Getal & Ruimte (13e editie) - vwo wiskunde B
'Sinus, cosinus en tangens'.
| 3 vwo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 41 \text{,}\) \(\angle A = 36\degree\) en \(\angle B = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle A) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\tan(36\degree) = {B\kern{-.8pt}C \over 41} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = 41 ⋅ \tan(36\degree) \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 29{,}8 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 28 \text{,}\) \(\angle P = 58\degree\) en \(\angle Q = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\tan(58\degree) = {28 \over P\kern{-.8pt}Q} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}Q = {28 \over \tan(58\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q ≈ 17{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 29 \text{,}\) \(K\kern{-.8pt}M = 26\) en \(\angle M = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle L) = {K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\tan(\angle L) = {26 \over 29} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \tan^{-1}({26 \over 29}) \text{.}\) 1p ○ Dus \(\angle L ≈ 41{,}9\degree \text{.}\) 1p |
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| 3 vwo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 71 \text{,}\) \(\angle P = 49\degree\) en \(\angle Q = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}R}\) ofwel \(\sin(49\degree) = {Q\kern{-.8pt}R \over 71} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = 71 ⋅ \sin(49\degree) \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 53{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 21 \text{,}\) \(\angle P = 50\degree\) en \(\angle Q = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}R}\) ofwel \(\sin(50\degree) = {21 \over P\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = {21 \over \sin(50\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 27{,}4 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 36 \text{,}\) \(L\kern{-.8pt}M = 62\) en \(\angle K = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle M) = {K\kern{-.8pt}L \over L\kern{-.8pt}M}\) ofwel \(\sin(\angle M) = {36 \over 62} \text{.}\) 1p ○ Hieruit volgt \(\angle M = \sin^{-1}({36 \over 62}) \text{.}\) 1p ○ Dus \(\angle M ≈ 35{,}5\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 69 \text{,}\) \(\angle M = 34\degree\) en \(\angle K = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle M) = {K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\cos(34\degree) = {K\kern{-.8pt}M \over 69} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = 69 ⋅ \cos(34\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 57{,}2 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 29 \text{,}\) \(\angle L = 42\degree\) en \(\angle M = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle L) = {L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\cos(42\degree) = {29 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {29 \over \cos(42\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 39{,}0 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 21 \text{,}\) \(L\kern{-.8pt}M = 31\) en \(\angle K = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle M) = {K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\cos(\angle M) = {21 \over 31} \text{.}\) 1p ○ Hieruit volgt \(\angle M = \cos^{-1}({21 \over 31}) \text{.}\) 1p ○ Dus \(\angle M ≈ 47{,}4\degree \text{.}\) 1p |