Getal & Ruimte (12e editie) - vwo wiskunde B
'Sinus, cosinus en tangens'.
| 3 vwo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 43 \text{,}\) \(\angle K = 37\degree\) en \(\angle L = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle K) = {L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\tan(37\degree) = {L\kern{-.8pt}M \over 43} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 43 ⋅ \tan(37\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 32{,}4 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 25 \text{,}\) \(\angle L = 49\degree\) en \(\angle M = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle L) = {K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\tan(49\degree) = {25 \over L\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = {25 \over \tan(49\degree)} \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 21{,}7 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 53 \text{,}\) \(Q\kern{-.8pt}R = 23\) en \(\angle Q = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\tan(\angle P) = {23 \over 53} \text{.}\) 1p ○ Hieruit volgt \(\angle P = \tan^{-1}({23 \over 53}) \text{.}\) 1p ○ Dus \(\angle P ≈ 23{,}5\degree \text{.}\) 1p |
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| 3 vwo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 65 \text{,}\) \(\angle C = 47\degree\) en \(\angle A = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle C) = {A\kern{-.8pt}B \over B\kern{-.8pt}C}\) ofwel \(\sin(47\degree) = {A\kern{-.8pt}B \over 65} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 65 ⋅ \sin(47\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 47{,}5 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 31 \text{,}\) \(\angle M = 50\degree\) en \(\angle K = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle M) = {K\kern{-.8pt}L \over L\kern{-.8pt}M}\) ofwel \(\sin(50\degree) = {31 \over L\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = {31 \over \sin(50\degree)} \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 40{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 49 \text{,}\) \(A\kern{-.8pt}B = 75\) en \(\angle C = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle B) = {A\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\sin(\angle B) = {49 \over 75} \text{.}\) 1p ○ Hieruit volgt \(\angle B = \sin^{-1}({49 \over 75}) \text{.}\) 1p ○ Dus \(\angle B ≈ 40{,}8\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 50 \text{,}\) \(\angle R = 53\degree\) 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 \(\cos(53\degree) = {P\kern{-.8pt}R \over 50} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = 50 ⋅ \cos(53\degree) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 30{,}1 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 58 \text{,}\) \(\angle B = 55\degree\) en \(\angle C = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle B) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\cos(55\degree) = {58 \over A\kern{-.8pt}B} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = {58 \over \cos(55\degree)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 101{,}1 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 55 \text{,}\) \(L\kern{-.8pt}M = 73\) 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 \(\cos(\angle M) = {55 \over 73} \text{.}\) 1p ○ Hieruit volgt \(\angle M = \cos^{-1}({55 \over 73}) \text{.}\) 1p ○ Dus \(\angle M ≈ 41{,}1\degree \text{.}\) 1p |