Getal & Ruimte (12e editie) - havo wiskunde B
'Sinus- en cosinusregel'.
| havo wiskunde B | 3.2 De sinusregel en de cosinusregel |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=28\text{,}\) \(\angle P=62\degree\) en \(\angle Q=78\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis a De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({Q\kern{-.8pt}R \over \sin(\angle P)}={P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}\text{.}\) 1p ○ Dus \(P\kern{-.8pt}R={Q\kern{-.8pt}R⋅\sin(\angle Q) \over \sin(\angle P)}={28⋅\sin(78\degree) \over \sin(62\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}R≈31{,}0\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=20\text{,}\) \(\angle R=28\degree\) en \(\angle P=121\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis b 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 ○ Dus \(Q\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle P) \over \sin(\angle R)}={20⋅\sin(121\degree) \over \sin(28\degree)}\text{.}\) 1p ○ \(Q\kern{-.8pt}R≈36{,}5\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=10\text{,}\) \(A\kern{-.8pt}B=15\) en \(\angle B=28\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis c 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 \(\sin(\angle C)={A\kern{-.8pt}B⋅\sin(\angle B) \over A\kern{-.8pt}C}={15⋅\sin(28\degree) \over 10}=0{,}704...\text{.}\) 1p ○ Dit geeft \(\angle C≈44{,}8\degree\) of \(\angle C≈135{,}2\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=19\text{,}\) \(A\kern{-.8pt}C=29\) en \(\angle A=33\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis d 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 \(\sin(\angle B)={A\kern{-.8pt}C⋅\sin(\angle A) \over B\kern{-.8pt}C}={29⋅\sin(33\degree) \over 19}=0{,}831...\text{.}\) 1p ○ Dit geeft \(\angle B≈56{,}2\degree\) of \(\angle B≈123{,}8\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=26\text{,}\) \(\angle K=35\degree\) en \(\angle M=58\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis a Uit \(\angle K+\angle L+\angle M=180\degree\) volgt \(\angle L=180\degree-\angle K-\angle M=180\degree-35\degree-58\degree=87\degree\text{.}\) 1p ○ 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 \(L\kern{-.8pt}M={K\kern{-.8pt}M⋅\sin(\angle K) \over \sin(\angle L)}={26⋅\sin(35\degree) \over \sin(87\degree)}\text{.}\) 1p ○ \(L\kern{-.8pt}M≈14{,}9\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=37\text{,}\) \(\angle A=48\degree\) en \(\angle C=32\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis b Uit \(\angle A+\angle B+\angle C=180\degree\) volgt \(\angle B=180\degree-\angle A-\angle C=180\degree-48\degree-32\degree=100\degree\text{.}\) 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 \(B\kern{-.8pt}C={A\kern{-.8pt}C⋅\sin(\angle A) \over \sin(\angle B)}={37⋅\sin(48\degree) \over \sin(100\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈27{,}9\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=27\text{,}\) \(Q\kern{-.8pt}R=44\) en \(\angle Q=84\degree\text{.}\) 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 \(P\kern{-.8pt}R^2=27^2+44^2-2⋅27⋅44⋅\cos(84\degree)=2416{,}640...\text{.}\) 1p ○ \(P\kern{-.8pt}R=\sqrt{2416{,}640...}≈49{,}2\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=28\text{,}\) \(A\kern{-.8pt}B=33\) en \(\angle A=118\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis d 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 ○ Dus \(B\kern{-.8pt}C^2=28^2+33^2-2⋅28⋅33⋅\cos(118\degree)=2740{,}583...\text{.}\) 1p ○ \(B\kern{-.8pt}C=\sqrt{2740{,}583...}≈52{,}4\text{.}\) 1p opgave 34p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=18\text{,}\) \(L\kern{-.8pt}M=17\) en \(K\kern{-.8pt}M=20\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis a De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(K\kern{-.8pt}M^2=K\kern{-.8pt}L^2+L\kern{-.8pt}M^2-2⋅K\kern{-.8pt}L⋅L\kern{-.8pt}M⋅\cos(\angle L)\text{.}\) 1p ○ Invullen geeft \(20^2=18^2+17^2-2⋅18⋅17⋅\cos(\angle L)\) 1p ○ Balansmethode geeft \(\cos(\angle L)={400-613 \over -612}=0{,}348...\) 1p ○ Hieruit volgt \(\angle L=\cos^{-1}(0{,}348...)≈69{,}6\degree\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=28\text{,}\) \(B\kern{-.8pt}C=28\) en \(A\kern{-.8pt}C=46\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis b 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 ○ Invullen geeft \(46^2=28^2+28^2-2⋅28⋅28⋅\cos(\angle B)\) 1p ○ Balansmethode geeft \(\cos(\angle B)={2\,116-1\,568 \over -1\,568}=-0{,}349...\) 1p ○ Hieruit volgt \(\angle B=\cos^{-1}(-0{,}349...)≈110{,}5\degree\text{.}\) 1p |