A, B, C, D pe un cerc
AD=R,<CBD=60⁰ și<BOC=80⁰
a)arcul mic AB
AC n BD=M=> <CMD
a) dacă AD=R =>∆ADO echilateral=><AOD=60⁰
<CBD=60⁰ unghiul pe cerc deci <DOC = 2×<CBD=2×60⁰=120⁰
observăm că <AOD+<DOC=60⁰+120⁰=180⁰
arcul ABC=180⁰
[tex]\it Folosim\ sensul\ trigonometric.\\ \\ \widehat{CBD} = 60^o\Rightarrow \stackrel\frown{DC}=120^o\\ \\ \widehat{BOC}=80^o\Rightarrow \stackrel\frown{CB}=80^o\\ \\ \Delta OAD-\ echilateral\ (OA=OD=AD=R) \Rightarrow \stackrel\frown{AD}=60^o[/tex]
[tex]\it \stackrel\frown{BA}=360^o-(80^o+120^o+60^o)=100^o\\ \\ \\ b)\ \ \stackrel\frown{BA}=100^o \Rightarrow \widehat{AOB}=100^o\\ \\ \widehat{DOB}=60^o+100^o=160^o\\ \\ \Delta DOB\ -\ isoscel\ (OB=OD=R) \Rightarrow \widehat{MDO}=(180^o-160^o):2=10^o\\ \\ Din\ \Delta MDO \Rightarrow \widehat{OMD}=180^o-(60^o+10^o)=110^o=\widehat{CMD}[/tex]
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A, B, C, D pe un cerc
AD=R,<CBD=60⁰ și<BOC=80⁰
a)arcul mic AB
AC n BD=M=> <CMD
demonstrație
a) dacă AD=R =>∆ADO echilateral=><AOD=60⁰
<CBD=60⁰ unghiul pe cerc deci <DOC = 2×<CBD=2×60⁰=120⁰
observăm că <AOD+<DOC=60⁰+120⁰=180⁰
(astfel încât AC diametru)
arcul ABC=180⁰
arcul mic AB=180⁰-80⁰=100⁰
b) <CMD=(arc DC+arc AB)/2=(120⁰+100⁰)/2=220/2=110⁰
[tex]\it Folosim\ sensul\ trigonometric.\\ \\ \widehat{CBD} = 60^o\Rightarrow \stackrel\frown{DC}=120^o\\ \\ \widehat{BOC}=80^o\Rightarrow \stackrel\frown{CB}=80^o\\ \\ \Delta OAD-\ echilateral\ (OA=OD=AD=R) \Rightarrow \stackrel\frown{AD}=60^o[/tex]
[tex]\it \stackrel\frown{BA}=360^o-(80^o+120^o+60^o)=100^o\\ \\ \\ b)\ \ \stackrel\frown{BA}=100^o \Rightarrow \widehat{AOB}=100^o\\ \\ \widehat{DOB}=60^o+100^o=160^o\\ \\ \Delta DOB\ -\ isoscel\ (OB=OD=R) \Rightarrow \widehat{MDO}=(180^o-160^o):2=10^o\\ \\ Din\ \Delta MDO \Rightarrow \widehat{OMD}=180^o-(60^o+10^o)=110^o=\widehat{CMD}[/tex]