Getal & Ruimte (13e editie) - 3 havo
'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 \(B\kern{-.8pt}C = 53 \text{,}\) \(\angle B = 59\degree\) en \(\angle C = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle B) = {A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\tan(59\degree) = {A\kern{-.8pt}C \over 53} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = 53 ⋅ \tan(59\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 88{,}2 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 23 \text{,}\) \(\angle M = 59\degree\) en \(\angle K = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle M) = {K\kern{-.8pt}L \over K\kern{-.8pt}M}\) ofwel \(\tan(59\degree) = {23 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {23 \over \tan(59\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 13{,}8 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 56 \text{,}\) \(K\kern{-.8pt}M = 41\) en \(\angle M = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle L) = {K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\tan(\angle L) = {41 \over 56} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \tan^{-1}({41 \over 56}) \text{.}\) 1p ○ Dus \(\angle L ≈ 36{,}2\degree \text{.}\) 1p |
|
| 3 havo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 59 \text{,}\) \(\angle R = 36\degree\) 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 \(\sin(36\degree) = {P\kern{-.8pt}Q \over 59} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}Q = 59 ⋅ \sin(36\degree) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q ≈ 34{,}7 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 21 \text{,}\) \(\angle Q = 35\degree\) en \(\angle R = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle Q) = {P\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\sin(35\degree) = {21 \over P\kern{-.8pt}Q} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}Q = {21 \over \sin(35\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q ≈ 36{,}6 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 55 \text{,}\) \(K\kern{-.8pt}L = 81\) en \(\angle M = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle L) = {K\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\sin(\angle L) = {55 \over 81} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \sin^{-1}({55 \over 81}) \text{.}\) 1p ○ Dus \(\angle L ≈ 42{,}8\degree \text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 71 \text{,}\) \(\angle C = 51\degree\) 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 \(\cos(51\degree) = {A\kern{-.8pt}C \over 71} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = 71 ⋅ \cos(51\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 44{,}7 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 43 \text{,}\) \(\angle L = 51\degree\) 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 \(\cos(51\degree) = {43 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {43 \over \cos(51\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 68{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 49 \text{,}\) \(K\kern{-.8pt}M = 62\) en \(\angle L = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle K) = {K\kern{-.8pt}L \over K\kern{-.8pt}M}\) ofwel \(\cos(\angle K) = {49 \over 62} \text{.}\) 1p ○ Hieruit volgt \(\angle K = \cos^{-1}({49 \over 62}) \text{.}\) 1p ○ Dus \(\angle K ≈ 37{,}8\degree \text{.}\) 1p |