Getal & Ruimte (13e editie) - vwo wiskunde B
'Sinus, cosinus en tangens'.
| 3 vwo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 54 \text{,}\) \(\angle P = 57\degree\) 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 \(\tan(57\degree) = {Q\kern{-.8pt}R \over 54} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = 54 ⋅ \tan(57\degree) \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 83{,}2 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 48 \text{,}\) \(\angle R = 46\degree\) en \(\angle P = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle R) = {P\kern{-.8pt}Q \over P\kern{-.8pt}R}\) ofwel \(\tan(46\degree) = {48 \over P\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = {48 \over \tan(46\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 46{,}4 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 34 \text{,}\) \(P\kern{-.8pt}Q = 23\) en \(\angle P = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle R) = {P\kern{-.8pt}Q \over P\kern{-.8pt}R}\) ofwel \(\tan(\angle R) = {23 \over 34} \text{.}\) 1p ○ Hieruit volgt \(\angle R = \tan^{-1}({23 \over 34}) \text{.}\) 1p ○ Dus \(\angle R ≈ 34{,}1\degree \text{.}\) 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 \(L\kern{-.8pt}M = 40 \text{,}\) \(\angle M = 56\degree\) en \(\angle K = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle M) = {K\kern{-.8pt}L \over L\kern{-.8pt}M}\) ofwel \(\sin(56\degree) = {K\kern{-.8pt}L \over 40} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = 40 ⋅ \sin(56\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 33{,}2 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 24 \text{,}\) \(\angle K = 33\degree\) en \(\angle L = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle K) = {L\kern{-.8pt}M \over K\kern{-.8pt}M}\) ofwel \(\sin(33\degree) = {24 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {24 \over \sin(33\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 44{,}1 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 34 \text{,}\) \(K\kern{-.8pt}L = 50\) 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 \(\sin(\angle L) = {34 \over 50} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \sin^{-1}({34 \over 50}) \text{.}\) 1p ○ Dus \(\angle L ≈ 42{,}8\degree \text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 72 \text{,}\) \(\angle B = 48\degree\) en \(\angle C = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle B) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\cos(48\degree) = {B\kern{-.8pt}C \over 72} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = 72 ⋅ \cos(48\degree) \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 48{,}2 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 22 \text{,}\) \(\angle K = 50\degree\) 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 \(\cos(50\degree) = {22 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {22 \over \cos(50\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 34{,}2 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 57 \text{,}\) \(K\kern{-.8pt}L = 73\) 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 \(\cos(\angle L) = {57 \over 73} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \cos^{-1}({57 \over 73}) \text{.}\) 1p ○ Dus \(\angle L ≈ 38{,}7\degree \text{.}\) 1p |