Getal & Ruimte (13e editie) - 3 vwo
'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 \(K\kern{-.8pt}L = 52 \text{,}\) \(\angle K = 55\degree\) 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 \(\tan(55\degree) = {L\kern{-.8pt}M \over 52} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 52 ⋅ \tan(55\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 74{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 50 \text{,}\) \(\angle P = 57\degree\) en \(\angle Q = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\tan(57\degree) = {50 \over P\kern{-.8pt}Q} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}Q = {50 \over \tan(57\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q ≈ 32{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 57 \text{,}\) \(A\kern{-.8pt}C = 56\) 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 \(\tan(\angle B) = {56 \over 57} \text{.}\) 1p ○ Hieruit volgt \(\angle B = \tan^{-1}({56 \over 57}) \text{.}\) 1p ○ Dus \(\angle B ≈ 44{,}5\degree \text{.}\) 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 \(A\kern{-.8pt}B = 68 \text{,}\) \(\angle B = 36\degree\) en \(\angle C = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle B) = {A\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\sin(36\degree) = {A\kern{-.8pt}C \over 68} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = 68 ⋅ \sin(36\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 40{,}0 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 27 \text{,}\) \(\angle B = 44\degree\) 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 \(\sin(44\degree) = {27 \over A\kern{-.8pt}B} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = {27 \over \sin(44\degree)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 38{,}9 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 46 \text{,}\) \(Q\kern{-.8pt}R = 62\) en \(\angle P = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle R) = {P\kern{-.8pt}Q \over Q\kern{-.8pt}R}\) ofwel \(\sin(\angle R) = {46 \over 62} \text{.}\) 1p ○ Hieruit volgt \(\angle R = \sin^{-1}({46 \over 62}) \text{.}\) 1p ○ Dus \(\angle R ≈ 47{,}9\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 57 \text{,}\) \(\angle M = 43\degree\) en \(\angle K = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle M) = {K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\cos(43\degree) = {K\kern{-.8pt}M \over 57} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = 57 ⋅ \cos(43\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 41{,}7 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 59 \text{,}\) \(\angle B = 54\degree\) 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 \(\cos(54\degree) = {59 \over A\kern{-.8pt}B} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = {59 \over \cos(54\degree)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 100{,}4 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 39 \text{,}\) \(B\kern{-.8pt}C = 46\) en \(\angle A = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle C) = {A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\cos(\angle C) = {39 \over 46} \text{.}\) 1p ○ Hieruit volgt \(\angle C = \cos^{-1}({39 \over 46}) \text{.}\) 1p ○ Dus \(\angle C ≈ 32{,}0\degree \text{.}\) 1p |