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 \(A\kern{-.8pt}C = 44 \text{,}\) \(\angle C = 43\degree\) 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 \(\tan(43\degree) = {A\kern{-.8pt}B \over 44} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 44 ⋅ \tan(43\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 41{,}0 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 58 \text{,}\) \(\angle B = 42\degree\) en \(\angle C = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle B) = {A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\tan(42\degree) = {58 \over B\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = {58 \over \tan(42\degree)} \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 64{,}4 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 33 \text{,}\) \(P\kern{-.8pt}Q = 48\) 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 \(\tan(\angle R) = {48 \over 33} \text{.}\) 1p ○ Hieruit volgt \(\angle R = \tan^{-1}({48 \over 33}) \text{.}\) 1p ○ Dus \(\angle R ≈ 55{,}5\degree \text{.}\) 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 \(B\kern{-.8pt}C = 55 \text{,}\) \(\angle C = 36\degree\) 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 \(\sin(36\degree) = {A\kern{-.8pt}B \over 55} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B = 55 ⋅ \sin(36\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B ≈ 32{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 36 \text{,}\) \(\angle C = 40\degree\) en \(\angle A = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle C) = {A\kern{-.8pt}B \over B\kern{-.8pt}C}\) ofwel \(\sin(40\degree) = {36 \over B\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = {36 \over \sin(40\degree)} \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 56{,}0 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 43 \text{,}\) \(B\kern{-.8pt}C = 57\) en \(\angle A = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle C) = {A\kern{-.8pt}B \over B\kern{-.8pt}C}\) ofwel \(\sin(\angle C) = {43 \over 57} \text{.}\) 1p ○ Hieruit volgt \(\angle C = \sin^{-1}({43 \over 57}) \text{.}\) 1p ○ Dus \(\angle C ≈ 49{,}0\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 54 \text{,}\) \(\angle L = 31\degree\) 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 \(\cos(31\degree) = {L\kern{-.8pt}M \over 54} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 54 ⋅ \cos(31\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 46{,}3 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 42 \text{,}\) \(\angle M = 35\degree\) 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 \(\cos(35\degree) = {42 \over L\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = {42 \over \cos(35\degree)} \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 51{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 36 \text{,}\) \(A\kern{-.8pt}C = 48\) en \(\angle B = 90\degree \text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle A) = {A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\cos(\angle A) = {36 \over 48} \text{.}\) 1p ○ Hieruit volgt \(\angle A = \cos^{-1}({36 \over 48}) \text{.}\) 1p ○ Dus \(\angle A ≈ 41{,}4\degree \text{.}\) 1p |