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8. < ΔDOC = 180 -(<BDC + <ACD)
< BCD = <DAB = 60
<ACD = <BCD - <ACB
= 60 - 35
= 25
< ADC = 360 - (2<BCD) / 2
= 180 - <BCD
= 180 - 60
= 120
< BDC = <ADC - <ADB
= 120 - 80
= 40
Shgga < DOC = 180 - (40+25)
= 115
9. Jika < garis AB dgn garis b = a⁰
maka <o = 1/2a°
maka sudut reflex = 180 - a⁰ yang nilainya sama dengan < garis AB dg a
sehingga < x = 1/2 (180-a°)
= 90 - 1/2 a
<APB = 180 - (<o + <x)
= 180 - (1/2a+90-1/2a)
= 180 - 1/2a + 90 + 1/2a
= 90 (terbukti)