Getal & Ruimte (13e editie) - 3 havo
'Sinus, cosinus en tangens'.
| 3 havo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 45 \text{,}\) \(\angle P = 39\degree\) 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 \(\tan(39\degree) = {Q\kern{-.8pt}R \over 45} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = 45 ⋅ \tan(39\degree) \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 36{,}4 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 22 \text{,}\) \(\angle C = 48\degree\) en \(\angle A = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle C) = {A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\tan(48\degree) = {22 \over A\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = {22 \over \tan(48\degree)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 19{,}8 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 58 \text{,}\) \(A\kern{-.8pt}C = 28\) en \(\angle C = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle B) = {A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\tan(\angle B) = {28 \over 58} \text{.}\) 1p ○ Hieruit volgt \(\angle B = \tan^{-1}({28 \over 58}) \text{.}\) 1p ○ Dus \(\angle B ≈ 25{,}8\degree \text{.}\) 1p |
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| 3 havo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 58 \text{,}\) \(\angle C = 31\degree\) 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 \(\sin(31\degree) = {A\kern{-.8pt}B \over 58} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 58 ⋅ \sin(31\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 29{,}9 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 34 \text{,}\) \(\angle K = 47\degree\) 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 \(\sin(47\degree) = {34 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {34 \over \sin(47\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 46{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 25 \text{,}\) \(L\kern{-.8pt}M = 42\) en \(\angle K = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle M) = {K\kern{-.8pt}L \over L\kern{-.8pt}M}\) ofwel \(\sin(\angle M) = {25 \over 42} \text{.}\) 1p ○ Hieruit volgt \(\angle M = \sin^{-1}({25 \over 42}) \text{.}\) 1p ○ Dus \(\angle M ≈ 36{,}5\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 51 \text{,}\) \(\angle K = 35\degree\) en \(\angle L = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle K) = {K\kern{-.8pt}L \over K\kern{-.8pt}M}\) ofwel \(\cos(35\degree) = {K\kern{-.8pt}L \over 51} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = 51 ⋅ \cos(35\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 41{,}8 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 60 \text{,}\) \(\angle A = 40\degree\) en \(\angle B = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle A) = {A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\cos(40\degree) = {60 \over A\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = {60 \over \cos(40\degree)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 78{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 42 \text{,}\) \(P\kern{-.8pt}R = 54\) 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 \(\cos(\angle P) = {42 \over 54} \text{.}\) 1p ○ Hieruit volgt \(\angle P = \cos^{-1}({42 \over 54}) \text{.}\) 1p ○ Dus \(\angle P ≈ 38{,}9\degree \text{.}\) 1p |