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 \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=11\text{,}\) \(\angle C=40\degree\) en \(\angle A=77\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms a 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 \(B\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle A) \over \sin(\angle C)}={11⋅\sin(77\degree) \over \sin(40\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈16{,}7\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=30\text{,}\) \(\angle C=29\degree\) en \(\angle A=92\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}\text{.}\) 1p ○ Dus \(B\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle A) \over \sin(\angle C)}={30⋅\sin(92\degree) \over \sin(29\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈61{,}8\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=7\text{,}\) \(K\kern{-.8pt}L=10\) en \(\angle L=38\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 5ms 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 \(\sin(\angle M)={K\kern{-.8pt}L⋅\sin(\angle L) \over K\kern{-.8pt}M}={10⋅\sin(38\degree) \over 7}=0{,}879...\text{.}\) 1p ○ Dit geeft \(\angle M≈61{,}6\degree\) of \(\angle M≈118{,}4\degree\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=18\text{,}\) \(K\kern{-.8pt}M=24\) en \(\angle K=46\degree\text{.}\) 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 \(\sin(\angle L)={K\kern{-.8pt}M⋅\sin(\angle K) \over L\kern{-.8pt}M}={24⋅\sin(46\degree) \over 18}=0{,}959...\text{.}\) 1p ○ Dit geeft \(\angle L≈73{,}6\degree\) of \(\angle L≈106{,}4\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=28\text{,}\) \(\angle R=65\degree\) en \(\angle Q=42\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle R+\angle P+\angle Q=180\degree\) volgt \(\angle P=180\degree-\angle R-\angle Q=180\degree-65\degree-42\degree=73\degree\text{.}\) 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 \(P\kern{-.8pt}Q={Q\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle P)}={28⋅\sin(65\degree) \over \sin(73\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}Q≈26{,}5\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=37\text{,}\) \(\angle A=59\degree\) en \(\angle C=30\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle A+\angle B+\angle C=180\degree\) volgt \(\angle B=180\degree-\angle A-\angle C=180\degree-59\degree-30\degree=91\degree\text{.}\) 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 \(B\kern{-.8pt}C={A\kern{-.8pt}C⋅\sin(\angle A) \over \sin(\angle B)}={37⋅\sin(59\degree) \over \sin(91\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈31{,}7\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=18\text{,}\) \(A\kern{-.8pt}C=17\) en \(\angle C=82\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 0ms c 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 ○ Dus \(A\kern{-.8pt}B^2=18^2+17^2-2⋅18⋅17⋅\cos(82\degree)=527{,}826...\text{.}\) 1p ○ \(A\kern{-.8pt}B=\sqrt{527{,}826...}≈23{,}0\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=18\text{,}\) \(K\kern{-.8pt}L=22\) en \(\angle K=121\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms d 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 \(L\kern{-.8pt}M^2=18^2+22^2-2⋅18⋅22⋅\cos(121\degree)=1215{,}910...\text{.}\) 1p ○ \(L\kern{-.8pt}M=\sqrt{1215{,}910...}≈34{,}9\text{.}\) 1p opgave 34p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=43\text{,}\) \(K\kern{-.8pt}L=30\) en \(L\kern{-.8pt}M=50\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 5ms a 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 ○ Invullen geeft \(50^2=43^2+30^2-2⋅43⋅30⋅\cos(\angle K)\) 1p ○ Balansmethode geeft \(\cos(\angle K)={2\,500-2\,749 \over -2\,580}=0{,}096...\) 1p ○ Hieruit volgt \(\angle K=\cos^{-1}(0{,}096...)≈84{,}5\degree\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=28\text{,}\) \(A\kern{-.8pt}C=23\) en \(A\kern{-.8pt}B=41\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms b 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 \(41^2=28^2+23^2-2⋅28⋅23⋅\cos(\angle C)\) 1p ○ Balansmethode geeft \(\cos(\angle C)={1\,681-1\,313 \over -1\,288}=-0{,}285...\) 1p ○ Hieruit volgt \(\angle C=\cos^{-1}(-0{,}285...)≈106{,}6\degree\text{.}\) 1p |