Getal & Ruimte (13e editie) - havo wiskunde B
'Sinus- en cosinusregel'.
| havo wiskunde B | 3.2 De sinusregel |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=18\text{,}\) \(\angle P=44\degree\) en \(\angle Q=78\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 1ms 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 \(P\kern{-.8pt}R={Q\kern{-.8pt}R⋅\sin(\angle Q) \over \sin(\angle P)}={18⋅\sin(78\degree) \over \sin(44\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}R≈25{,}3\text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=28\text{,}\) \(\angle M=31\degree\) en \(\angle K=91\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b 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 ○ Dus \(L\kern{-.8pt}M={K\kern{-.8pt}L⋅\sin(\angle K) \over \sin(\angle M)}={28⋅\sin(91\degree) \over \sin(31\degree)}\text{.}\) 1p ○ \(L\kern{-.8pt}M≈54{,}4\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=13\text{,}\) \(L\kern{-.8pt}M=21\) en \(\angle M=25\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 9ms 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 \(\sin(\angle K)={L\kern{-.8pt}M⋅\sin(\angle M) \over K\kern{-.8pt}L}={21⋅\sin(25\degree) \over 13}=0{,}682...\text{.}\) 1p ○ Dit geeft \(\angle K≈43{,}1\degree\) of \(\angle K≈136{,}9\degree\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=10\text{,}\) \(K\kern{-.8pt}M=15\) en \(\angle K=37\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d 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 ○ Daaruit volgt \(\sin(\angle L)={K\kern{-.8pt}M⋅\sin(\angle K) \over L\kern{-.8pt}M}={15⋅\sin(37\degree) \over 10}=0{,}902...\text{.}\) 1p ○ Dit geeft \(\angle L≈64{,}5\degree\) of \(\angle L≈115{,}5\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=24\text{,}\) \(\angle B=65\degree\) en \(\angle A=32\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 1ms a Uit \(\angle B+\angle C+\angle A=180\degree\) volgt \(\angle C=180\degree-\angle B-\angle A=180\degree-65\degree-32\degree=83\degree\text{.}\) 1p ○ 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 \(A\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle B) \over \sin(\angle C)}={24⋅\sin(65\degree) \over \sin(83\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}C≈21{,}9\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=25\text{,}\) \(\angle B=32\degree\) en \(\angle A=41\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle B+\angle C+\angle A=180\degree\) volgt \(\angle C=180\degree-\angle B-\angle A=180\degree-32\degree-41\degree=107\degree\text{.}\) 1p ○ 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 \(A\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle B) \over \sin(\angle C)}={25⋅\sin(32\degree) \over \sin(107\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}C≈13{,}9\text{.}\) 1p |
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| havo wiskunde B | 3.3 De cosinusregel |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=17\text{,}\) \(B\kern{-.8pt}C=15\) en \(\angle B=88\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 1ms a 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 \(A\kern{-.8pt}C^2=17^2+15^2-2⋅17⋅15⋅\cos(88\degree)=496{,}201...\text{.}\) 1p ○ \(A\kern{-.8pt}C=\sqrt{496{,}201...}≈22{,}3\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=36\text{,}\) \(P\kern{-.8pt}R=19\) en \(\angle R=97\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms 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 ○ Dus \(P\kern{-.8pt}Q^2=36^2+19^2-2⋅36⋅19⋅\cos(97\degree)=1823{,}717...\text{.}\) 1p ○ \(P\kern{-.8pt}Q=\sqrt{1823{,}717...}≈42{,}7\text{.}\) 1p 4p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=19\text{,}\) \(A\kern{-.8pt}B=21\) en \(B\kern{-.8pt}C=26\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 8ms 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 ○ Invullen geeft \(26^2=19^2+21^2-2⋅19⋅21⋅\cos(\angle A)\) 1p ○ Balansmethode geeft \(\cos(\angle A)={676-802 \over -798}=0{,}157...\) 1p ○ Hieruit volgt \(\angle A=\cos^{-1}(0{,}157...)≈80{,}9\degree\text{.}\) 1p 4p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=14\text{,}\) \(A\kern{-.8pt}C=14\) en \(A\kern{-.8pt}B=25\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms d De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(A\kern{-.8pt}B^2=B\kern{-.8pt}C^2+A\kern{-.8pt}C^2-2⋅B\kern{-.8pt}C⋅A\kern{-.8pt}C⋅\cos(\angle C)\text{.}\) 1p ○ Invullen geeft \(25^2=14^2+14^2-2⋅14⋅14⋅\cos(\angle C)\) 1p ○ Balansmethode geeft \(\cos(\angle C)={625-392 \over -392}=-0{,}594...\) 1p ○ Hieruit volgt \(\angle C=\cos^{-1}(-0{,}594...)≈126{,}5\degree\text{.}\) 1p |