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 \(B\kern{-.8pt}C=57\text{,}\) \(\angle B=56\degree\) en \(\angle C=90\degree\text{.}\) Tangens (1) 007m - Sinus, cosinus en tangens - basis - 0ms a Tangens in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\tan(\angle B)={A\kern{-.8pt}C \over B\kern{-.8pt}C}\) ofwel \(\tan(56\degree)={A\kern{-.8pt}C \over 57}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C=57⋅\tan(56\degree)\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C≈84{,}5\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=25\text{,}\) \(\angle B=44\degree\) 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 \(\tan(44\degree)={25 \over B\kern{-.8pt}C}\text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C={25 \over \tan(44\degree)}\text{.}\) 1p ○ Dus \(B\kern{-.8pt}C≈25{,}9\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=28\text{,}\) \(P\kern{-.8pt}R=32\) 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 \(\tan(\angle Q)={32 \over 28}\text{.}\) 1p ○ Hieruit volgt \(\angle Q=\tan^{-1}({32 \over 28})\text{.}\) 1p ○ Dus \(\angle Q≈48{,}8\degree\text{.}\) 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 \(L\kern{-.8pt}M=47\text{,}\) \(\angle M=31\degree\) en \(\angle K=90\degree\text{.}\) Sinus (1) 007g - Sinus, cosinus en tangens - basis - 0ms a Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle M)={K\kern{-.8pt}L \over L\kern{-.8pt}M}\) ofwel \(\sin(31\degree)={K\kern{-.8pt}L \over 47}\text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L=47⋅\sin(31\degree)\text{.}\) 1p ○ Dus \(K\kern{-.8pt}L≈24{,}2\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=30\text{,}\) \(\angle A=48\degree\) en \(\angle B=90\degree\text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\sin(\angle A)={B\kern{-.8pt}C \over A\kern{-.8pt}C}\) ofwel \(\sin(48\degree)={30 \over A\kern{-.8pt}C}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C={30 \over \sin(48\degree)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C≈40{,}4\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=24\text{,}\) \(B\kern{-.8pt}C=61\) 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 \(\sin(\angle C)={24 \over 61}\text{.}\) 1p ○ Hieruit volgt \(\angle C=\sin^{-1}({24 \over 61})\text{.}\) 1p ○ Dus \(\angle C≈23{,}2\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=68\text{,}\) \(\angle C=47\degree\) 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 \(\cos(47\degree)={A\kern{-.8pt}C \over 68}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}C=68⋅\cos(47\degree)\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C≈46{,}4\text{.}\) 1p opgave 23p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=51\text{,}\) \(\angle B=56\degree\) en \(\angle C=90\degree\text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle B)={B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\cos(56\degree)={51 \over A\kern{-.8pt}B}\text{.}\) 1p ○ Hieruit volgt \(A\kern{-.8pt}B={51 \over \cos(56\degree)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}B≈91{,}2\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=40\text{,}\) \(A\kern{-.8pt}B=69\) en \(\angle C=90\degree\text{.}\) Cosinus (3) 007l - Sinus, cosinus en tangens - basis - 1ms 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 \(\cos(\angle B)={40 \over 69}\text{.}\) 1p ○ Hieruit volgt \(\angle B=\cos^{-1}({40 \over 69})\text{.}\) 1p ○ Dus \(\angle B≈54{,}6\degree\text{.}\) 1p |