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 \(K\kern{-.8pt}M=58\text{,}\) \(\angle M=47\degree\) en \(\angle K=90\degree\text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle M)={K\kern{-.8pt}L \over K\kern{-.8pt}M}\) ofwel \(\tan(47\degree)={K\kern{-.8pt}L \over 58}\text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L=58⋅\tan(47\degree)\text{.}\) 1p ○ Dus \(K\kern{-.8pt}L≈62{,}2\text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=37\text{,}\) \(\angle L=54\degree\) en \(\angle M=90\degree\text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle L)={K\kern{-.8pt}M \over L\kern{-.8pt}M}\) ofwel \(\tan(54\degree)={37 \over L\kern{-.8pt}M}\text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M={37 \over \tan(54\degree)}\text{.}\) 1p ○ Dus \(L\kern{-.8pt}M≈26{,}9\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=60\text{,}\) \(P\kern{-.8pt}R=34\) en \(\angle R=90\degree\text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle Q)={P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\tan(\angle Q)={34 \over 60}\text{.}\) 1p ○ Hieruit volgt \(\angle Q=\tan^{-1}({34 \over 60})\text{.}\) 1p ○ Dus \(\angle Q≈29{,}5\degree\text{.}\) 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 \(K\kern{-.8pt}L=46\text{,}\) \(\angle L=50\degree\) en \(\angle M=90\degree\text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle L)={K\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\sin(50\degree)={K\kern{-.8pt}M \over 46}\text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M=46⋅\sin(50\degree)\text{.}\) 1p ○ Dus \(K\kern{-.8pt}M≈35{,}2\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=39\text{,}\) \(\angle C=42\degree\) en \(\angle A=90\degree\text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle C)={A\kern{-.8pt}B \over B\kern{-.8pt}C}\) ofwel \(\sin(42\degree)={39 \over B\kern{-.8pt}C}\text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C={39 \over \sin(42\degree)}\text{.}\) 1p ○ Dus \(B\kern{-.8pt}C≈58{,}3\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=38\text{,}\) \(L\kern{-.8pt}M=48\) en \(\angle K=90\degree\text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms 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 \(\sin(\angle M)={38 \over 48}\text{.}\) 1p ○ Hieruit volgt \(\angle M=\sin^{-1}({38 \over 48})\text{.}\) 1p ○ Dus \(\angle M≈52{,}3\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=50\text{,}\) \(\angle C=55\degree\) 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 \(\cos(55\degree)={A\kern{-.8pt}C \over 50}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C=50⋅\cos(55\degree)\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C≈28{,}7\text{.}\) 1p opgave 23p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=54\text{,}\) \(\angle A=54\degree\) en \(\angle B=90\degree\text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle A)={A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\cos(54\degree)={54 \over A\kern{-.8pt}C}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C={54 \over \cos(54\degree)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C≈91{,}9\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=43\text{,}\) \(A\kern{-.8pt}B=48\) en \(\angle C=90\degree\text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle B)={B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\cos(\angle B)={43 \over 48}\text{.}\) 1p ○ Hieruit volgt \(\angle B=\cos^{-1}({43 \over 48})\text{.}\) 1p ○ Dus \(\angle B≈26{,}4\degree\text{.}\) 1p |