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 \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=17\text{,}\) \(\angle A=57\degree\) en \(\angle B=62\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms a 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 \(A\kern{-.8pt}C={B\kern{-.8pt}C⋅\sin(\angle B) \over \sin(\angle A)}={17⋅\sin(62\degree) \over \sin(57\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}C≈17{,}9\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=21\text{,}\) \(\angle A=32\degree\) en \(\angle B=98\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms 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 \(A\kern{-.8pt}C={B\kern{-.8pt}C⋅\sin(\angle B) \over \sin(\angle A)}={21⋅\sin(98\degree) \over \sin(32\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}C≈39{,}2\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=13\text{,}\) \(P\kern{-.8pt}R=18\) en \(\angle P=46\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 8ms c 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 ○ Daaruit volgt \(\sin(\angle Q)={P\kern{-.8pt}R⋅\sin(\angle P) \over Q\kern{-.8pt}R}={18⋅\sin(46\degree) \over 13}=0{,}996...\text{.}\) 1p ○ Dit geeft \(\angle Q≈84{,}9\degree\) of \(\angle Q≈95{,}1\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=18\text{,}\) \(A\kern{-.8pt}B=26\) en \(\angle B=41\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms 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 \(\sin(\angle C)={A\kern{-.8pt}B⋅\sin(\angle B) \over A\kern{-.8pt}C}={26⋅\sin(41\degree) \over 18}=0{,}947...\text{.}\) 1p ○ Dit geeft \(\angle C≈71{,}4\degree\) of \(\angle C≈108{,}6\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=16\text{,}\) \(\angle A=53\degree\) en \(\angle C=58\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle A+\angle B+\angle C=180\degree\) volgt \(\angle B=180\degree-\angle A-\angle C=180\degree-53\degree-58\degree=69\degree\text{.}\) 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 \(B\kern{-.8pt}C={A\kern{-.8pt}C⋅\sin(\angle A) \over \sin(\angle B)}={16⋅\sin(53\degree) \over \sin(69\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈13{,}7\text{.}\) 1p 4p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=40\text{,}\) \(\angle K=27\degree\) en \(\angle M=35\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle K+\angle L+\angle M=180\degree\) volgt \(\angle L=180\degree-\angle K-\angle M=180\degree-27\degree-35\degree=118\degree\text{.}\) 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 \(L\kern{-.8pt}M={K\kern{-.8pt}M⋅\sin(\angle K) \over \sin(\angle L)}={40⋅\sin(27\degree) \over \sin(118\degree)}\text{.}\) 1p ○ \(L\kern{-.8pt}M≈20{,}6\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=21\text{,}\) \(B\kern{-.8pt}C=22\) en \(\angle B=69\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 1ms c 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 \(A\kern{-.8pt}C^2=21^2+22^2-2⋅21⋅22⋅\cos(69\degree)=593{,}868...\text{.}\) 1p ○ \(A\kern{-.8pt}C=\sqrt{593{,}868...}≈24{,}4\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=18\text{,}\) \(A\kern{-.8pt}B=16\) en \(\angle A=110\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms 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 \(B\kern{-.8pt}C^2=18^2+16^2-2⋅18⋅16⋅\cos(110\degree)=777{,}003...\text{.}\) 1p ○ \(B\kern{-.8pt}C=\sqrt{777{,}003...}≈27{,}9\text{.}\) 1p opgave 34p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=21\text{,}\) \(Q\kern{-.8pt}R=25\) en \(P\kern{-.8pt}R=27\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 5ms a 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 ○ Invullen geeft \(27^2=21^2+25^2-2⋅21⋅25⋅\cos(\angle Q)\) 1p ○ Balansmethode geeft \(\cos(\angle Q)={729-1\,066 \over -1\,050}=0{,}320...\) 1p ○ Hieruit volgt \(\angle Q=\cos^{-1}(0{,}320...)≈71{,}3\degree\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=37\text{,}\) \(B\kern{-.8pt}C=20\) en \(A\kern{-.8pt}C=43\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms 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 \(43^2=37^2+20^2-2⋅37⋅20⋅\cos(\angle B)\) 1p ○ Balansmethode geeft \(\cos(\angle B)={1\,849-1\,769 \over -1\,480}=-0{,}054...\) 1p ○ Hieruit volgt \(\angle B=\cos^{-1}(-0{,}054...)≈93{,}1\degree\text{.}\) 1p |