Getal & Ruimte (12e editie) - havo wiskunde B
'Sinus, cosinus en tangens'.
| 3 havo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 37 \text{,}\) \(\angle Q = 54\degree\) en \(\angle R = 90\degree \text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle Q) = {P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\tan(54\degree) = {P\kern{-.8pt}R \over 37} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = 37 ⋅ \tan(54\degree) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 50{,}9 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 37 \text{,}\) \(\angle B = 58\degree\) 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 \(\tan(58\degree) = {37 \over B\kern{-.8pt}C} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = {37 \over \tan(58\degree)} \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 23{,}1 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 42 \text{,}\) \(Q\kern{-.8pt}R = 25\) en \(\angle Q = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle P) = {Q\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\tan(\angle P) = {25 \over 42} \text{.}\) 1p ○ Hieruit volgt \(\angle P = \tan^{-1}({25 \over 42}) \text{.}\) 1p ○ Dus \(\angle P ≈ 30{,}8\degree \text{.}\) 1p |
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| 3 havo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 73 \text{,}\) \(\angle M = 40\degree\) en \(\angle K = 90\degree \text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle M) = {K\kern{-.8pt}L \over L\kern{-.8pt}M}\) ofwel \(\sin(40\degree) = {K\kern{-.8pt}L \over 73} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = 73 ⋅ \sin(40\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 46{,}9 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 51 \text{,}\) \(\angle R = 33\degree\) en \(\angle P = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle R) = {P\kern{-.8pt}Q \over Q\kern{-.8pt}R}\) ofwel \(\sin(33\degree) = {51 \over Q\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R = {51 \over \sin(33\degree)} \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R ≈ 93{,}6 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 34 \text{,}\) \(A\kern{-.8pt}C = 60\) en \(\angle B = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle A) = {B\kern{-.8pt}C \over A\kern{-.8pt}C}\) ofwel \(\sin(\angle A) = {34 \over 60} \text{.}\) 1p ○ Hieruit volgt \(\angle A = \sin^{-1}({34 \over 60}) \text{.}\) 1p ○ Dus \(\angle A ≈ 34{,}5\degree \text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 61 \text{,}\) \(\angle C = 37\degree\) 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 \(\cos(37\degree) = {A\kern{-.8pt}C \over 61} \text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C = 61 ⋅ \cos(37\degree) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C ≈ 48{,}7 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 52 \text{,}\) \(\angle K = 59\degree\) 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 \(\cos(59\degree) = {52 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {52 \over \cos(59\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 101{,}0 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 44 \text{,}\) \(L\kern{-.8pt}M = 49\) 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 \(\cos(\angle M) = {44 \over 49} \text{.}\) 1p ○ Hieruit volgt \(\angle M = \cos^{-1}({44 \over 49}) \text{.}\) 1p ○ Dus \(\angle M ≈ 26{,}1\degree \text{.}\) 1p |