Getal & Ruimte (13e editie) - havo wiskunde B
'Sinus- en cosinusregel'.
| havo wiskunde B | 3.2 De sinusregel |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=28\text{,}\) \(\angle A=35\degree\) en \(\angle B=85\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 1ms 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 \(A\kern{-.8pt}C={B\kern{-.8pt}C⋅\sin(\angle B) \over \sin(\angle A)}={28⋅\sin(85\degree) \over \sin(35\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}C≈48{,}6\text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=24\text{,}\) \(\angle K=32\degree\) en \(\angle L=119\degree\text{.}\) 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 \(K\kern{-.8pt}M={L\kern{-.8pt}M⋅\sin(\angle L) \over \sin(\angle K)}={24⋅\sin(119\degree) \over \sin(32\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}M≈39{,}6\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=8\text{,}\) \(K\kern{-.8pt}L=11\) en \(\angle L=46\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 9ms c De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}M \over \sin(\angle L)}={K\kern{-.8pt}L \over \sin(\angle M)}={L\kern{-.8pt}M \over \sin(\angle K)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle M)={K\kern{-.8pt}L⋅\sin(\angle L) \over K\kern{-.8pt}M}={11⋅\sin(46\degree) \over 8}=0{,}989...\text{.}\) 1p ○ Dit geeft \(\angle M≈81{,}5\degree\) of \(\angle M≈98{,}5\degree\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=6\text{,}\) \(K\kern{-.8pt}L=12\) en \(\angle L=27\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}M \over \sin(\angle L)}={K\kern{-.8pt}L \over \sin(\angle M)}={L\kern{-.8pt}M \over \sin(\angle K)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle M)={K\kern{-.8pt}L⋅\sin(\angle L) \over K\kern{-.8pt}M}={12⋅\sin(27\degree) \over 6}=0{,}907...\text{.}\) 1p ○ Dit geeft \(\angle M≈65{,}2\degree\) of \(\angle M≈114{,}8\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=42\text{,}\) \(\angle M=41\degree\) en \(\angle L=55\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 1ms a Uit \(\angle M+\angle K+\angle L=180\degree\) volgt \(\angle K=180\degree-\angle M-\angle L=180\degree-41\degree-55\degree=84\degree\text{.}\) 1p ○ 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 \(K\kern{-.8pt}L={L\kern{-.8pt}M⋅\sin(\angle M) \over \sin(\angle K)}={42⋅\sin(41\degree) \over \sin(84\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}L≈27{,}7\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=40\text{,}\) \(\angle C=28\degree\) en \(\angle B=26\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle C+\angle A+\angle B=180\degree\) volgt \(\angle A=180\degree-\angle C-\angle B=180\degree-28\degree-26\degree=126\degree\text{.}\) 1p ○ De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}B={B\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle A)}={40⋅\sin(28\degree) \over \sin(126\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}B≈23{,}2\text{.}\) 1p |
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| havo wiskunde B | 3.3 De cosinusregel |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=15\text{,}\) \(Q\kern{-.8pt}R=15\) en \(\angle Q=69\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 1ms 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 ○ Dus \(P\kern{-.8pt}R^2=15^2+15^2-2⋅15⋅15⋅\cos(69\degree)=288{,}734...\text{.}\) 1p ○ \(P\kern{-.8pt}R=\sqrt{288{,}734...}≈17{,}0\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=21\text{,}\) \(Q\kern{-.8pt}R=39\) en \(\angle Q=91\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms b De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}R^2=P\kern{-.8pt}Q^2+Q\kern{-.8pt}R^2-2⋅P\kern{-.8pt}Q⋅Q\kern{-.8pt}R⋅\cos(\angle Q)\text{.}\) 1p ○ Dus \(P\kern{-.8pt}R^2=21^2+39^2-2⋅21⋅39⋅\cos(91\degree)=1990{,}587...\text{.}\) 1p ○ \(P\kern{-.8pt}R=\sqrt{1990{,}587...}≈44{,}6\text{.}\) 1p 4p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=14\text{,}\) \(A\kern{-.8pt}B=14\) en \(B\kern{-.8pt}C=20\text{.}\) 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 \(20^2=14^2+14^2-2⋅14⋅14⋅\cos(\angle A)\) 1p ○ Balansmethode geeft \(\cos(\angle A)={400-392 \over -392}=-0{,}020...\) 1p ○ Hieruit volgt \(\angle A=\cos^{-1}(-0{,}020...)≈91{,}2\degree\text{.}\) 1p 4p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=25\text{,}\) \(Q\kern{-.8pt}R=14\) en \(P\kern{-.8pt}R=30\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms d 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 \(30^2=25^2+14^2-2⋅25⋅14⋅\cos(\angle Q)\) 1p ○ Balansmethode geeft \(\cos(\angle Q)={900-821 \over -700}=-0{,}112...\) 1p ○ Hieruit volgt \(\angle Q=\cos^{-1}(-0{,}112...)≈96{,}5\degree\text{.}\) 1p |