Getal & Ruimte (12e editie) - vwo wiskunde B
'Sinus, cosinus en tangens'.
| 3 vwo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 24 \text{,}\) \(\angle A = 33\degree\) 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 \(\tan(33\degree) = {B\kern{-.8pt}C \over 24} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = 24 ⋅ \tan(33\degree) \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 15{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 59 \text{,}\) \(\angle L = 53\degree\) 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 \(\tan(53\degree) = {59 \over L\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = {59 \over \tan(53\degree)} \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 44{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 47 \text{,}\) \(A\kern{-.8pt}C = 51\) 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 \(\tan(\angle B) = {51 \over 47} \text{.}\) 1p ○ Hieruit volgt \(\angle B = \tan^{-1}({51 \over 47}) \text{.}\) 1p ○ Dus \(\angle B ≈ 47{,}3\degree \text{.}\) 1p |
|
| 3 vwo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 69 \text{,}\) \(\angle Q = 46\degree\) en \(\angle R = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle Q) = {P\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\sin(46\degree) = {P\kern{-.8pt}R \over 69} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = 69 ⋅ \sin(46\degree) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 49{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 20 \text{,}\) \(\angle K = 57\degree\) en \(\angle L = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle K) = {L\kern{-.8pt}M \over K\kern{-.8pt}M}\) ofwel \(\sin(57\degree) = {20 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {20 \over \sin(57\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 23{,}8 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 47 \text{,}\) \(K\kern{-.8pt}M = 62\) en \(\angle L = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle K) = {L\kern{-.8pt}M \over K\kern{-.8pt}M}\) ofwel \(\sin(\angle K) = {47 \over 62} \text{.}\) 1p ○ Hieruit volgt \(\angle K = \sin^{-1}({47 \over 62}) \text{.}\) 1p ○ Dus \(\angle K ≈ 49{,}3\degree \text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 73 \text{,}\) \(\angle C = 57\degree\) en \(\angle A = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle C) = {A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\cos(57\degree) = {A\kern{-.8pt}C \over 73} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = 73 ⋅ \cos(57\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 39{,}8 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 41 \text{,}\) \(\angle L = 43\degree\) 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 \(\cos(43\degree) = {41 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {41 \over \cos(43\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 56{,}1 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 29 \text{,}\) \(A\kern{-.8pt}B = 51\) en \(\angle C = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle B) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\cos(\angle B) = {29 \over 51} \text{.}\) 1p ○ Hieruit volgt \(\angle B = \cos^{-1}({29 \over 51}) \text{.}\) 1p ○ Dus \(\angle B ≈ 55{,}3\degree \text{.}\) 1p |