Getal & Ruimte (13e editie) - 3 havo
'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 \(K\kern{-.8pt}L = 27 \text{,}\) \(\angle K = 35\degree\) 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 \(\tan(35\degree) = {L\kern{-.8pt}M \over 27} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 27 ⋅ \tan(35\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 18{,}9 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 57 \text{,}\) \(\angle R = 33\degree\) en \(\angle P = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle R) = {P\kern{-.8pt}Q \over P\kern{-.8pt}R}\) ofwel \(\tan(33\degree) = {57 \over P\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = {57 \over \tan(33\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 87{,}8 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 47 \text{,}\) \(Q\kern{-.8pt}R = 21\) en \(\angle Q = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\tan(\angle P) = {21 \over 47} \text{.}\) 1p ○ Hieruit volgt \(\angle P = \tan^{-1}({21 \over 47}) \text{.}\) 1p ○ Dus \(\angle P ≈ 24{,}1\degree \text{.}\) 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 \(A\kern{-.8pt}C = 52 \text{,}\) \(\angle A = 50\degree\) 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 \(\sin(50\degree) = {B\kern{-.8pt}C \over 52} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = 52 ⋅ \sin(50\degree) \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 39{,}8 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 33 \text{,}\) \(\angle P = 48\degree\) 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 \(\sin(48\degree) = {33 \over P\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = {33 \over \sin(48\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 44{,}4 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 54 \text{,}\) \(P\kern{-.8pt}Q = 60\) en \(\angle R = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle Q) = {P\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\sin(\angle Q) = {54 \over 60} \text{.}\) 1p ○ Hieruit volgt \(\angle Q = \sin^{-1}({54 \over 60}) \text{.}\) 1p ○ Dus \(\angle Q ≈ 64{,}2\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 44 \text{,}\) \(\angle M = 31\degree\) 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 \(\cos(31\degree) = {K\kern{-.8pt}M \over 44} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = 44 ⋅ \cos(31\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 37{,}7 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 29 \text{,}\) \(\angle L = 54\degree\) en \(\angle M = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle L) = {L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\cos(54\degree) = {29 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {29 \over \cos(54\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 49{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 50 \text{,}\) \(Q\kern{-.8pt}R = 65\) 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 \(\cos(\angle R) = {50 \over 65} \text{.}\) 1p ○ Hieruit volgt \(\angle R = \cos^{-1}({50 \over 65}) \text{.}\) 1p ○ Dus \(\angle R ≈ 39{,}7\degree \text{.}\) 1p |