Getal & Ruimte (12e editie) - havo wiskunde B
'Sinus, cosinus en tangens'.
| 3 havo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=20\text{,}\) \(\angle L=48\degree\) en \(\angle M=90\degree\text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle L)={K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\tan(48\degree)={K\kern{-.8pt}M \over 20}\text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M=20⋅\tan(48\degree)\text{.}\) 1p ○ Dus \(K\kern{-.8pt}M≈22{,}2\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=49\text{,}\) \(\angle B=44\degree\) 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 \(\tan(44\degree)={49 \over B\kern{-.8pt}C}\text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C={49 \over \tan(44\degree)}\text{.}\) 1p ○ Dus \(B\kern{-.8pt}C≈50{,}7\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=50\text{,}\) \(A\kern{-.8pt}C=22\) en \(\angle C=90\degree\text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle B)={A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\tan(\angle B)={22 \over 50}\text{.}\) 1p ○ Hieruit volgt \(\angle B=\tan^{-1}({22 \over 50})\text{.}\) 1p ○ Dus \(\angle B≈23{,}7\degree\text{.}\) 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 \(K\kern{-.8pt}M=50\text{,}\) \(\angle K=36\degree\) en \(\angle L=90\degree\text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle K)={L\kern{-.8pt}M \over K\kern{-.8pt}M}\) ofwel \(\sin(36\degree)={L\kern{-.8pt}M \over 50}\text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M=50⋅\sin(36\degree)\text{.}\) 1p ○ Dus \(L\kern{-.8pt}M≈29{,}4\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=50\text{,}\) \(\angle P=59\degree\) 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 \(\sin(59\degree)={50 \over P\kern{-.8pt}R}\text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R={50 \over \sin(59\degree)}\text{.}\) 1p ○ Dus \(P\kern{-.8pt}R≈58{,}3\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=33\text{,}\) \(B\kern{-.8pt}C=43\) 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 \(\sin(\angle C)={33 \over 43}\text{.}\) 1p ○ Hieruit volgt \(\angle C=\sin^{-1}({33 \over 43})\text{.}\) 1p ○ Dus \(\angle C≈50{,}1\degree\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=52\text{,}\) \(\angle M=41\degree\) 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 \(\cos(41\degree)={K\kern{-.8pt}M \over 52}\text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M=52⋅\cos(41\degree)\text{.}\) 1p ○ Dus \(K\kern{-.8pt}M≈39{,}2\text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=45\text{,}\) \(\angle M=46\degree\) 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 \(\cos(46\degree)={45 \over L\kern{-.8pt}M}\text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M={45 \over \cos(46\degree)}\text{.}\) 1p ○ Dus \(L\kern{-.8pt}M≈64{,}8\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=52\text{,}\) \(A\kern{-.8pt}C=59\) 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 \(\cos(\angle A)={52 \over 59}\text{.}\) 1p ○ Hieruit volgt \(\angle A=\cos^{-1}({52 \over 59})\text{.}\) 1p ○ Dus \(\angle A≈28{,}2\degree\text{.}\) 1p |