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 \(Q\kern{-.8pt}R = 24 \text{,}\) \(\angle Q = 42\degree\) 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 \(\tan(42\degree) = {P\kern{-.8pt}R \over 24} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = 24 ⋅ \tan(42\degree) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 21{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 53 \text{,}\) \(\angle M = 36\degree\) 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 \(\tan(36\degree) = {53 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {53 \over \tan(36\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 72{,}9 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 60 \text{,}\) \(K\kern{-.8pt}M = 27\) 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 \(\tan(\angle L) = {27 \over 60} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \tan^{-1}({27 \over 60}) \text{.}\) 1p ○ Dus \(\angle L ≈ 24{,}2\degree \text{.}\) 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 \(P\kern{-.8pt}R = 61 \text{,}\) \(\angle P = 59\degree\) 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 \(\sin(59\degree) = {Q\kern{-.8pt}R \over 61} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = 61 ⋅ \sin(59\degree) \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 52{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 37 \text{,}\) \(\angle C = 46\degree\) en \(\angle A = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle C) = {A\kern{-.8pt}B \over B\kern{-.8pt}C}\) ofwel \(\sin(46\degree) = {37 \over B\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = {37 \over \sin(46\degree)} \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 51{,}4 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 55 \text{,}\) \(A\kern{-.8pt}C = 61\) en \(\angle B = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle A) = {B\kern{-.8pt}C \over A\kern{-.8pt}C}\) ofwel \(\sin(\angle A) = {55 \over 61} \text{.}\) 1p ○ Hieruit volgt \(\angle A = \sin^{-1}({55 \over 61}) \text{.}\) 1p ○ Dus \(\angle A ≈ 64{,}4\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 64 \text{,}\) \(\angle Q = 49\degree\) en \(\angle R = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle Q) = {Q\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\cos(49\degree) = {Q\kern{-.8pt}R \over 64} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = 64 ⋅ \cos(49\degree) \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 42{,}0 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 45 \text{,}\) \(\angle L = 52\degree\) 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 \(\cos(52\degree) = {45 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {45 \over \cos(52\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 73{,}1 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 44 \text{,}\) \(B\kern{-.8pt}C = 50\) 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 \(\cos(\angle C) = {44 \over 50} \text{.}\) 1p ○ Hieruit volgt \(\angle C = \cos^{-1}({44 \over 50}) \text{.}\) 1p ○ Dus \(\angle C ≈ 28{,}4\degree \text{.}\) 1p |