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Sagot :
1. so since we assume that CF is a straight line that means that the angle CAF is 180 angle CAD is 90 and FAE is 65 so 180-(65+90)=DAE=25
2. so since angle QPR= angle TPS (both are signified by f) and we assume that QT is straight so then QPT=180
180-154=2f=26
angle f=13
QPR=13
3. so we assume that BG is straight so it is 180
we add up the known angles (BQD+DQE+FQG=10+103+12=125) 180-125=DQC+EQF=55=2x+3x=5x
5x=55
x=11
CQD=2x=22
EQF=3x=33
4. we know that x=x and the whole is 360 so
x+x+x+x+71+71+71+71=360=4x+284=360
4x=76
x=19
8. so we assume that EF and CD are straight lines so EQD=CQF so 155=56+3x so
99=3x
CQG=99
9. so adjacent and the ratio is 4:5 so a line is 180 so 4+5=9 180/9=20 20*4=80 20*5=100 4:5 =80:100
a. 80 and 100
b. you should be able to do this
10. so on a line is 180
a. 3:4:5 3+4+5=12 180/12=15
3*15=45 4*15=60 5*15=75
3:4:5 =45:60:75
the angles are 45 60 75
b. you should be able to do this from my work
2. so since angle QPR= angle TPS (both are signified by f) and we assume that QT is straight so then QPT=180
180-154=2f=26
angle f=13
QPR=13
3. so we assume that BG is straight so it is 180
we add up the known angles (BQD+DQE+FQG=10+103+12=125) 180-125=DQC+EQF=55=2x+3x=5x
5x=55
x=11
CQD=2x=22
EQF=3x=33
4. we know that x=x and the whole is 360 so
x+x+x+x+71+71+71+71=360=4x+284=360
4x=76
x=19
8. so we assume that EF and CD are straight lines so EQD=CQF so 155=56+3x so
99=3x
CQG=99
9. so adjacent and the ratio is 4:5 so a line is 180 so 4+5=9 180/9=20 20*4=80 20*5=100 4:5 =80:100
a. 80 and 100
b. you should be able to do this
10. so on a line is 180
a. 3:4:5 3+4+5=12 180/12=15
3*15=45 4*15=60 5*15=75
3:4:5 =45:60:75
the angles are 45 60 75
b. you should be able to do this from my work
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