Getal & Ruimte (12e editie) - vwo wiskunde B
'Sinus- en cosinusregel'.
| vwo wiskunde B | 3.5 De sinusregel en de cosinusregel |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=18\text{,}\) \(\angle Q=57\degree\) en \(\angle R=86\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 1ms a De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}={Q\kern{-.8pt}R \over \sin(\angle P)}\text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q={P\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle Q)}={18⋅\sin(86\degree) \over \sin(57\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}Q≈21{,}4\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=13\text{,}\) \(\angle C=33\degree\) en \(\angle A=91\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}\text{.}\) 1p ○ Dus \(B\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle A) \over \sin(\angle C)}={13⋅\sin(91\degree) \over \sin(33\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈23{,}9\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=16\text{,}\) \(A\kern{-.8pt}C=20\) en \(\angle A=50\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 9ms c De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}={A\kern{-.8pt}B \over \sin(\angle C)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle B)={A\kern{-.8pt}C⋅\sin(\angle A) \over B\kern{-.8pt}C}={20⋅\sin(50\degree) \over 16}=0{,}957...\text{.}\) 1p ○ Dit geeft \(\angle B≈73{,}2\degree\) of \(\angle B≈106{,}8\degree\text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=9\text{,}\) \(Q\kern{-.8pt}R=16\) en \(\angle R=27\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}Q \over \sin(\angle R)}={Q\kern{-.8pt}R \over \sin(\angle P)}={P\kern{-.8pt}R \over \sin(\angle Q)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle P)={Q\kern{-.8pt}R⋅\sin(\angle R) \over P\kern{-.8pt}Q}={16⋅\sin(27\degree) \over 9}=0{,}807...\text{.}\) 1p ○ Dit geeft \(\angle P≈53{,}8\degree\) of \(\angle P≈126{,}2\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=24\text{,}\) \(\angle A=62\degree\) en \(\angle C=37\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 1ms a Uit \(\angle A+\angle B+\angle C=180\degree\) volgt \(\angle B=180\degree-\angle A-\angle C=180\degree-62\degree-37\degree=81\degree\text{.}\) 1p ○ De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}={A\kern{-.8pt}B \over \sin(\angle C)}\text{.}\) 1p ○ Dus \(B\kern{-.8pt}C={A\kern{-.8pt}C⋅\sin(\angle A) \over \sin(\angle B)}={24⋅\sin(62\degree) \over \sin(81\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈21{,}5\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=39\text{,}\) \(\angle C=28\degree\) en \(\angle B=35\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle C+\angle A+\angle B=180\degree\) volgt \(\angle A=180\degree-\angle C-\angle B=180\degree-28\degree-35\degree=117\degree\text{.}\) 1p ○ De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}B={B\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle A)}={39⋅\sin(28\degree) \over \sin(117\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}B≈20{,}5\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=22\text{,}\) \(Q\kern{-.8pt}R=34\) en \(\angle Q=80\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 1ms c De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}R^2=P\kern{-.8pt}Q^2+Q\kern{-.8pt}R^2-2⋅P\kern{-.8pt}Q⋅Q\kern{-.8pt}R⋅\cos(\angle Q)\text{.}\) 1p ○ Dus \(P\kern{-.8pt}R^2=22^2+34^2-2⋅22⋅34⋅\cos(80\degree)=1380{,}222...\text{.}\) 1p ○ \(P\kern{-.8pt}R=\sqrt{1380{,}222...}≈37{,}2\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=15\text{,}\) \(B\kern{-.8pt}C=21\) en \(\angle B=107\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms d De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(A\kern{-.8pt}C^2=A\kern{-.8pt}B^2+B\kern{-.8pt}C^2-2⋅A\kern{-.8pt}B⋅B\kern{-.8pt}C⋅\cos(\angle B)\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C^2=15^2+21^2-2⋅15⋅21⋅\cos(107\degree)=850{,}194...\text{.}\) 1p ○ \(A\kern{-.8pt}C=\sqrt{850{,}194...}≈29{,}2\text{.}\) 1p opgave 34p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=12\text{,}\) \(P\kern{-.8pt}Q=12\) en \(Q\kern{-.8pt}R=13\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 8ms a De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(Q\kern{-.8pt}R^2=P\kern{-.8pt}R^2+P\kern{-.8pt}Q^2-2⋅P\kern{-.8pt}R⋅P\kern{-.8pt}Q⋅\cos(\angle P)\text{.}\) 1p ○ Invullen geeft \(13^2=12^2+12^2-2⋅12⋅12⋅\cos(\angle P)\) 1p ○ Balansmethode geeft \(\cos(\angle P)={169-288 \over -288}=0{,}413...\) 1p ○ Hieruit volgt \(\angle P=\cos^{-1}(0{,}413...)≈65{,}6\degree\text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=12\text{,}\) \(P\kern{-.8pt}R=17\) en \(P\kern{-.8pt}Q=22\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms b De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}Q^2=Q\kern{-.8pt}R^2+P\kern{-.8pt}R^2-2⋅Q\kern{-.8pt}R⋅P\kern{-.8pt}R⋅\cos(\angle R)\text{.}\) 1p ○ Invullen geeft \(22^2=12^2+17^2-2⋅12⋅17⋅\cos(\angle R)\) 1p ○ Balansmethode geeft \(\cos(\angle R)={484-433 \over -408}=-0{,}125\) 1p ○ Hieruit volgt \(\angle R=\cos^{-1}(-0{,}125)≈97{,}2\degree\text{.}\) 1p |