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 \(B\kern{-.8pt}C = 23 \text{,}\) \(\angle B = 48\degree\) 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 \(\tan(48\degree) = {A\kern{-.8pt}C \over 23} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = 23 ⋅ \tan(48\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 25{,}5 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 28 \text{,}\) \(\angle M = 58\degree\) 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 \(\tan(58\degree) = {28 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {28 \over \tan(58\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 17{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 44 \text{,}\) \(K\kern{-.8pt}M = 49\) 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 \(\tan(\angle L) = {49 \over 44} \text{.}\) 1p ○ Hieruit volgt \(\angle L = \tan^{-1}({49 \over 44}) \text{.}\) 1p ○ Dus \(\angle L ≈ 48{,}1\degree \text{.}\) 1p |
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| 3 vwo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 79 \text{,}\) \(\angle B = 49\degree\) 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 \(\sin(49\degree) = {A\kern{-.8pt}C \over 79} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = 79 ⋅ \sin(49\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 59{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 24 \text{,}\) \(\angle P = 42\degree\) en \(\angle Q = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}R}\) ofwel \(\sin(42\degree) = {24 \over P\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = {24 \over \sin(42\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 35{,}9 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 29 \text{,}\) \(Q\kern{-.8pt}R = 45\) en \(\angle P = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle R) = {P\kern{-.8pt}Q \over Q\kern{-.8pt}R}\) ofwel \(\sin(\angle R) = {29 \over 45} \text{.}\) 1p ○ Hieruit volgt \(\angle R = \sin^{-1}({29 \over 45}) \text{.}\) 1p ○ Dus \(\angle R ≈ 40{,}1\degree \text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 70 \text{,}\) \(\angle M = 57\degree\) en \(\angle K = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle M) = {K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\cos(57\degree) = {K\kern{-.8pt}M \over 70} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = 70 ⋅ \cos(57\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 38{,}1 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 49 \text{,}\) \(\angle C = 36\degree\) en \(\angle A = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle C) = {A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\cos(36\degree) = {49 \over B\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = {49 \over \cos(36\degree)} \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 60{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 49 \text{,}\) \(L\kern{-.8pt}M = 61\) 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 \(\cos(\angle M) = {49 \over 61} \text{.}\) 1p ○ Hieruit volgt \(\angle M = \cos^{-1}({49 \over 61}) \text{.}\) 1p ○ Dus \(\angle M ≈ 36{,}6\degree \text{.}\) 1p |