Hoeken berekenen deel 2

Leerdoelen

•Je kunt in een figuur hoeken uitrekenen gebruikmakend van de regels om hoeken te berekenen

Voorbeeldopgave

Stel, je hebt een figuur met\angle A=52\degree,\,\angle C=48\degree\text{ en }\angle D_2=27\degree\angle A=52\degree,\,\angle C=48\degree\text{ en }\angle D=27\degree\angle A=52\degree,\,\angle=48\degree\text{ en }\angle D=27\degree\angle A=52\degree,\,\angle B=48\degree\text{ en }\angle D=27\degree\angle A=52\degree,\,\angle B=48\degree\text{ en }\angle=27\degree\angle A=52\degree,\,\angle B=48\degree\text{ en }\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\text{ en }\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree\,\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree,\,\angle C=27\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree,\,\angle C=2\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree,\,\angle C=52\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree,\,\angle=52\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=48\degree,\,\angle A=52\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=4\degree,\,\angle A=52\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=\degree,\,\angle A=52\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=5\degree,\,\angle A=52\degree\triangle ABC,\,\angle A=52\degree,\,\angle B=52\degree,\,\angle A=52\degree\triangle ABC,\,\angle A=52\degree,\,\angle=52\degree,\,\angle A=52\degree\triangle ABC,\,\angle A=52\degree,\,\angle A=52\degree,\,\angle A=52\degree\triangle ABC,\,\angle A=52\degree,\,\angle A=52\degree\triangle ABC,\,\angle A=52\degree\triangle ABC,\,\angle A=52\triangle ABC,\,\angle A=52\triangle ABC,\,\angle A=52\triangle ABC,\,\angle A=52\triangle ABC,\,\angle A=52\triangle ABC,\,\angle A=5\triangle ABC,\,\angle A=\triangle ABC,\,\angle A\triangle ABC,\,\angle\triangle ABC,\,\triangle ABC,\,\triangle ABC,\,\triangle ABC,\,\triangle ABC,\triangle ABC,\triangle ABC,\triangle ABC,\triangle ABC,\triangle ABCABCABCABCABC.

Daarnaast is gegeven dat lijnevenwijdig is met lijn, en lijnevenwijdig is met. De uitdaging is om de waarden van\angle D_3,\,\angle E_3,\,\text{ en }\angle S_1\angle D_{},\,\angle E_3,\,\text{ en }\angle S_1\angle D_2,\,\angle E_3,\,\text{ en }\angle S_1\angle D,\,\angle E_3,\,\text{ en }\angle S_1\angle,\,\angle E_3,\,\text{ en }\angle S_1\angle A,\,\angle E_3,\,\text{ en }\angle S_1\angle A,\,\angle E,\,\text{ en }\angle S_1\angle A,\,\angle,\,\text{ en }\angle S_1\angle A,\,\angle A,\,\text{ en }\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S_1\angle A,\,\angle A,\,\angle S\angle A,\,\angle A,\,\angle\angle A,\,\angle A,\,\angle A\angle A,\,\angle A,\,\angle A,\angle A,\,\angle A,\,\angle A,\,\angle A,\,\angle A,\,\angle A,\,\angle A,\angle A,\angle A,\angle A\angle A=\angle A=5\angle A=52\angle A=52\degree,\,\angle A=52\degreete berekenen.

Zoeken naar F- en Z-hoeken

Een goede start is om eerst te kijken of je F-hoeken of Z-hoeken kunt vinden. Deze hoeken ontstaan wanneer lijnen parallel aan elkaar lopen.

1.Omdat lijnen lijnevenwijdig zijn, kun je F-hoeken identificeren. Hieruit volgt dat\angle F_1=\angle A=52\degree\angle F=\angle A=52\degree(F-hoeken).

2.Omdat lijn AC en DE evenwijdig zijn, kun je Z-hoeken identificeren. Hieruit volgt dat\angle C_1=\angle D_2=27\degree\angle C_1=\angle D=27\degree\angle C_1=\angle D2=27\degree\angle C=\angle D2=27\degree(Z-hoeken).

