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 \(A\kern{-.8pt}C = 25 \text{,}\) \(\angle C = 57\degree\) 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 \(\tan(57\degree) = {A\kern{-.8pt}B \over 25} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 25 ⋅ \tan(57\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 38{,}5 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 33 \text{,}\) \(\angle C = 35\degree\) en \(\angle A = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle C) = {A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\tan(35\degree) = {33 \over A\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = {33 \over \tan(35\degree)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 47{,}1 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 54 \text{,}\) \(A\kern{-.8pt}B = 37\) en \(\angle A = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle C) = {A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\tan(\angle C) = {37 \over 54} \text{.}\) 1p ○ Hieruit volgt \(\angle C = \tan^{-1}({37 \over 54}) \text{.}\) 1p ○ Dus \(\angle C ≈ 34{,}4\degree \text{.}\) 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 \(A\kern{-.8pt}B = 57 \text{,}\) \(\angle B = 37\degree\) 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 \(\sin(37\degree) = {A\kern{-.8pt}C \over 57} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = 57 ⋅ \sin(37\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 34{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 26 \text{,}\) \(\angle L = 43\degree\) en \(\angle M = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle L) = {K\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\sin(43\degree) = {26 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {26 \over \sin(43\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 38{,}1 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 27 \text{,}\) \(P\kern{-.8pt}Q = 49\) 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 \(\sin(\angle Q) = {27 \over 49} \text{.}\) 1p ○ Hieruit volgt \(\angle Q = \sin^{-1}({27 \over 49}) \text{.}\) 1p ○ Dus \(\angle Q ≈ 33{,}4\degree \text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 55 \text{,}\) \(\angle B = 56\degree\) en \(\angle C = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle B) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\cos(56\degree) = {B\kern{-.8pt}C \over 55} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = 55 ⋅ \cos(56\degree) \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 30{,}8 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 50 \text{,}\) \(\angle K = 50\degree\) en \(\angle L = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle K) = {K\kern{-.8pt}L \over K\kern{-.8pt}M}\) ofwel \(\cos(50\degree) = {50 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {50 \over \cos(50\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 77{,}8 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 30 \text{,}\) \(L\kern{-.8pt}M = 67\) 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 \(\cos(\angle M) = {30 \over 67} \text{.}\) 1p ○ Hieruit volgt \(\angle M = \cos^{-1}({30 \over 67}) \text{.}\) 1p ○ Dus \(\angle M ≈ 63{,}4\degree \text{.}\) 1p |