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 \(K\kern{-.8pt}M = 50 \text{,}\) \(\angle M = 32\degree\) en \(\angle K = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle M) = {K\kern{-.8pt}L \over K\kern{-.8pt}M}\) ofwel \(\tan(32\degree) = {K\kern{-.8pt}L \over 50} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = 50 ⋅ \tan(32\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 31{,}2 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 53 \text{,}\) \(\angle Q = 57\degree\) en \(\angle R = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle Q) = {P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\tan(57\degree) = {53 \over Q\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = {53 \over \tan(57\degree)} \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 34{,}4 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 29 \text{,}\) \(L\kern{-.8pt}M = 20\) 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 \(\tan(\angle K) = {20 \over 29} \text{.}\) 1p ○ Hieruit volgt \(\angle K = \tan^{-1}({20 \over 29}) \text{.}\) 1p ○ Dus \(\angle K ≈ 34{,}6\degree \text{.}\) 1p |
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| 3 vwo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 46 \text{,}\) \(\angle R = 52\degree\) en \(\angle P = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle R) = {P\kern{-.8pt}Q \over Q\kern{-.8pt}R}\) ofwel \(\sin(52\degree) = {P\kern{-.8pt}Q \over 46} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}Q = 46 ⋅ \sin(52\degree) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q ≈ 36{,}2 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 51 \text{,}\) \(\angle C = 53\degree\) en \(\angle A = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle C) = {A\kern{-.8pt}B \over B\kern{-.8pt}C}\) ofwel \(\sin(53\degree) = {51 \over B\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = {51 \over \sin(53\degree)} \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 63{,}9 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 32 \text{,}\) \(P\kern{-.8pt}R = 41\) en \(\angle Q = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}R}\) ofwel \(\sin(\angle P) = {32 \over 41} \text{.}\) 1p ○ Hieruit volgt \(\angle P = \sin^{-1}({32 \over 41}) \text{.}\) 1p ○ Dus \(\angle P ≈ 51{,}3\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 50 \text{,}\) \(\angle M = 37\degree\) 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 \(\cos(37\degree) = {K\kern{-.8pt}M \over 50} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = 50 ⋅ \cos(37\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 39{,}9 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 35 \text{,}\) \(\angle B = 35\degree\) 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 \(\cos(35\degree) = {35 \over A\kern{-.8pt}B} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = {35 \over \cos(35\degree)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 42{,}7 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 25 \text{,}\) \(Q\kern{-.8pt}R = 34\) en \(\angle P = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle R) = {P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\cos(\angle R) = {25 \over 34} \text{.}\) 1p ○ Hieruit volgt \(\angle R = \cos^{-1}({25 \over 34}) \text{.}\) 1p ○ Dus \(\angle R ≈ 42{,}7\degree \text{.}\) 1p |