Getal & Ruimte (13e editie) - 3 havo
'Sinus, cosinus en tangens'.
| 3 havo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 29 \text{,}\) \(\angle A = 40\degree\) en \(\angle B = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle A) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\tan(40\degree) = {B\kern{-.8pt}C \over 29} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = 29 ⋅ \tan(40\degree) \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 24{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 21 \text{,}\) \(\angle M = 57\degree\) 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 \(\tan(57\degree) = {21 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {21 \over \tan(57\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 13{,}6 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 27 \text{,}\) \(L\kern{-.8pt}M = 51\) 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 \(\tan(\angle K) = {51 \over 27} \text{.}\) 1p ○ Hieruit volgt \(\angle K = \tan^{-1}({51 \over 27}) \text{.}\) 1p ○ Dus \(\angle K ≈ 62{,}1\degree \text{.}\) 1p |
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| 3 havo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 71 \text{,}\) \(\angle K = 48\degree\) 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 \(\sin(48\degree) = {L\kern{-.8pt}M \over 71} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 71 ⋅ \sin(48\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 52{,}8 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 38 \text{,}\) \(\angle K = 43\degree\) 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 \(\sin(43\degree) = {38 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {38 \over \sin(43\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 55{,}7 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 44 \text{,}\) \(B\kern{-.8pt}C = 48\) 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 \(\sin(\angle C) = {44 \over 48} \text{.}\) 1p ○ Hieruit volgt \(\angle C = \sin^{-1}({44 \over 48}) \text{.}\) 1p ○ Dus \(\angle C ≈ 66{,}4\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 68 \text{,}\) \(\angle P = 54\degree\) en \(\angle Q = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle P) = {P\kern{-.8pt}Q \over P\kern{-.8pt}R}\) ofwel \(\cos(54\degree) = {P\kern{-.8pt}Q \over 68} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}Q = 68 ⋅ \cos(54\degree) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q ≈ 40{,}0 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 34 \text{,}\) \(\angle P = 35\degree\) en \(\angle Q = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle P) = {P\kern{-.8pt}Q \over P\kern{-.8pt}R}\) ofwel \(\cos(35\degree) = {34 \over P\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = {34 \over \cos(35\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 41{,}5 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 41 \text{,}\) \(P\kern{-.8pt}R = 59\) 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 \(\cos(\angle P) = {41 \over 59} \text{.}\) 1p ○ Hieruit volgt \(\angle P = \cos^{-1}({41 \over 59}) \text{.}\) 1p ○ Dus \(\angle P ≈ 46{,}0\degree \text{.}\) 1p |