Getal & Ruimte (13e editie) - 3 vwo
'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 \(A\kern{-.8pt}C = 54 \text{,}\) \(\angle C = 39\degree\) en \(\angle A = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle C) = {A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\tan(39\degree) = {A\kern{-.8pt}B \over 54} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 54 ⋅ \tan(39\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 43{,}7 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 36 \text{,}\) \(\angle R = 36\degree\) 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 \(\tan(36\degree) = {36 \over P\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = {36 \over \tan(36\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 49{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 54 \text{,}\) \(B\kern{-.8pt}C = 26\) en \(\angle B = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle A) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\tan(\angle A) = {26 \over 54} \text{.}\) 1p ○ Hieruit volgt \(\angle A = \tan^{-1}({26 \over 54}) \text{.}\) 1p ○ Dus \(\angle A ≈ 25{,}7\degree \text{.}\) 1p |
|
| 3 vwo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 74 \text{,}\) \(\angle B = 40\degree\) en \(\angle C = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle B) = {A\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\sin(40\degree) = {A\kern{-.8pt}C \over 74} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = 74 ⋅ \sin(40\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 47{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 51 \text{,}\) \(\angle K = 32\degree\) 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 \(\sin(32\degree) = {51 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {51 \over \sin(32\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 96{,}2 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 30 \text{,}\) \(P\kern{-.8pt}R = 64\) 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 \(\sin(\angle P) = {30 \over 64} \text{.}\) 1p ○ Hieruit volgt \(\angle P = \sin^{-1}({30 \over 64}) \text{.}\) 1p ○ Dus \(\angle P ≈ 28{,}0\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 54 \text{,}\) \(\angle L = 32\degree\) 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 \(\cos(32\degree) = {L\kern{-.8pt}M \over 54} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 54 ⋅ \cos(32\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 45{,}8 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 20 \text{,}\) \(\angle L = 40\degree\) 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 \(\cos(40\degree) = {20 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {20 \over \cos(40\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 26{,}1 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 28 \text{,}\) \(A\kern{-.8pt}B = 50\) 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 \(\cos(\angle B) = {28 \over 50} \text{.}\) 1p ○ Hieruit volgt \(\angle B = \cos^{-1}({28 \over 50}) \text{.}\) 1p ○ Dus \(\angle B ≈ 55{,}9\degree \text{.}\) 1p |