Getal & Ruimte (13e editie) - havo wiskunde B
'Sinus, cosinus en tangens'.
| 3 havo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 52 \text{,}\) \(\angle A = 52\degree\) en \(\angle B = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle A) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\tan(52\degree) = {B\kern{-.8pt}C \over 52} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = 52 ⋅ \tan(52\degree) \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 66{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 38 \text{,}\) \(\angle Q = 49\degree\) 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 \(\tan(49\degree) = {38 \over Q\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = {38 \over \tan(49\degree)} \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 33{,}0 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 54 \text{,}\) \(A\kern{-.8pt}B = 36\) en \(\angle A = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle C) = {A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\tan(\angle C) = {36 \over 54} \text{.}\) 1p ○ Hieruit volgt \(\angle C = \tan^{-1}({36 \over 54}) \text{.}\) 1p ○ Dus \(\angle C ≈ 33{,}7\degree \text{.}\) 1p |
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| 3 havo | 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 \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 27 \text{,}\) \(\angle L = 41\degree\) 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 \(\sin(41\degree) = {27 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {27 \over \sin(41\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 41{,}2 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 43 \text{,}\) \(P\kern{-.8pt}R = 52\) 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 \(\sin(\angle P) = {43 \over 52} \text{.}\) 1p ○ Hieruit volgt \(\angle P = \sin^{-1}({43 \over 52}) \text{.}\) 1p ○ Dus \(\angle P ≈ 55{,}8\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 48 \text{,}\) \(\angle Q = 53\degree\) 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 \(\cos(53\degree) = {Q\kern{-.8pt}R \over 48} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = 48 ⋅ \cos(53\degree) \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 28{,}9 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 23 \text{,}\) \(\angle R = 40\degree\) en \(\angle P = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle R) = {P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\cos(40\degree) = {23 \over Q\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = {23 \over \cos(40\degree)} \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 30{,}0 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 29 \text{,}\) \(Q\kern{-.8pt}R = 36\) 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 \(\cos(\angle R) = {29 \over 36} \text{.}\) 1p ○ Hieruit volgt \(\angle R = \cos^{-1}({29 \over 36}) \text{.}\) 1p ○ Dus \(\angle R ≈ 36{,}3\degree \text{.}\) 1p |