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 \(Q\kern{-.8pt}R=16\text{,}\) \(\angle P=63\degree\) en \(\angle Q=55\degree\text{.}\) 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 \(P\kern{-.8pt}R={Q\kern{-.8pt}R⋅\sin(\angle Q) \over \sin(\angle P)}={16⋅\sin(55\degree) \over \sin(63\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}R≈14{,}7\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=17\text{,}\) \(\angle R=36\degree\) en \(\angle P=103\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b 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 \(Q\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle P) \over \sin(\angle R)}={17⋅\sin(103\degree) \over \sin(36\degree)}\text{.}\) 1p ○ \(Q\kern{-.8pt}R≈28{,}2\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=13\text{,}\) \(B\kern{-.8pt}C=18\) en \(\angle C=37\degree\text{.}\) 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 \(\sin(\angle A)={B\kern{-.8pt}C⋅\sin(\angle C) \over A\kern{-.8pt}B}={18⋅\sin(37\degree) \over 13}=0{,}833...\text{.}\) 1p ○ Dit geeft \(\angle A≈56{,}4\degree\) of \(\angle A≈123{,}6\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=13\text{,}\) \(B\kern{-.8pt}C=20\) en \(\angle C=25\degree\text{.}\) 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 \(\sin(\angle A)={B\kern{-.8pt}C⋅\sin(\angle C) \over A\kern{-.8pt}B}={20⋅\sin(25\degree) \over 13}=0{,}650...\text{.}\) 1p ○ Dit geeft \(\angle A≈40{,}6\degree\) of \(\angle A≈139{,}4\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=24\text{,}\) \(\angle C=39\degree\) en \(\angle B=65\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 1ms a Uit \(\angle C+\angle A+\angle B=180\degree\) volgt \(\angle A=180\degree-\angle C-\angle B=180\degree-39\degree-65\degree=76\degree\text{.}\) 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 \(A\kern{-.8pt}B={B\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle A)}={24⋅\sin(39\degree) \over \sin(76\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}B≈15{,}6\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=55\text{,}\) \(\angle A=36\degree\) en \(\angle C=28\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle A+\angle B+\angle C=180\degree\) volgt \(\angle B=180\degree-\angle A-\angle C=180\degree-36\degree-28\degree=116\degree\text{.}\) 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 \(B\kern{-.8pt}C={A\kern{-.8pt}C⋅\sin(\angle A) \over \sin(\angle B)}={55⋅\sin(36\degree) \over \sin(116\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈36{,}0\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=29\text{,}\) \(L\kern{-.8pt}M=24\) en \(\angle L=68\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 1ms c De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(K\kern{-.8pt}M^2=K\kern{-.8pt}L^2+L\kern{-.8pt}M^2-2⋅K\kern{-.8pt}L⋅L\kern{-.8pt}M⋅\cos(\angle L)\text{.}\) 1p ○ Dus \(K\kern{-.8pt}M^2=29^2+24^2-2⋅29⋅24⋅\cos(68\degree)=895{,}547...\text{.}\) 1p ○ \(K\kern{-.8pt}M=\sqrt{895{,}547...}≈29{,}9\text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=32\text{,}\) \(Q\kern{-.8pt}R=24\) en \(\angle Q=115\degree\text{.}\) CosinusregelZijdeInStomp 007w - 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 ○ Dus \(P\kern{-.8pt}R^2=32^2+24^2-2⋅32⋅24⋅\cos(115\degree)=2249{,}141...\text{.}\) 1p ○ \(P\kern{-.8pt}R=\sqrt{2249{,}141...}≈47{,}4\text{.}\) 1p opgave 34p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=26\text{,}\) \(K\kern{-.8pt}M=26\) en \(K\kern{-.8pt}L=29\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 8ms a 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 \(29^2=26^2+26^2-2⋅26⋅26⋅\cos(\angle M)\) 1p ○ Balansmethode geeft \(\cos(\angle M)={841-1\,352 \over -1\,352}=0{,}377...\) 1p ○ Hieruit volgt \(\angle M=\cos^{-1}(0{,}377...)≈67{,}8\degree\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=27\text{,}\) \(A\kern{-.8pt}C=20\) en \(A\kern{-.8pt}B=35\text{.}\) 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 \(35^2=27^2+20^2-2⋅27⋅20⋅\cos(\angle C)\) 1p ○ Balansmethode geeft \(\cos(\angle C)={1\,225-1\,129 \over -1\,080}=-0{,}088...\) 1p ○ Hieruit volgt \(\angle C=\cos^{-1}(-0{,}088...)≈95{,}1\degree\text{.}\) 1p |