Berekenen van onbekende hoeken

Nu je de basis hebt gelegd, kun je beginnen met het berekenen van de gevraagde hoeken.

\angle S_1\angle S\angle\angle A\angle A,\angle A,\,:

•Nu je weet dat\angle F_1=52\degree\angle F_1=52\angle F_1=52\angle F_1=52\angle F_1=52\angle F_1=52\angle F_1=5\angle F_1=\angle F_1\angle F\angle\angle E\angle E_{}\angle E_3en\angle C_1=27\degree\angle C_1=27\angle C_1=27\angle C_1=27\angle C_1=27\angle C_1=27\angle C_1=27graden\angle C_1=h27graden\angle C_1=ho27graden\angle C_1=hoe27graden\angle C_1=hoek27graden\angle C_1=hoekC27graden\angle C_1=hoekC127graden\angle C_{}=hoekC127graden\angle C_3=hoekC127graden\angle_3=hoekC127graden\angle E_3=hoekC127graden\angle E_3hoekC127graden, kun je\angle S_1vinden door de hoekensom van\triangle FSC\triangle FS\triangle CFS\triangle CF\triangle C\trianglete gebruiken. Deze is altijd180\degree180180180180180180.\angle S_1\angle S\angle\angle E\angle E_{}\angle E_3is dan gelijk aan180-52-27=101\degree180-52-27=101180-52-27=101180-52-27=101180-52-27=101180-52-27=101180-52-27=101.

\angle D_3\angle D\angle\angle d\angle\angle S\angle S_{}\angle S_1:

•Gebruik de kennis van een gestrekte hoek om\angle D_3te vinden. Je weet al\angle D_1\angle D_{}\angle D_3(gelijk aan\angle S_1door F-hoeken) en\angle D_2\angle D_{}\angle D_3. Een gestrekte hoek is gelijk aan180\degree180180180180.\angle D_3is dan gelijk aan180-101-27=52\degree180-101-27=52180-101-27=52180-101-27=52180-101-27=52180-101-27=52.

\angle E_3\angle_3\angle D_3:

•Dit kan lastig lijken, maar door eerst\angle B\angle\angle E\angle E_{}\angle E_3te berekenen via de hoekensom van\triangle ABCABCABCABCABC, kun je vervolgens\angle E_3uitrekenen.\angle B\angle\angle E\angle E_{}\angle E_3is gelijk aan180-52-48=80\degree180-52-48=80180-52-48=80180-52-48=80180-52-48=80180-52-48=80.\angle E_3is dan gelijk aan180-52-80=48\degree180-52-80=48180-52-80=48180-52-80=48180-52-80=48180-52-80=48180-52-80=48g(de hoekensom van\triangle BDE\triangle BD\triangle B\triangle).

•Tip: je had\angle E_3\angle Eook sneller kunnen vinden. Omdat\angle E_3\,\text{en }\angle C\angle E_3\,\text{en }\angle C\angle E_3\,\text{en}\angle C\angle E_3\,\angle C\angle E_3\,\angle C\angle E_3\,\angle C\angle E_3\,\angle C\angle E_3\,\angle C\angle E_3\,\angle C\angle E_3\,\angle C\angle E_3\,\angle C\angle E_3\,\angle C\angle E_3\,\angle C\angle E_3\,\angle E_3\angle E_3\angle E_3\angle E_3\,\angle E_3\,\text{e}\angle E_3\,\text{en}\angle E_3\,\text{end}\angle E_3\,\text{en\$\$ \textbackslash angle C \$\$}\angle E_3\,\text{en\$\$ \textbackslash angle C \$\$}\angle E_3\,\text{en}\angle E_3\,\text{end}\angle E_3\,\text{endx}\angle E_3\,\text{endxd}\angle E_3\,\text{endx}\angle E_3\,\text{endx}\angle E_3\,\text{en}\angle E_3\,\text{en}\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\,\angle E_3\angle E_3\angle E_3\angle EF-hoeken zijn, geldt\angle E_3=\angle C=48\degree\angle E=\angle C=48\degree.