Question

Buna seara! Va rog problema 11 pct e.

Answer 1

BH fiind bisectoare y=3x-12°
x°+x°+17°+3x-12°+3x-12°=180°
8×x°=180°+12°+12°-17°
8×x°=187°
8×x°=184°180'
8×x°=184°176'240''
x°=184°176'240'' :8
x°23°22'30''
m<ABE=23°22'30''
m<EBD=23°22'30''
m<JBH=3×23°22'30''-12°=69°66'90''-12°=69°67'30''-12°=70°7'30''-12°=58°7'30''=m<HBC
f)x°+x°=90°
x=45°
180°-45°-45°-20°=y°+y°
135°-45°-20°=2×y°
90°-20°=2×y°
70°=2×y°
y°=35°
m<EAB=m<EBD=45°
m<JBH=m<CBH=35°