Getal & Ruimte (13e editie) - 3 vwo
'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 \(Q\kern{-.8pt}R = 42 \text{,}\) \(\angle Q = 52\degree\) en \(\angle R = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle Q) = {P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\tan(52\degree) = {P\kern{-.8pt}R \over 42} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = 42 ⋅ \tan(52\degree) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 53{,}8 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 49 \text{,}\) \(\angle M = 44\degree\) en \(\angle K = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle M) = {K\kern{-.8pt}L \over K\kern{-.8pt}M}\) ofwel \(\tan(44\degree) = {49 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {49 \over \tan(44\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 50{,}7 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 28 \text{,}\) \(L\kern{-.8pt}M = 37\) en \(\angle L = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle K) = {L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\tan(\angle K) = {37 \over 28} \text{.}\) 1p ○ Hieruit volgt \(\angle K = \tan^{-1}({37 \over 28}) \text{.}\) 1p ○ Dus \(\angle K ≈ 52{,}9\degree \text{.}\) 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 \(Q\kern{-.8pt}R = 66 \text{,}\) \(\angle R = 46\degree\) en \(\angle P = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle R) = {P\kern{-.8pt}Q \over Q\kern{-.8pt}R}\) ofwel \(\sin(46\degree) = {P\kern{-.8pt}Q \over 66} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}Q = 66 ⋅ \sin(46\degree) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q ≈ 47{,}5 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 51 \text{,}\) \(\angle Q = 33\degree\) 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 \(\sin(33\degree) = {51 \over P\kern{-.8pt}Q} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}Q = {51 \over \sin(33\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q ≈ 93{,}6 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 21 \text{,}\) \(B\kern{-.8pt}C = 50\) en \(\angle A = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle C) = {A\kern{-.8pt}B \over B\kern{-.8pt}C}\) ofwel \(\sin(\angle C) = {21 \over 50} \text{.}\) 1p ○ Hieruit volgt \(\angle C = \sin^{-1}({21 \over 50}) \text{.}\) 1p ○ Dus \(\angle C ≈ 24{,}8\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 40 \text{,}\) \(\angle R = 43\degree\) en \(\angle P = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle R) = {P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\cos(43\degree) = {P\kern{-.8pt}R \over 40} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = 40 ⋅ \cos(43\degree) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 29{,}3 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 49 \text{,}\) \(\angle C = 54\degree\) en \(\angle A = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle C) = {A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\cos(54\degree) = {49 \over B\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = {49 \over \cos(54\degree)} \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 83{,}4 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 36 \text{,}\) \(K\kern{-.8pt}L = 55\) en \(\angle M = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle L) = {L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\cos(\angle L) = {36 \over 55} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \cos^{-1}({36 \over 55}) \text{.}\) 1p ○ Dus \(\angle L ≈ 49{,}1\degree \text{.}\) 1p |