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 \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=10\text{,}\) \(\angle K=27\degree\) en \(\angle L=89\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis a De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({L\kern{-.8pt}M \over \sin(\angle K)}={K\kern{-.8pt}M \over \sin(\angle L)}={K\kern{-.8pt}L \over \sin(\angle M)}\text{.}\) 1p ○ Dus \(K\kern{-.8pt}M={L\kern{-.8pt}M⋅\sin(\angle L) \over \sin(\angle K)}={10⋅\sin(89\degree) \over \sin(27\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}M≈22{,}0\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=45\text{,}\) \(\angle A=47\degree\) en \(\angle B=107\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis b 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 \(A\kern{-.8pt}C={B\kern{-.8pt}C⋅\sin(\angle B) \over \sin(\angle A)}={45⋅\sin(107\degree) \over \sin(47\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}C≈58{,}8\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=8\text{,}\) \(L\kern{-.8pt}M=12\) en \(\angle M=40\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis c De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}L \over \sin(\angle M)}={L\kern{-.8pt}M \over \sin(\angle K)}={K\kern{-.8pt}M \over \sin(\angle L)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle K)={L\kern{-.8pt}M⋅\sin(\angle M) \over K\kern{-.8pt}L}={12⋅\sin(40\degree) \over 8}=0{,}964...\text{.}\) 1p ○ Dit geeft \(\angle K≈74{,}6\degree\) of \(\angle K≈105{,}4\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=11\text{,}\) \(A\kern{-.8pt}B=19\) en \(\angle B=26\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis d De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}C \over \sin(\angle B)}={A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle C)={A\kern{-.8pt}B⋅\sin(\angle B) \over A\kern{-.8pt}C}={19⋅\sin(26\degree) \over 11}=0{,}757...\text{.}\) 1p ○ Dit geeft \(\angle C≈49{,}2\degree\) of \(\angle C≈130{,}8\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=13\text{,}\) \(\angle L=65\degree\) en \(\angle K=64\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis a Uit \(\angle L+\angle M+\angle K=180\degree\) volgt \(\angle M=180\degree-\angle L-\angle K=180\degree-65\degree-64\degree=51\degree\text{.}\) 1p ○ De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}M \over \sin(\angle L)}={K\kern{-.8pt}L \over \sin(\angle M)}={L\kern{-.8pt}M \over \sin(\angle K)}\text{.}\) 1p ○ Dus \(K\kern{-.8pt}M={K\kern{-.8pt}L⋅\sin(\angle L) \over \sin(\angle M)}={13⋅\sin(65\degree) \over \sin(51\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}M≈15{,}2\text{.}\) 1p 4p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=45\text{,}\) \(\angle L=39\degree\) en \(\angle K=40\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis b Uit \(\angle L+\angle M+\angle K=180\degree\) volgt \(\angle M=180\degree-\angle L-\angle K=180\degree-39\degree-40\degree=101\degree\text{.}\) 1p ○ De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}M \over \sin(\angle L)}={K\kern{-.8pt}L \over \sin(\angle M)}={L\kern{-.8pt}M \over \sin(\angle K)}\text{.}\) 1p ○ Dus \(K\kern{-.8pt}M={K\kern{-.8pt}L⋅\sin(\angle L) \over \sin(\angle M)}={45⋅\sin(39\degree) \over \sin(101\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}M≈28{,}8\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=18\text{,}\) \(Q\kern{-.8pt}R=23\) en \(\angle Q=80\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis 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 \(P\kern{-.8pt}R^2=18^2+23^2-2⋅18⋅23⋅\cos(80\degree)=709{,}219...\text{.}\) 1p ○ \(P\kern{-.8pt}R=\sqrt{709{,}219...}≈26{,}6\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=43\text{,}\) \(K\kern{-.8pt}L=25\) en \(\angle K=100\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis d De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(L\kern{-.8pt}M^2=K\kern{-.8pt}M^2+K\kern{-.8pt}L^2-2⋅K\kern{-.8pt}M⋅K\kern{-.8pt}L⋅\cos(\angle K)\text{.}\) 1p ○ Dus \(L\kern{-.8pt}M^2=43^2+25^2-2⋅43⋅25⋅\cos(100\degree)=2847{,}343...\text{.}\) 1p ○ \(L\kern{-.8pt}M=\sqrt{2847{,}343...}≈53{,}4\text{.}\) 1p opgave 34p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=45\text{,}\) \(A\kern{-.8pt}B=24\) en \(B\kern{-.8pt}C=51\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis a De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(B\kern{-.8pt}C^2=A\kern{-.8pt}C^2+A\kern{-.8pt}B^2-2⋅A\kern{-.8pt}C⋅A\kern{-.8pt}B⋅\cos(\angle A)\text{.}\) 1p ○ Invullen geeft \(51^2=45^2+24^2-2⋅45⋅24⋅\cos(\angle A)\) 1p ○ Balansmethode geeft \(\cos(\angle A)={2\,601-2\,601 \over -2\,160}=-0\) 1p ○ Hieruit volgt \(\angle A=\cos^{-1}(-0)=90{,}0\degree\text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=30\text{,}\) \(P\kern{-.8pt}R=26\) en \(P\kern{-.8pt}Q=48\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis 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 \(48^2=30^2+26^2-2⋅30⋅26⋅\cos(\angle R)\) 1p ○ Balansmethode geeft \(\cos(\angle R)={2\,304-1\,576 \over -1\,560}=-0{,}466...\) 1p ○ Hieruit volgt \(\angle R=\cos^{-1}(-0{,}466...)≈117{,}8\degree\text{.}\) 1p |