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 \(K\kern{-.8pt}M=13\text{,}\) \(\angle L=53\degree\) en \(\angle M=80\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms a 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 \(K\kern{-.8pt}L={K\kern{-.8pt}M⋅\sin(\angle M) \over \sin(\angle L)}={13⋅\sin(80\degree) \over \sin(53\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}L≈16{,}0\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=25\text{,}\) \(\angle B=36\degree\) en \(\angle C=114\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}C \over \sin(\angle B)}={A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}B={A\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle B)}={25⋅\sin(114\degree) \over \sin(36\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}B≈38{,}9\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=17\text{,}\) \(P\kern{-.8pt}R=28\) en \(\angle P=35\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 5ms 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 \(\sin(\angle Q)={P\kern{-.8pt}R⋅\sin(\angle P) \over Q\kern{-.8pt}R}={28⋅\sin(35\degree) \over 17}=0{,}944...\text{.}\) 1p ○ Dit geeft \(\angle Q≈70{,}9\degree\) of \(\angle Q≈109{,}1\degree\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=10\text{,}\) \(L\kern{-.8pt}M=14\) en \(\angle M=34\degree\text{.}\) 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 \(\sin(\angle K)={L\kern{-.8pt}M⋅\sin(\angle M) \over K\kern{-.8pt}L}={14⋅\sin(34\degree) \over 10}=0{,}782...\text{.}\) 1p ○ Dit geeft \(\angle K≈51{,}5\degree\) of \(\angle K≈128{,}5\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=22\text{,}\) \(\angle K=58\degree\) en \(\angle M=59\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle K+\angle L+\angle M=180\degree\) volgt \(\angle L=180\degree-\angle K-\angle M=180\degree-58\degree-59\degree=63\degree\text{.}\) 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 \(L\kern{-.8pt}M={K\kern{-.8pt}M⋅\sin(\angle K) \over \sin(\angle L)}={22⋅\sin(58\degree) \over \sin(63\degree)}\text{.}\) 1p ○ \(L\kern{-.8pt}M≈20{,}9\text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=45\text{,}\) \(\angle Q=29\degree\) en \(\angle P=44\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle Q+\angle R+\angle P=180\degree\) volgt \(\angle R=180\degree-\angle Q-\angle P=180\degree-29\degree-44\degree=107\degree\text{.}\) 1p ○ 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 \(P\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle Q) \over \sin(\angle R)}={45⋅\sin(29\degree) \over \sin(107\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}R≈22{,}8\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=30\text{,}\) \(A\kern{-.8pt}B=25\) en \(\angle A=81\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 0ms c 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 \(B\kern{-.8pt}C^2=30^2+25^2-2⋅30⋅25⋅\cos(81\degree)=1290{,}348...\text{.}\) 1p ○ \(B\kern{-.8pt}C=\sqrt{1290{,}348...}≈35{,}9\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=19\text{,}\) \(A\kern{-.8pt}B=20\) en \(\angle A=105\degree\text{.}\) 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 \(B\kern{-.8pt}C^2=19^2+20^2-2⋅19⋅20⋅\cos(105\degree)=957{,}702...\text{.}\) 1p ○ \(B\kern{-.8pt}C=\sqrt{957{,}702...}≈30{,}9\text{.}\) 1p opgave 34p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=28\text{,}\) \(A\kern{-.8pt}B=33\) en \(B\kern{-.8pt}C=39\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 5ms 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 \(39^2=28^2+33^2-2⋅28⋅33⋅\cos(\angle A)\) 1p ○ Balansmethode geeft \(\cos(\angle A)={1\,521-1\,873 \over -1\,848}=0{,}190...\) 1p ○ Hieruit volgt \(\angle A=\cos^{-1}(0{,}190...)≈79{,}0\degree\text{.}\) 1p 4p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=21\text{,}\) \(K\kern{-.8pt}M=27\) en \(K\kern{-.8pt}L=41\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms b De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(K\kern{-.8pt}L^2=L\kern{-.8pt}M^2+K\kern{-.8pt}M^2-2⋅L\kern{-.8pt}M⋅K\kern{-.8pt}M⋅\cos(\angle M)\text{.}\) 1p ○ Invullen geeft \(41^2=21^2+27^2-2⋅21⋅27⋅\cos(\angle M)\) 1p ○ Balansmethode geeft \(\cos(\angle M)={1\,681-1\,170 \over -1\,134}=-0{,}450...\) 1p ○ Hieruit volgt \(\angle M=\cos^{-1}(-0{,}450...)≈116{,}8\degree\text{.}\) 1p |