Getal & Ruimte (12e editie) - havo wiskunde B
'Sinus, cosinus en tangens'.
| 3 havo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=37\text{,}\) \(\angle C=44\degree\) 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 \(\tan(44\degree)={A\kern{-.8pt}B \over 37}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B=37⋅\tan(44\degree)\text{.}\) 1p ○ Dus \(A\kern{-.8pt}B≈35{,}7\text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=27\text{,}\) \(\angle K=59\degree\) en \(\angle L=90\degree\text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle K)={L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\tan(59\degree)={27 \over K\kern{-.8pt}L}\text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L={27 \over \tan(59\degree)}\text{.}\) 1p ○ Dus \(K\kern{-.8pt}L≈16{,}2\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=30\text{,}\) \(Q\kern{-.8pt}R=40\) 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 \(\tan(\angle P)={40 \over 30}\text{.}\) 1p ○ Hieruit volgt \(\angle P=\tan^{-1}({40 \over 30})\text{.}\) 1p ○ Dus \(\angle P≈53{,}1\degree\text{.}\) 1p |
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| 3 havo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=49\text{,}\) \(\angle P=51\degree\) en \(\angle Q=90\degree\text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle P)={Q\kern{-.8pt}R \over P\kern{-.8pt}R}\) ofwel \(\sin(51\degree)={Q\kern{-.8pt}R \over 49}\text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R=49⋅\sin(51\degree)\text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R≈38{,}1\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=31\text{,}\) \(\angle B=40\degree\) en \(\angle C=90\degree\text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle B)={A\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\sin(40\degree)={31 \over A\kern{-.8pt}B}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B={31 \over \sin(40\degree)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}B≈48{,}2\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=41\text{,}\) \(P\kern{-.8pt}R=53\) en \(\angle Q=90\degree\text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle P)={Q\kern{-.8pt}R \over P\kern{-.8pt}R}\) ofwel \(\sin(\angle P)={41 \over 53}\text{.}\) 1p ○ Hieruit volgt \(\angle P=\sin^{-1}({41 \over 53})\text{.}\) 1p ○ Dus \(\angle P≈50{,}7\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=40\text{,}\) \(\angle A=37\degree\) en \(\angle B=90\degree\text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle A)={A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\cos(37\degree)={A\kern{-.8pt}B \over 40}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B=40⋅\cos(37\degree)\text{.}\) 1p ○ Dus \(A\kern{-.8pt}B≈31{,}9\text{.}\) 1p opgave 23p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=40\text{,}\) \(\angle R=40\degree\) en \(\angle P=90\degree\text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle R)={P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\cos(40\degree)={40 \over Q\kern{-.8pt}R}\text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R={40 \over \cos(40\degree)}\text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R≈52{,}2\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=26\text{,}\) \(P\kern{-.8pt}R=52\) en \(\angle Q=90\degree\text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle P)={P\kern{-.8pt}Q \over P\kern{-.8pt}R}\) ofwel \(\cos(\angle P)={26 \over 52}\text{.}\) 1p ○ Hieruit volgt \(\angle P=\cos^{-1}({26 \over 52})\text{.}\) 1p ○ Dus \(\angle P=60{,}0\degree\text{.}\) 1p |