Getal & Ruimte (12e editie) - havo wiskunde B
'Sinus, cosinus en tangens'.
| 3 havo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 26 \text{,}\) \(\angle M = 59\degree\) 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 \(\tan(59\degree) = {K\kern{-.8pt}L \over 26} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = 26 ⋅ \tan(59\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 43{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 56 \text{,}\) \(\angle L = 50\degree\) en \(\angle M = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle L) = {K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\tan(50\degree) = {56 \over L\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = {56 \over \tan(50\degree)} \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 47{,}0 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 45 \text{,}\) \(P\kern{-.8pt}R = 20\) en \(\angle R = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle Q) = {P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\tan(\angle Q) = {20 \over 45} \text{.}\) 1p ○ Hieruit volgt \(\angle Q = \tan^{-1}({20 \over 45}) \text{.}\) 1p ○ Dus \(\angle Q ≈ 24{,}0\degree \text{.}\) 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 \(B\kern{-.8pt}C = 76 \text{,}\) \(\angle C = 40\degree\) 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 \(\sin(40\degree) = {A\kern{-.8pt}B \over 76} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 76 ⋅ \sin(40\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 48{,}9 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 20 \text{,}\) \(\angle P = 36\degree\) en \(\angle Q = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}R}\) ofwel \(\sin(36\degree) = {20 \over P\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = {20 \over \sin(36\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 34{,}0 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 38 \text{,}\) \(B\kern{-.8pt}C = 47\) 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 \(\sin(\angle C) = {38 \over 47} \text{.}\) 1p ○ Hieruit volgt \(\angle C = \sin^{-1}({38 \over 47}) \text{.}\) 1p ○ Dus \(\angle C ≈ 54{,}0\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 65 \text{,}\) \(\angle L = 46\degree\) en \(\angle M = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle L) = {L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\cos(46\degree) = {L\kern{-.8pt}M \over 65} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 65 ⋅ \cos(46\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 45{,}2 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 38 \text{,}\) \(\angle P = 39\degree\) 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 \(\cos(39\degree) = {38 \over P\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = {38 \over \cos(39\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 48{,}9 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 48 \text{,}\) \(K\kern{-.8pt}M = 63\) en \(\angle L = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle K) = {K\kern{-.8pt}L \over K\kern{-.8pt}M}\) ofwel \(\cos(\angle K) = {48 \over 63} \text{.}\) 1p ○ Hieruit volgt \(\angle K = \cos^{-1}({48 \over 63}) \text{.}\) 1p ○ Dus \(\angle K ≈ 40{,}4\degree \text{.}\) 1p |