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 \(P\kern{-.8pt}Q=21\text{,}\) \(\angle R=62\degree\) en \(\angle P=60\degree\text{.}\) 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 \(Q\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle P) \over \sin(\angle R)}={21⋅\sin(60\degree) \over \sin(62\degree)}\text{.}\) 1p ○ \(Q\kern{-.8pt}R≈20{,}6\text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=31\text{,}\) \(\angle K=44\degree\) en \(\angle L=99\degree\text{.}\) 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 \(K\kern{-.8pt}M={L\kern{-.8pt}M⋅\sin(\angle L) \over \sin(\angle K)}={31⋅\sin(99\degree) \over \sin(44\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}M≈44{,}1\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=16\text{,}\) \(K\kern{-.8pt}L=22\) en \(\angle L=39\degree\text{.}\) 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 \(\sin(\angle M)={K\kern{-.8pt}L⋅\sin(\angle L) \over K\kern{-.8pt}M}={22⋅\sin(39\degree) \over 16}=0{,}865...\text{.}\) 1p ○ Dit geeft \(\angle M≈59{,}9\degree\) of \(\angle M≈120{,}1\degree\text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=24\text{,}\) \(P\kern{-.8pt}Q=30\) en \(\angle Q=51\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}={Q\kern{-.8pt}R \over \sin(\angle P)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle R)={P\kern{-.8pt}Q⋅\sin(\angle Q) \over P\kern{-.8pt}R}={30⋅\sin(51\degree) \over 24}=0{,}971...\text{.}\) 1p ○ Dit geeft \(\angle R≈76{,}3\degree\) of \(\angle R≈103{,}7\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=40\text{,}\) \(\angle Q=33\degree\) en \(\angle P=63\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 1ms a Uit \(\angle Q+\angle R+\angle P=180\degree\) volgt \(\angle R=180\degree-\angle Q-\angle P=180\degree-33\degree-63\degree=84\degree\text{.}\) 1p ○ De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}={Q\kern{-.8pt}R \over \sin(\angle P)}\text{.}\) 1p ○ Dus \(P\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle Q) \over \sin(\angle R)}={40⋅\sin(33\degree) \over \sin(84\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}R≈21{,}9\text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=36\text{,}\) \(\angle P=39\degree\) en \(\angle R=47\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle P+\angle Q+\angle R=180\degree\) volgt \(\angle Q=180\degree-\angle P-\angle R=180\degree-39\degree-47\degree=94\degree\text{.}\) 1p ○ 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 \(Q\kern{-.8pt}R={P\kern{-.8pt}R⋅\sin(\angle P) \over \sin(\angle Q)}={36⋅\sin(39\degree) \over \sin(94\degree)}\text{.}\) 1p ○ \(Q\kern{-.8pt}R≈22{,}7\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=17\text{,}\) \(A\kern{-.8pt}C=12\) en \(\angle C=77\degree\text{.}\) 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 \(A\kern{-.8pt}B^2=17^2+12^2-2⋅17⋅12⋅\cos(77\degree)=341{,}219...\text{.}\) 1p ○ \(A\kern{-.8pt}B=\sqrt{341{,}219...}≈18{,}5\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=24\text{,}\) \(K\kern{-.8pt}M=35\) en \(\angle M=102\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms d 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 ○ Dus \(K\kern{-.8pt}L^2=24^2+35^2-2⋅24⋅35⋅\cos(102\degree)=2150{,}291...\text{.}\) 1p ○ \(K\kern{-.8pt}L=\sqrt{2150{,}291...}≈46{,}4\text{.}\) 1p opgave 34p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=30\text{,}\) \(L\kern{-.8pt}M=35\) en \(K\kern{-.8pt}M=42\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 8ms a 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 ○ Invullen geeft \(42^2=30^2+35^2-2⋅30⋅35⋅\cos(\angle L)\) 1p ○ Balansmethode geeft \(\cos(\angle L)={1\,764-2\,125 \over -2\,100}=0{,}171...\) 1p ○ Hieruit volgt \(\angle L=\cos^{-1}(0{,}171...)≈80{,}1\degree\text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=21\text{,}\) \(P\kern{-.8pt}R=17\) en \(P\kern{-.8pt}Q=33\text{.}\) CosinusregelHoekInStomp 007y - 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 ○ Invullen geeft \(33^2=21^2+17^2-2⋅21⋅17⋅\cos(\angle R)\) 1p ○ Balansmethode geeft \(\cos(\angle R)={1\,089-730 \over -714}=-0{,}502...\) 1p ○ Hieruit volgt \(\angle R=\cos^{-1}(-0{,}502...)≈120{,}2\degree\text{.}\) 1p |