Getal & Ruimte (13e editie) - 3 vwo
'Sinus, cosinus en tangens'.
| 3 vwo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 38 \text{,}\) \(\angle C = 40\degree\) en \(\angle A = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle C) = {A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\tan(40\degree) = {A\kern{-.8pt}B \over 38} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 38 ⋅ \tan(40\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 31{,}9 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 57 \text{,}\) \(\angle K = 49\degree\) en \(\angle L = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle K) = {L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\tan(49\degree) = {57 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {57 \over \tan(49\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 49{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 31 \text{,}\) \(K\kern{-.8pt}M = 44\) 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 \(\tan(\angle L) = {44 \over 31} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \tan^{-1}({44 \over 31}) \text{.}\) 1p ○ Dus \(\angle L ≈ 54{,}8\degree \text{.}\) 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 \(K\kern{-.8pt}M = 48 \text{,}\) \(\angle K = 57\degree\) en \(\angle L = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle K) = {L\kern{-.8pt}M \over K\kern{-.8pt}M}\) ofwel \(\sin(57\degree) = {L\kern{-.8pt}M \over 48} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 48 ⋅ \sin(57\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 40{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 22 \text{,}\) \(\angle B = 46\degree\) en \(\angle C = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle B) = {A\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\sin(46\degree) = {22 \over A\kern{-.8pt}B} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = {22 \over \sin(46\degree)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 30{,}6 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 50 \text{,}\) \(P\kern{-.8pt}Q = 55\) en \(\angle R = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle Q) = {P\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\sin(\angle Q) = {50 \over 55} \text{.}\) 1p ○ Hieruit volgt \(\angle Q = \sin^{-1}({50 \over 55}) \text{.}\) 1p ○ Dus \(\angle Q ≈ 65{,}4\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 68 \text{,}\) \(\angle Q = 44\degree\) 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 \(\cos(44\degree) = {Q\kern{-.8pt}R \over 68} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = 68 ⋅ \cos(44\degree) \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 48{,}9 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 40 \text{,}\) \(\angle R = 56\degree\) en \(\angle P = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle R) = {P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\cos(56\degree) = {40 \over Q\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = {40 \over \cos(56\degree)} \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 71{,}5 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 39 \text{,}\) \(P\kern{-.8pt}R = 49\) en \(\angle Q = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle P) = {P\kern{-.8pt}Q \over P\kern{-.8pt}R}\) ofwel \(\cos(\angle P) = {39 \over 49} \text{.}\) 1p ○ Hieruit volgt \(\angle P = \cos^{-1}({39 \over 49}) \text{.}\) 1p ○ Dus \(\angle P ≈ 37{,}3\degree \text{.}\) 1p |