Getal & Ruimte (13e editie) - vwo wiskunde B
'Sinus- en cosinusregel'.
| vwo wiskunde B | 3.4 De sinusregel en de cosinusregel |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=21\text{,}\) \(\angle K=55\degree\) en \(\angle L=64\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms 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 \(K\kern{-.8pt}M={L\kern{-.8pt}M⋅\sin(\angle L) \over \sin(\angle K)}={21⋅\sin(64\degree) \over \sin(55\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}M≈23{,}0\text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=24\text{,}\) \(\angle K=63\degree\) en \(\angle L=91\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b 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 \(K\kern{-.8pt}M={L\kern{-.8pt}M⋅\sin(\angle L) \over \sin(\angle K)}={24⋅\sin(91\degree) \over \sin(63\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}M≈26{,}9\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=7\text{,}\) \(L\kern{-.8pt}M=11\) en \(\angle M=27\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 8ms 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 \(\sin(\angle K)={L\kern{-.8pt}M⋅\sin(\angle M) \over K\kern{-.8pt}L}={11⋅\sin(27\degree) \over 7}=0{,}713...\text{.}\) 1p ○ Dit geeft \(\angle K≈45{,}5\degree\) of \(\angle K≈134{,}5\degree\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=7\text{,}\) \(L\kern{-.8pt}M=10\) en \(\angle M=36\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d 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 \(\sin(\angle K)={L\kern{-.8pt}M⋅\sin(\angle M) \over K\kern{-.8pt}L}={10⋅\sin(36\degree) \over 7}=0{,}839...\text{.}\) 1p ○ Dit geeft \(\angle K≈57{,}1\degree\) of \(\angle K≈122{,}9\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=39\text{,}\) \(\angle K=48\degree\) en \(\angle M=45\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle K+\angle L+\angle M=180\degree\) volgt \(\angle L=180\degree-\angle K-\angle M=180\degree-48\degree-45\degree=87\degree\text{.}\) 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 \(L\kern{-.8pt}M={K\kern{-.8pt}M⋅\sin(\angle K) \over \sin(\angle L)}={39⋅\sin(48\degree) \over \sin(87\degree)}\text{.}\) 1p ○ \(L\kern{-.8pt}M≈29{,}0\text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=43\text{,}\) \(\angle R=42\degree\) en \(\angle Q=44\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle R+\angle P+\angle Q=180\degree\) volgt \(\angle P=180\degree-\angle R-\angle Q=180\degree-42\degree-44\degree=94\degree\text{.}\) 1p ○ 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 \(P\kern{-.8pt}Q={Q\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle P)}={43⋅\sin(42\degree) \over \sin(94\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}Q≈28{,}8\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=21\text{,}\) \(K\kern{-.8pt}L=23\) en \(\angle K=87\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 1ms c 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 \(L\kern{-.8pt}M^2=21^2+23^2-2⋅21⋅23⋅\cos(87\degree)=919{,}443...\text{.}\) 1p ○ \(L\kern{-.8pt}M=\sqrt{919{,}443...}≈30{,}3\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=35\text{,}\) \(B\kern{-.8pt}C=26\) en \(\angle B=105\degree\text{.}\) 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 \(A\kern{-.8pt}C^2=35^2+26^2-2⋅35⋅26⋅\cos(105\degree)=2372{,}050...\text{.}\) 1p ○ \(A\kern{-.8pt}C=\sqrt{2372{,}050...}≈48{,}7\text{.}\) 1p opgave 34p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=10\text{,}\) \(Q\kern{-.8pt}R=10\) en \(P\kern{-.8pt}R=10\text{.}\) 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 \(10^2=10^2+10^2-2⋅10⋅10⋅\cos(\angle Q)\) 1p ○ Balansmethode geeft \(\cos(\angle Q)={100-200 \over -200}=0{,}5\) 1p ○ Hieruit volgt \(\angle Q=\cos^{-1}(0{,}5)=60{,}0\degree\text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=25\text{,}\) \(P\kern{-.8pt}Q=14\) en \(Q\kern{-.8pt}R=32\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms b 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 \(32^2=25^2+14^2-2⋅25⋅14⋅\cos(\angle P)\) 1p ○ Balansmethode geeft \(\cos(\angle P)={1\,024-821 \over -700}=-0{,}29\) 1p ○ Hieruit volgt \(\angle P=\cos^{-1}(-0{,}29)≈106{,}9\degree\text{.}\) 1p |