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 \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=16\text{,}\) \(\angle R=34\degree\) en \(\angle P=81\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 1ms a 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 \(Q\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle P) \over \sin(\angle R)}={16⋅\sin(81\degree) \over \sin(34\degree)}\text{.}\) 1p ○ \(Q\kern{-.8pt}R≈28{,}3\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=21\text{,}\) \(\angle P=47\degree\) en \(\angle Q=100\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b 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 \(P\kern{-.8pt}R={Q\kern{-.8pt}R⋅\sin(\angle Q) \over \sin(\angle P)}={21⋅\sin(100\degree) \over \sin(47\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}R≈28{,}3\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=19\text{,}\) \(B\kern{-.8pt}C=25\) en \(\angle C=46\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 9ms c 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 ○ Daaruit volgt \(\sin(\angle A)={B\kern{-.8pt}C⋅\sin(\angle C) \over A\kern{-.8pt}B}={25⋅\sin(46\degree) \over 19}=0{,}946...\text{.}\) 1p ○ Dit geeft \(\angle A≈71{,}2\degree\) of \(\angle A≈108{,}8\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=22\text{,}\) \(B\kern{-.8pt}C=28\) en \(\angle C=49\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d 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 ○ Daaruit volgt \(\sin(\angle A)={B\kern{-.8pt}C⋅\sin(\angle C) \over A\kern{-.8pt}B}={28⋅\sin(49\degree) \over 22}=0{,}960...\text{.}\) 1p ○ Dit geeft \(\angle A≈73{,}9\degree\) of \(\angle A≈106{,}1\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=16\text{,}\) \(\angle B=58\degree\) en \(\angle A=41\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 1ms a Uit \(\angle B+\angle C+\angle A=180\degree\) volgt \(\angle C=180\degree-\angle B-\angle A=180\degree-58\degree-41\degree=81\degree\text{.}\) 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 \(A\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle B) \over \sin(\angle C)}={16⋅\sin(58\degree) \over \sin(81\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}C≈13{,}7\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=45\text{,}\) \(\angle C=50\degree\) en \(\angle B=25\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle C+\angle A+\angle B=180\degree\) volgt \(\angle A=180\degree-\angle C-\angle B=180\degree-50\degree-25\degree=105\degree\text{.}\) 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 \(A\kern{-.8pt}B={B\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle A)}={45⋅\sin(50\degree) \over \sin(105\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}B≈35{,}7\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=18\text{,}\) \(A\kern{-.8pt}C=19\) en \(\angle C=65\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 1ms 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 \(A\kern{-.8pt}B^2=18^2+19^2-2⋅18⋅19⋅\cos(65\degree)=395{,}929...\text{.}\) 1p ○ \(A\kern{-.8pt}B=\sqrt{395{,}929...}≈19{,}9\text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=11\text{,}\) \(P\kern{-.8pt}Q=18\) en \(\angle P=94\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms d 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 ○ Dus \(Q\kern{-.8pt}R^2=11^2+18^2-2⋅11⋅18⋅\cos(94\degree)=472{,}623...\text{.}\) 1p ○ \(Q\kern{-.8pt}R=\sqrt{472{,}623...}≈21{,}7\text{.}\) 1p opgave 34p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=28\text{,}\) \(A\kern{-.8pt}B=27\) en \(B\kern{-.8pt}C=34\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 8ms 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 \(34^2=28^2+27^2-2⋅28⋅27⋅\cos(\angle A)\) 1p ○ Balansmethode geeft \(\cos(\angle A)={1\,156-1\,513 \over -1\,512}=0{,}236...\) 1p ○ Hieruit volgt \(\angle A=\cos^{-1}(0{,}236...)≈76{,}3\degree\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=34\text{,}\) \(A\kern{-.8pt}B=21\) en \(B\kern{-.8pt}C=44\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms b 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 \(44^2=34^2+21^2-2⋅34⋅21⋅\cos(\angle A)\) 1p ○ Balansmethode geeft \(\cos(\angle A)={1\,936-1\,597 \over -1\,428}=-0{,}237...\) 1p ○ Hieruit volgt \(\angle A=\cos^{-1}(-0{,}237...)≈103{,}7\degree\text{.}\) 1p |