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 \(A\kern{-.8pt}C = 51 \text{,}\) \(\angle C = 32\degree\) 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 \(\tan(32\degree) = {A\kern{-.8pt}B \over 51} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 51 ⋅ \tan(32\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 31{,}9 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 52 \text{,}\) \(\angle K = 49\degree\) en \(\angle L = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle K) = {L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\tan(49\degree) = {52 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {52 \over \tan(49\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 45{,}2 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 20 \text{,}\) \(P\kern{-.8pt}Q = 56\) en \(\angle P = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle R) = {P\kern{-.8pt}Q \over P\kern{-.8pt}R}\) ofwel \(\tan(\angle R) = {56 \over 20} \text{.}\) 1p ○ Hieruit volgt \(\angle R = \tan^{-1}({56 \over 20}) \text{.}\) 1p ○ Dus \(\angle R ≈ 70{,}3\degree \text{.}\) 1p |
|
| 3 havo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 56 \text{,}\) \(\angle C = 31\degree\) 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 \(\sin(31\degree) = {A\kern{-.8pt}B \over 56} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 56 ⋅ \sin(31\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 28{,}8 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 22 \text{,}\) \(\angle K = 53\degree\) 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 \(\sin(53\degree) = {22 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {22 \over \sin(53\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 27{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 44 \text{,}\) \(P\kern{-.8pt}Q = 53\) 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 \(\sin(\angle Q) = {44 \over 53} \text{.}\) 1p ○ Hieruit volgt \(\angle Q = \sin^{-1}({44 \over 53}) \text{.}\) 1p ○ Dus \(\angle Q ≈ 56{,}1\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 53 \text{,}\) \(\angle L = 46\degree\) 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 \(\cos(46\degree) = {L\kern{-.8pt}M \over 53} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 53 ⋅ \cos(46\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 36{,}8 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 54 \text{,}\) \(\angle M = 50\degree\) en \(\angle K = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle M) = {K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\cos(50\degree) = {54 \over L\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = {54 \over \cos(50\degree)} \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 84{,}0 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 28 \text{,}\) \(L\kern{-.8pt}M = 56\) en \(\angle K = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle M) = {K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\cos(\angle M) = {28 \over 56} \text{.}\) 1p ○ Hieruit volgt \(\angle M = \cos^{-1}({28 \over 56}) \text{.}\) 1p ○ Dus \(\angle M = 60{,}0\degree \text{.}\) 1p |