Getal & Ruimte (13e editie) - havo wiskunde B
'Sinus, cosinus en tangens'.
| 3 havo | 6.3 Berekeningen met de tangens |
opgave 13p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=49\text{,}\) \(\angle Q=50\degree\) 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 \(\tan(50\degree)={P\kern{-.8pt}R \over 49}\text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R=49⋅\tan(50\degree)\text{.}\) 1p ○ Dus \(P\kern{-.8pt}R≈58{,}4\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=51\text{,}\) \(\angle C=48\degree\) en \(\angle A=90\degree\text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle C)={A\kern{-.8pt}B \over A\kern{-.8pt}C}\) ofwel \(\tan(48\degree)={51 \over A\kern{-.8pt}C}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C={51 \over \tan(48\degree)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C≈45{,}9\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=42\text{,}\) \(Q\kern{-.8pt}R=28\) 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 \(\tan(\angle P)={28 \over 42}\text{.}\) 1p ○ Hieruit volgt \(\angle P=\tan^{-1}({28 \over 42})\text{.}\) 1p ○ Dus \(\angle P≈33{,}7\degree\text{.}\) 1p |
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| 3 havo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=69\text{,}\) \(\angle C=34\degree\) en \(\angle A=90\degree\text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle C)={A\kern{-.8pt}B \over B\kern{-.8pt}C}\) ofwel \(\sin(34\degree)={A\kern{-.8pt}B \over 69}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B=69⋅\sin(34\degree)\text{.}\) 1p ○ Dus \(A\kern{-.8pt}B≈38{,}6\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=29\text{,}\) \(\angle P=49\degree\) 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 \(\sin(49\degree)={29 \over P\kern{-.8pt}R}\text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R={29 \over \sin(49\degree)}\text{.}\) 1p ○ Dus \(P\kern{-.8pt}R≈38{,}4\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=39\text{,}\) \(P\kern{-.8pt}Q=65\) en \(\angle R=90\degree\text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle Q)={P\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\sin(\angle Q)={39 \over 65}\text{.}\) 1p ○ Hieruit volgt \(\angle Q=\sin^{-1}({39 \over 65})\text{.}\) 1p ○ Dus \(\angle Q≈36{,}9\degree\text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=49\text{,}\) \(\angle Q=53\degree\) en \(\angle R=90\degree\text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle Q)={Q\kern{-.8pt}R \over P\kern{-.8pt}Q}\) ofwel \(\cos(53\degree)={Q\kern{-.8pt}R \over 49}\text{.}\) 1p ○ Hieruit volgt \(Q\kern{-.8pt}R=49⋅\cos(53\degree)\text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R≈29{,}5\text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=50\text{,}\) \(\angle K=50\degree\) 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 \(\cos(50\degree)={50 \over K\kern{-.8pt}M}\text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M={50 \over \cos(50\degree)}\text{.}\) 1p ○ Dus \(K\kern{-.8pt}M≈77{,}8\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=60\text{,}\) \(Q\kern{-.8pt}R=77\) en \(\angle P=90\degree\text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 0ms b Cosinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\cos(\angle R)={P\kern{-.8pt}R \over Q\kern{-.8pt}R}\) ofwel \(\cos(\angle R)={60 \over 77}\text{.}\) 1p ○ Hieruit volgt \(\angle R=\cos^{-1}({60 \over 77})\text{.}\) 1p ○ Dus \(\angle R≈38{,}8\degree\text{.}\) 1p |