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 \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 23 \text{,}\) \(\angle K = 59\degree\) en \(\angle L = 83\degree \text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms a 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 \(K\kern{-.8pt}M = {L\kern{-.8pt}M ⋅ \sin(\angle L) \over \sin(\angle K)} = {23 ⋅ \sin(83\degree) \over \sin(59\degree)} \text{.}\) 1p ○ \(K\kern{-.8pt}M ≈ 26{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 21 \text{,}\) \(\angle M = 28\degree\) en \(\angle K = 99\degree \text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b 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 \(L\kern{-.8pt}M = {K\kern{-.8pt}L ⋅ \sin(\angle K) \over \sin(\angle M)} = {21 ⋅ \sin(99\degree) \over \sin(28\degree)} \text{.}\) 1p ○ \(L\kern{-.8pt}M ≈ 44{,}2 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 8 \text{,}\) \(L\kern{-.8pt}M = 12\) en \(\angle M = 29\degree \text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 4ms c 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 ○ Daaruit volgt \(\sin(\angle K) = {L\kern{-.8pt}M ⋅ \sin(\angle M) \over K\kern{-.8pt}L} = {12 ⋅ \sin(29\degree) \over 8} = 0{,}727... \text{.}\) 1p ○ Dit geeft \(\angle K ≈ 46{,}7\degree\) of \(\angle K ≈ 133{,}3\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 17 \text{,}\) \(K\kern{-.8pt}L = 24\) en \(\angle L = 34\degree \text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d 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 \(\sin(\angle M) = {K\kern{-.8pt}L ⋅ \sin(\angle L) \over K\kern{-.8pt}M} = {24 ⋅ \sin(34\degree) \over 17} = 0{,}789... \text{.}\) 1p ○ Dit geeft \(\angle M ≈ 52{,}1\degree\) of \(\angle M ≈ 127{,}9\degree \text{.}\) 1p opgave 24p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 29 \text{,}\) \(\angle A = 48\degree\) en \(\angle C = 43\degree \text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle A + \angle B + \angle C = 180\degree\) volgt \(\angle B = 180\degree - \angle A - \angle C = 180\degree - 48\degree - 43\degree = 89\degree \text{.}\) 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 \(B\kern{-.8pt}C = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over \sin(\angle B)} = {29 ⋅ \sin(48\degree) \over \sin(89\degree)} \text{.}\) 1p ○ \(B\kern{-.8pt}C ≈ 21{,}6 \text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 50 \text{,}\) \(\angle A = 50\degree\) en \(\angle C = 32\degree \text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle A + \angle B + \angle C = 180\degree\) volgt \(\angle B = 180\degree - \angle A - \angle C = 180\degree - 50\degree - 32\degree = 98\degree \text{.}\) 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 \(B\kern{-.8pt}C = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over \sin(\angle B)} = {50 ⋅ \sin(50\degree) \over \sin(98\degree)} \text{.}\) 1p ○ \(B\kern{-.8pt}C ≈ 38{,}7 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 20 \text{,}\) \(L\kern{-.8pt}M = 20\) en \(\angle L = 82\degree \text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 0ms 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 \(K\kern{-.8pt}M^{2} = 20^{2} + 20^{2} - 2 ⋅ 20 ⋅ 20 ⋅ \cos(82\degree) = 688{,}661... \text{.}\) 1p ○ \(K\kern{-.8pt}M = \sqrt{688{,}661...} ≈ 26{,}2 \text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 17 \text{,}\) \(B\kern{-.8pt}C = 12\) en \(\angle B = 97\degree \text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms d 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 ○ Dus \(A\kern{-.8pt}C^{2} = 17^{2} + 12^{2} - 2 ⋅ 17 ⋅ 12 ⋅ \cos(97\degree) = 482{,}722... \text{.}\) 1p ○ \(A\kern{-.8pt}C = \sqrt{482{,}722...} ≈ 22{,}0 \text{.}\) 1p opgave 34p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 29 \text{,}\) \(K\kern{-.8pt}L = 19\) en \(L\kern{-.8pt}M = 33 \text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 4ms 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 \(33^{2} = 29^{2} + 19^{2} - 2 ⋅ 29 ⋅ 19 ⋅ \cos(\angle K)\) 1p ○ Balansmethode geeft \(\cos(\angle K) = {1\,089 - 1\,202 \over -1\,102} = 0{,}102...\) 1p ○ Hieruit volgt \(\angle K = \cos^{-1}(0{,}102...) ≈ 84{,}1\degree \text{.}\) 1p 4p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 15 \text{,}\) \(K\kern{-.8pt}M = 27\) en \(K\kern{-.8pt}L = 31 \text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms b 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 \(31^{2} = 15^{2} + 27^{2} - 2 ⋅ 15 ⋅ 27 ⋅ \cos(\angle M)\) 1p ○ Balansmethode geeft \(\cos(\angle M) = {961 - 954 \over -810} = -0{,}008...\) 1p ○ Hieruit volgt \(\angle M = \cos^{-1}(-0{,}008...) ≈ 90{,}5\degree \text{.}\) 1p |