Getal & Ruimte (12e editie) - vwo wiskunde B
'Sinus, cosinus en tangens'.
| 3 vwo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 23 \text{,}\) \(\angle P = 51\degree\) en \(\angle Q = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\tan(51\degree) = {Q\kern{-.8pt}R \over 23} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = 23 ⋅ \tan(51\degree) \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 28{,}4 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 25 \text{,}\) \(\angle Q = 54\degree\) en \(\angle R = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle Q) = {P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\tan(54\degree) = {25 \over Q\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = {25 \over \tan(54\degree)} \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 18{,}2 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 55 \text{,}\) \(P\kern{-.8pt}R = 54\) en \(\angle R = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle Q) = {P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\tan(\angle Q) = {54 \over 55} \text{.}\) 1p ○ Hieruit volgt \(\angle Q = \tan^{-1}({54 \over 55}) \text{.}\) 1p ○ Dus \(\angle Q ≈ 44{,}5\degree \text{.}\) 1p |
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| 3 vwo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 61 \text{,}\) \(\angle L = 51\degree\) en \(\angle M = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle L) = {K\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\sin(51\degree) = {K\kern{-.8pt}M \over 61} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = 61 ⋅ \sin(51\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 47{,}4 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 59 \text{,}\) \(\angle Q = 40\degree\) en \(\angle R = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle Q) = {P\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\sin(40\degree) = {59 \over P\kern{-.8pt}Q} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}Q = {59 \over \sin(40\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q ≈ 91{,}8 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 42 \text{,}\) \(K\kern{-.8pt}L = 65\) en \(\angle M = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle L) = {K\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\sin(\angle L) = {42 \over 65} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \sin^{-1}({42 \over 65}) \text{.}\) 1p ○ Dus \(\angle L ≈ 40{,}3\degree \text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 55 \text{,}\) \(\angle A = 46\degree\) en \(\angle B = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle A) = {A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\cos(46\degree) = {A\kern{-.8pt}B \over 55} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 55 ⋅ \cos(46\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 38{,}2 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 27 \text{,}\) \(\angle K = 33\degree\) en \(\angle L = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle K) = {K\kern{-.8pt}L \over K\kern{-.8pt}M}\) ofwel \(\cos(33\degree) = {27 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {27 \over \cos(33\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 32{,}2 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 55 \text{,}\) \(B\kern{-.8pt}C = 69\) en \(\angle A = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle C) = {A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\cos(\angle C) = {55 \over 69} \text{.}\) 1p ○ Hieruit volgt \(\angle C = \cos^{-1}({55 \over 69}) \text{.}\) 1p ○ Dus \(\angle C ≈ 37{,}1\degree \text{.}\) 1p |