Getal & Ruimte (12e editie) - vwo wiskunde B
'Sinus- en cosinusregel'.
| vwo wiskunde B | 3.5 De sinusregel en de cosinusregel |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 21 \text{,}\) \(\angle M = 65\degree\) en \(\angle K = 84\degree \text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms 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 \(L\kern{-.8pt}M = {K\kern{-.8pt}L ⋅ \sin(\angle K) \over \sin(\angle M)} = {21 ⋅ \sin(84\degree) \over \sin(65\degree)} \text{.}\) 1p ○ \(L\kern{-.8pt}M ≈ 23{,}0 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 51 \text{,}\) \(\angle A = 57\degree\) en \(\angle B = 97\degree \text{.}\) 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 \(A\kern{-.8pt}C = {B\kern{-.8pt}C ⋅ \sin(\angle B) \over \sin(\angle A)} = {51 ⋅ \sin(97\degree) \over \sin(57\degree)} \text{.}\) 1p ○ \(A\kern{-.8pt}C ≈ 60{,}4 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 12 \text{,}\) \(A\kern{-.8pt}C = 20\) en \(\angle A = 34\degree \text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 5ms 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 \(\sin(\angle B) = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over B\kern{-.8pt}C} = {20 ⋅ \sin(34\degree) \over 12} = 0{,}931... \text{.}\) 1p ○ Dit geeft \(\angle B ≈ 68{,}7\degree\) of \(\angle B ≈ 111{,}3\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 11 \text{,}\) \(P\kern{-.8pt}Q = 14\) en \(\angle Q = 48\degree \text{.}\) 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 \(\sin(\angle R) = {P\kern{-.8pt}Q ⋅ \sin(\angle Q) \over P\kern{-.8pt}R} = {14 ⋅ \sin(48\degree) \over 11} = 0{,}945... \text{.}\) 1p ○ Dit geeft \(\angle R ≈ 71{,}1\degree\) of \(\angle R ≈ 108{,}9\degree \text{.}\) 1p opgave 24p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 45 \text{,}\) \(\angle K = 51\degree\) en \(\angle M = 40\degree \text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle K + \angle L + \angle M = 180\degree\) volgt \(\angle L = 180\degree - \angle K - \angle M = 180\degree - 51\degree - 40\degree = 89\degree \text{.}\) 1p ○ 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 \(L\kern{-.8pt}M = {K\kern{-.8pt}M ⋅ \sin(\angle K) \over \sin(\angle L)} = {45 ⋅ \sin(51\degree) \over \sin(89\degree)} \text{.}\) 1p ○ \(L\kern{-.8pt}M ≈ 35{,}0 \text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 44 \text{,}\) \(\angle C = 37\degree\) en \(\angle B = 43\degree \text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle C + \angle A + \angle B = 180\degree\) volgt \(\angle A = 180\degree - \angle C - \angle B = 180\degree - 37\degree - 43\degree = 100\degree \text{.}\) 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 \(A\kern{-.8pt}B = {B\kern{-.8pt}C ⋅ \sin(\angle C) \over \sin(\angle A)} = {44 ⋅ \sin(37\degree) \over \sin(100\degree)} \text{.}\) 1p ○ \(A\kern{-.8pt}B ≈ 26{,}9 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 17 \text{,}\) \(K\kern{-.8pt}L = 15\) en \(\angle K = 85\degree \text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 0ms c 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 ○ Dus \(L\kern{-.8pt}M^{2} = 17^{2} + 15^{2} - 2 ⋅ 17 ⋅ 15 ⋅ \cos(85\degree) = 469{,}550... \text{.}\) 1p ○ \(L\kern{-.8pt}M = \sqrt{469{,}550...} ≈ 21{,}7 \text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 14 \text{,}\) \(A\kern{-.8pt}B = 17\) en \(\angle A = 110\degree \text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms d 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 \(B\kern{-.8pt}C^{2} = 14^{2} + 17^{2} - 2 ⋅ 14 ⋅ 17 ⋅ \cos(110\degree) = 647{,}801... \text{.}\) 1p ○ \(B\kern{-.8pt}C = \sqrt{647{,}801...} ≈ 25{,}5 \text{.}\) 1p opgave 34p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 41 \text{,}\) \(K\kern{-.8pt}L = 29\) en \(L\kern{-.8pt}M = 44 \text{.}\) 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 \(44^{2} = 41^{2} + 29^{2} - 2 ⋅ 41 ⋅ 29 ⋅ \cos(\angle K)\) 1p ○ Balansmethode geeft \(\cos(\angle K) = {1\,936 - 2\,522 \over -2\,378} = 0{,}246...\) 1p ○ Hieruit volgt \(\angle K = \cos^{-1}(0{,}246...) ≈ 75{,}7\degree \text{.}\) 1p 4p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 13 \text{,}\) \(K\kern{-.8pt}M = 14\) en \(K\kern{-.8pt}L = 21 \text{.}\) 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 \(21^{2} = 13^{2} + 14^{2} - 2 ⋅ 13 ⋅ 14 ⋅ \cos(\angle M)\) 1p ○ Balansmethode geeft \(\cos(\angle M) = {441 - 365 \over -364} = -0{,}208...\) 1p ○ Hieruit volgt \(\angle M = \cos^{-1}(-0{,}208...) ≈ 102{,}1\degree \text{.}\) 1p |