Getal & Ruimte (13e editie) - havo wiskunde B
'Sinus, cosinus en tangens'.
| 3 havo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=40\text{,}\) \(\angle K=37\degree\) en \(\angle L=90\degree\text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis a Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle K)={L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\tan(37\degree)={L\kern{-.8pt}M \over 40}\text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M=40⋅\tan(37\degree)\text{.}\) 1p ○ Dus \(L\kern{-.8pt}M≈30{,}1\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=23\text{,}\) \(\angle C=55\degree\) en \(\angle A=90\degree\text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis 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 \(\tan(55\degree)={23 \over A\kern{-.8pt}C}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C={23 \over \tan(55\degree)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C≈16{,}1\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=39\text{,}\) \(K\kern{-.8pt}M=47\) en \(\angle M=90\degree\text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis 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 \(\tan(\angle L)={47 \over 39}\text{.}\) 1p ○ Hieruit volgt \(\angle L=\tan^{-1}({47 \over 39})\text{.}\) 1p ○ Dus \(\angle L≈50{,}3\degree\text{.}\) 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 \(K\kern{-.8pt}L=41\text{,}\) \(\angle L=33\degree\) en \(\angle M=90\degree\text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis a Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle L)={K\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\sin(33\degree)={K\kern{-.8pt}M \over 41}\text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M=41⋅\sin(33\degree)\text{.}\) 1p ○ Dus \(K\kern{-.8pt}M≈22{,}3\text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=40\text{,}\) \(\angle M=33\degree\) en \(\angle K=90\degree\text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis b Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle M)={K\kern{-.8pt}L \over L\kern{-.8pt}M}\) ofwel \(\sin(33\degree)={40 \over L\kern{-.8pt}M}\text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M={40 \over \sin(33\degree)}\text{.}\) 1p ○ Dus \(L\kern{-.8pt}M≈73{,}4\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=21\text{,}\) \(L\kern{-.8pt}M=60\) en \(\angle K=90\degree\text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis c Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle M)={K\kern{-.8pt}L \over L\kern{-.8pt}M}\) ofwel \(\sin(\angle M)={21 \over 60}\text{.}\) 1p ○ Hieruit volgt \(\angle M=\sin^{-1}({21 \over 60})\text{.}\) 1p ○ Dus \(\angle M≈20{,}5\degree\text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=71\text{,}\) \(\angle P=40\degree\) en \(\angle Q=90\degree\text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis d Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle P)={P\kern{-.8pt}Q \over P\kern{-.8pt}R}\) ofwel \(\cos(40\degree)={P\kern{-.8pt}Q \over 71}\text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}Q=71⋅\cos(40\degree)\text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q≈54{,}4\text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=31\text{,}\) \(\angle L=59\degree\) en \(\angle M=90\degree\text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis a Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle L)={L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\cos(59\degree)={31 \over K\kern{-.8pt}L}\text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L={31 \over \cos(59\degree)}\text{.}\) 1p ○ Dus \(K\kern{-.8pt}L≈60{,}2\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=51\text{,}\) \(A\kern{-.8pt}C=64\) en \(\angle B=90\degree\text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis 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 \(\cos(\angle A)={51 \over 64}\text{.}\) 1p ○ Hieruit volgt \(\angle A=\cos^{-1}({51 \over 64})\text{.}\) 1p ○ Dus \(\angle A≈37{,}2\degree\text{.}\) 1p |