Getal & Ruimte (13e editie) - havo wiskunde B

'Sinus- en cosinusregel'.

havo wiskunde B	3.2 De sinusregel
Sinus- en cosinusregel (6)	opgave 1 3p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 25 \text{,}\) \(\angle R = 33\degree\) en \(\angle P = 83\degree \text{.}\) Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms 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 \(Q\kern{-.8pt}R = {P\kern{-.8pt}Q ⋅ \sin(\angle P) \over \sin(\angle R)} = {25 ⋅ \sin(83\degree) \over \sin(33\degree)} \text{.}\) 1p ○ \(Q\kern{-.8pt}R ≈ 45{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 30 \text{,}\) \(\angle B = 44\degree\) en \(\angle C = 96\degree \text{.}\) Bereken de lengte van zijde \(A\kern{-.8pt}B \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b 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 ○ Dus \(A\kern{-.8pt}B = {A\kern{-.8pt}C ⋅ \sin(\angle C) \over \sin(\angle B)} = {30 ⋅ \sin(96\degree) \over \sin(44\degree)} \text{.}\) 1p ○ \(A\kern{-.8pt}B ≈ 43{,}0 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 11 \text{,}\) \(P\kern{-.8pt}Q = 21\) en \(\angle Q = 25\degree \text{.}\) Bereken \(\angle \text{R} \text{.}\) Rond indien nodig af op één decimaal. SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 5ms c 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 \(\sin(\angle R) = {P\kern{-.8pt}Q ⋅ \sin(\angle Q) \over P\kern{-.8pt}R} = {21 ⋅ \sin(25\degree) \over 11} = 0{,}806... \text{.}\) 1p ○ Dit geeft \(\angle R ≈ 53{,}8\degree\) of \(\angle R ≈ 126{,}2\degree \text{.}\) Uit de afbeelding volgt dat \(\angle R\) een scherpe hoek is, dus \(\angle R ≈ 53{,}8\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 13 \text{,}\) \(P\kern{-.8pt}R = 20\) en \(\angle P = 32\degree \text{.}\) Bereken \(\angle \text{Q} \text{.}\) Rond indien nodig af op één decimaal. SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d 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 ○ Daaruit volgt \(\sin(\angle Q) = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over Q\kern{-.8pt}R} = {20 ⋅ \sin(32\degree) \over 13} = 0{,}815... \text{.}\) 1p ○ Dit geeft \(\angle Q ≈ 54{,}6\degree\) of \(\angle Q ≈ 125{,}4\degree \text{.}\) Uit de afbeelding volgt dat \(\angle Q\) een stompe hoek is, dus \(\angle Q ≈ 125{,}4\degree \text{.}\) 1p opgave 2 4p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 15 \text{,}\) \(\angle P = 48\degree\) en \(\angle R = 55\degree \text{.}\) Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle P + \angle Q + \angle R = 180\degree\) volgt \(\angle Q = 180\degree - \angle P - \angle R = 180\degree - 48\degree - 55\degree = 77\degree \text{.}\) 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 \(Q\kern{-.8pt}R = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over \sin(\angle Q)} = {15 ⋅ \sin(48\degree) \over \sin(77\degree)} \text{.}\) 1p ○ \(Q\kern{-.8pt}R ≈ 11{,}4 \text{.}\) 1p 4p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 42 \text{,}\) \(\angle L = 25\degree\) en \(\angle K = 57\degree \text{.}\) Bereken de lengte van zijde \(K\kern{-.8pt}M \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle L + \angle M + \angle K = 180\degree\) volgt \(\angle M = 180\degree - \angle L - \angle K = 180\degree - 25\degree - 57\degree = 98\degree \text{.}\) 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 \(K\kern{-.8pt}M = {K\kern{-.8pt}L ⋅ \sin(\angle L) \over \sin(\angle M)} = {42 ⋅ \sin(25\degree) \over \sin(98\degree)} \text{.}\) 1p ○ \(K\kern{-.8pt}M ≈ 17{,}9 \text{.}\) 1p
havo wiskunde B	3.3 De cosinusregel
Sinus- en cosinusregel (4)	opgave 1 3p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 32 \text{,}\) \(P\kern{-.8pt}Q = 23\) en \(\angle P = 85\degree \text{.}\) Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\) Rond indien nodig af op één decimaal. CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 0ms a 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 \(Q\kern{-.8pt}R^{2} = 32^{2} + 23^{2} - 2 ⋅ 32 ⋅ 23 ⋅ \cos(85\degree) = 1424{,}706... \text{.}\) 1p ○ \(Q\kern{-.8pt}R = \sqrt{1424{,}706...} ≈ 37{,}7 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 20 \text{,}\) \(P\kern{-.8pt}R = 26\) en \(\angle R = 94\degree \text{.}\) Bereken de lengte van zijde \(P\kern{-.8pt}Q \text{.}\) Rond indien nodig af op één decimaal. CosinusregelZijdeInStomp 007w - 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 ○ Dus \(P\kern{-.8pt}Q^{2} = 20^{2} + 26^{2} - 2 ⋅ 20 ⋅ 26 ⋅ \cos(94\degree) = 1148{,}546... \text{.}\) 1p ○ \(P\kern{-.8pt}Q = \sqrt{1148{,}546...} ≈ 33{,}9 \text{.}\) 1p 4p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 23 \text{,}\) \(P\kern{-.8pt}Q = 24\) en \(Q\kern{-.8pt}R = 28 \text{.}\) Bereken \(\angle \text{P} \text{.}\) Rond indien nodig af op één decimaal. CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 5ms 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 ○ Invullen geeft \(28^{2} = 23^{2} + 24^{2} - 2 ⋅ 23 ⋅ 24 ⋅ \cos(\angle P)\) dus \(784 = 1\,105 - 1\,104 ⋅ \cos(\angle P) \text{.}\) 1p ○ Balansmethode geeft \(\cos(\angle P) = {784 - 1\,105 \over -1\,104} = 0{,}290...\) 1p ○ Hieruit volgt \(\angle P = \cos^{-1}(0{,}290...) ≈ 73{,}1\degree \text{.}\) 1p 4p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 14 \text{,}\) \(P\kern{-.8pt}Q = 11\) en \(Q\kern{-.8pt}R = 22 \text{.}\) Bereken \(\angle \text{P} \text{.}\) Rond indien nodig af op één decimaal. CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms d 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 ○ Invullen geeft \(22^{2} = 14^{2} + 11^{2} - 2 ⋅ 14 ⋅ 11 ⋅ \cos(\angle P)\) dus \(484 = 317 - 308 ⋅ \cos(\angle P) \text{.}\) 1p ○ Balansmethode geeft \(\cos(\angle P) = {484 - 317 \over -308} = -0{,}542...\) 1p ○ Hieruit volgt \(\angle P = \cos^{-1}(-0{,}542...) ≈ 122{,}8\degree \text{.}\) 1p

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