYQ+Z=180°(TS is a straight line )

36°+Z=180°

Z=180°-36°

Z=144°