Getal & Ruimte (12e editie) - havo wiskunde B
'Sinus- en cosinusregel'.
| havo wiskunde B | 3.2 De sinusregel en de cosinusregel |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=22\text{,}\) \(\angle M=54\degree\) en \(\angle K=82\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 1ms a 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 \(L\kern{-.8pt}M={K\kern{-.8pt}L⋅\sin(\angle K) \over \sin(\angle M)}={22⋅\sin(82\degree) \over \sin(54\degree)}\text{.}\) 1p ○ \(L\kern{-.8pt}M≈26{,}9\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=19\text{,}\) \(\angle R=46\degree\) en \(\angle P=93\degree\text{.}\) 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 \(Q\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle P) \over \sin(\angle R)}={19⋅\sin(93\degree) \over \sin(46\degree)}\text{.}\) 1p ○ \(Q\kern{-.8pt}R≈26{,}4\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=22\text{,}\) \(A\kern{-.8pt}C=29\) en \(\angle A=43\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 9ms c 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 ○ Daaruit volgt \(\sin(\angle B)={A\kern{-.8pt}C⋅\sin(\angle A) \over B\kern{-.8pt}C}={29⋅\sin(43\degree) \over 22}=0{,}898...\text{.}\) 1p ○ Dit geeft \(\angle B≈64{,}0\degree\) of \(\angle B≈116{,}0\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=8\text{,}\) \(A\kern{-.8pt}B=11\) en \(\angle B=37\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}C \over \sin(\angle B)}={A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle C)={A\kern{-.8pt}B⋅\sin(\angle B) \over A\kern{-.8pt}C}={11⋅\sin(37\degree) \over 8}=0{,}827...\text{.}\) 1p ○ Dit geeft \(\angle C≈55{,}8\degree\) of \(\angle C≈124{,}2\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=35\text{,}\) \(\angle L=38\degree\) en \(\angle K=53\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 1ms a Uit \(\angle L+\angle M+\angle K=180\degree\) volgt \(\angle M=180\degree-\angle L-\angle K=180\degree-38\degree-53\degree=89\degree\text{.}\) 1p ○ 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 ○ Dus \(K\kern{-.8pt}M={K\kern{-.8pt}L⋅\sin(\angle L) \over \sin(\angle M)}={35⋅\sin(38\degree) \over \sin(89\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}M≈21{,}6\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=31\text{,}\) \(\angle C=28\degree\) en \(\angle B=50\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle C+\angle A+\angle B=180\degree\) volgt \(\angle A=180\degree-\angle C-\angle B=180\degree-28\degree-50\degree=102\degree\text{.}\) 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 \(A\kern{-.8pt}B={B\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle A)}={31⋅\sin(28\degree) \over \sin(102\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}B≈14{,}9\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=18\text{,}\) \(P\kern{-.8pt}Q=19\) en \(\angle P=80\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 1ms c 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 \(Q\kern{-.8pt}R^2=18^2+19^2-2⋅18⋅19⋅\cos(80\degree)=566{,}224...\text{.}\) 1p ○ \(Q\kern{-.8pt}R=\sqrt{566{,}224...}≈23{,}8\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=29\text{,}\) \(K\kern{-.8pt}M=43\) en \(\angle M=116\degree\text{.}\) 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 \(K\kern{-.8pt}L^2=29^2+43^2-2⋅29⋅43⋅\cos(116\degree)=3783{,}297...\text{.}\) 1p ○ \(K\kern{-.8pt}L=\sqrt{3783{,}297...}≈61{,}5\text{.}\) 1p opgave 34p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=26\text{,}\) \(L\kern{-.8pt}M=27\) en \(K\kern{-.8pt}M=30\text{.}\) 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 \(30^2=26^2+27^2-2⋅26⋅27⋅\cos(\angle L)\) 1p ○ Balansmethode geeft \(\cos(\angle L)={900-1\,405 \over -1\,404}=0{,}359...\) 1p ○ Hieruit volgt \(\angle L=\cos^{-1}(0{,}359...)≈68{,}9\degree\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=11\text{,}\) \(B\kern{-.8pt}C=17\) en \(A\kern{-.8pt}C=21\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms b De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(A\kern{-.8pt}C^2=A\kern{-.8pt}B^2+B\kern{-.8pt}C^2-2⋅A\kern{-.8pt}B⋅B\kern{-.8pt}C⋅\cos(\angle B)\text{.}\) 1p ○ Invullen geeft \(21^2=11^2+17^2-2⋅11⋅17⋅\cos(\angle B)\) 1p ○ Balansmethode geeft \(\cos(\angle B)={441-410 \over -374}=-0{,}082...\) 1p ○ Hieruit volgt \(\angle B=\cos^{-1}(-0{,}082...)≈94{,}8\degree\text{.}\) 1p |