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 \(K\kern{-.8pt}L = 25 \text{,}\) \(\angle K = 36\degree\) 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 \(\tan(36\degree) = {L\kern{-.8pt}M \over 25} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 25 ⋅ \tan(36\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 18{,}2 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 38 \text{,}\) \(\angle A = 39\degree\) en \(\angle B = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle A) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\tan(39\degree) = {38 \over A\kern{-.8pt}B} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = {38 \over \tan(39\degree)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 46{,}9 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 32 \text{,}\) \(B\kern{-.8pt}C = 41\) en \(\angle B = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle A) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\tan(\angle A) = {41 \over 32} \text{.}\) 1p ○ Hieruit volgt \(\angle A = \tan^{-1}({41 \over 32}) \text{.}\) 1p ○ Dus \(\angle A ≈ 52{,}0\degree \text{.}\) 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 \(A\kern{-.8pt}C = 64 \text{,}\) \(\angle A = 59\degree\) en \(\angle B = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle A) = {B\kern{-.8pt}C \over A\kern{-.8pt}C}\) ofwel \(\sin(59\degree) = {B\kern{-.8pt}C \over 64} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = 64 ⋅ \sin(59\degree) \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 54{,}9 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 31 \text{,}\) \(\angle L = 57\degree\) en \(\angle M = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle L) = {K\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\sin(57\degree) = {31 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {31 \over \sin(57\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 37{,}0 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 54 \text{,}\) \(K\kern{-.8pt}L = 75\) 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 \(\sin(\angle L) = {54 \over 75} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \sin^{-1}({54 \over 75}) \text{.}\) 1p ○ Dus \(\angle L ≈ 46{,}1\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 59 \text{,}\) \(\angle Q = 47\degree\) 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 \(\cos(47\degree) = {Q\kern{-.8pt}R \over 59} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = 59 ⋅ \cos(47\degree) \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 40{,}2 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 29 \text{,}\) \(\angle C = 42\degree\) en \(\angle A = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle C) = {A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\cos(42\degree) = {29 \over B\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = {29 \over \cos(42\degree)} \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 39{,}0 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 54 \text{,}\) \(L\kern{-.8pt}M = 64\) 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 \(\cos(\angle M) = {54 \over 64} \text{.}\) 1p ○ Hieruit volgt \(\angle M = \cos^{-1}({54 \over 64}) \text{.}\) 1p ○ Dus \(\angle M ≈ 32{,}5\degree \text{.}\) 1p |