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 \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 25 \text{,}\) \(\angle C = 52\degree\) en \(\angle A = 64\degree \text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms a 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 \(B\kern{-.8pt}C = {A\kern{-.8pt}B ⋅ \sin(\angle A) \over \sin(\angle C)} = {25 ⋅ \sin(64\degree) \over \sin(52\degree)} \text{.}\) 1p ○ \(B\kern{-.8pt}C ≈ 28{,}5 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 30 \text{,}\) \(\angle A = 29\degree\) en \(\angle B = 106\degree \text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b 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 \(A\kern{-.8pt}C = {B\kern{-.8pt}C ⋅ \sin(\angle B) \over \sin(\angle A)} = {30 ⋅ \sin(106\degree) \over \sin(29\degree)} \text{.}\) 1p ○ \(A\kern{-.8pt}C ≈ 59{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 17 \text{,}\) \(B\kern{-.8pt}C = 28\) en \(\angle C = 30\degree \text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 5ms 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 \(\sin(\angle A) = {B\kern{-.8pt}C ⋅ \sin(\angle C) \over A\kern{-.8pt}B} = {28 ⋅ \sin(30\degree) \over 17} = 0{,}823... \text{.}\) 1p ○ Dit geeft \(\angle A ≈ 55{,}4\degree\) of \(\angle A ≈ 124{,}6\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 15 \text{,}\) \(L\kern{-.8pt}M = 25\) en \(\angle M = 26\degree \text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}L \over \sin(\angle M)} = {L\kern{-.8pt}M \over \sin(\angle K)} = {K\kern{-.8pt}M \over \sin(\angle L)} \text{.}\) 1p ○ Daaruit volgt \(\sin(\angle K) = {L\kern{-.8pt}M ⋅ \sin(\angle M) \over K\kern{-.8pt}L} = {25 ⋅ \sin(26\degree) \over 15} = 0{,}730... \text{.}\) 1p ○ Dit geeft \(\angle K ≈ 46{,}9\degree\) of \(\angle K ≈ 133{,}1\degree \text{.}\) 1p opgave 24p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 31 \text{,}\) \(\angle A = 44\degree\) en \(\angle C = 52\degree \text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle A + \angle B + \angle C = 180\degree\) volgt \(\angle B = 180\degree - \angle A - \angle C = 180\degree - 44\degree - 52\degree = 84\degree \text{.}\) 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 \(B\kern{-.8pt}C = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over \sin(\angle B)} = {31 ⋅ \sin(44\degree) \over \sin(84\degree)} \text{.}\) 1p ○ \(B\kern{-.8pt}C ≈ 21{,}7 \text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 29 \text{,}\) \(\angle A = 49\degree\) en \(\angle C = 40\degree \text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle A + \angle B + \angle C = 180\degree\) volgt \(\angle B = 180\degree - \angle A - \angle C = 180\degree - 49\degree - 40\degree = 91\degree \text{.}\) 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 \(B\kern{-.8pt}C = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over \sin(\angle B)} = {29 ⋅ \sin(49\degree) \over \sin(91\degree)} \text{.}\) 1p ○ \(B\kern{-.8pt}C ≈ 21{,}9 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 27 \text{,}\) \(A\kern{-.8pt}B = 20\) en \(\angle A = 77\degree \text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 0ms c 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 ○ Dus \(B\kern{-.8pt}C^{2} = 27^{2} + 20^{2} - 2 ⋅ 27 ⋅ 20 ⋅ \cos(77\degree) = 886{,}052... \text{.}\) 1p ○ \(B\kern{-.8pt}C = \sqrt{886{,}052...} ≈ 29{,}8 \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 12 \text{,}\) \(P\kern{-.8pt}R = 15\) en \(\angle R = 113\degree \text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms d De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}Q^{2} = Q\kern{-.8pt}R^{2} + P\kern{-.8pt}R^{2} - 2 ⋅ Q\kern{-.8pt}R ⋅ P\kern{-.8pt}R ⋅ \cos(\angle R) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q^{2} = 12^{2} + 15^{2} - 2 ⋅ 12 ⋅ 15 ⋅ \cos(113\degree) = 509{,}663... \text{.}\) 1p ○ \(P\kern{-.8pt}Q = \sqrt{509{,}663...} ≈ 22{,}6 \text{.}\) 1p opgave 34p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 19 \text{,}\) \(K\kern{-.8pt}L = 23\) en \(L\kern{-.8pt}M = 28 \text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 5ms 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 \(28^{2} = 19^{2} + 23^{2} - 2 ⋅ 19 ⋅ 23 ⋅ \cos(\angle K)\) 1p ○ Balansmethode geeft \(\cos(\angle K) = {784 - 890 \over -874} = 0{,}121...\) 1p ○ Hieruit volgt \(\angle K = \cos^{-1}(0{,}121...) ≈ 83{,}0\degree \text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 27 \text{,}\) \(Q\kern{-.8pt}R = 37\) en \(P\kern{-.8pt}R = 47 \text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms b 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 ○ Invullen geeft \(47^{2} = 27^{2} + 37^{2} - 2 ⋅ 27 ⋅ 37 ⋅ \cos(\angle Q)\) 1p ○ Balansmethode geeft \(\cos(\angle Q) = {2\,209 - 2\,098 \over -1\,998} = -0{,}055...\) 1p ○ Hieruit volgt \(\angle Q = \cos^{-1}(-0{,}055...) ≈ 93{,}2\degree \text{.}\) 1p |