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 \(K\kern{-.8pt}M = 28 \text{,}\) \(\angle M = 57\degree\) 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 \(\tan(57\degree) = {K\kern{-.8pt}L \over 28} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = 28 ⋅ \tan(57\degree) \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 43{,}1 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 44 \text{,}\) \(\angle R = 42\degree\) en \(\angle P = 90\degree \text{.}\) Tangens (2) 007n - Sinus, cosinus en tangens - basis - 0ms b Tangens in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\tan(\angle R) = {P\kern{-.8pt}Q \over P\kern{-.8pt}R}\) ofwel \(\tan(42\degree) = {44 \over P\kern{-.8pt}R} \text{.}\) 1p ○ Hieruit volgt \(P\kern{-.8pt}R = {44 \over \tan(42\degree)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}R ≈ 48{,}9 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 54 \text{,}\) \(K\kern{-.8pt}L = 40\) en \(\angle K = 90\degree \text{.}\) Tangens (3) 007o - Sinus, cosinus en tangens - basis - 0ms c Tangens in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\tan(\angle M) = {K\kern{-.8pt}L \over K\kern{-.8pt}M}\) ofwel \(\tan(\angle M) = {40 \over 54} \text{.}\) 1p ○ Hieruit volgt \(\angle M = \tan^{-1}({40 \over 54}) \text{.}\) 1p ○ Dus \(\angle M ≈ 36{,}5\degree \text{.}\) 1p |
|
| 3 havo | 6.4 De sinus en de cosinus |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 72 \text{,}\) \(\angle K = 40\degree\) 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 \(\sin(40\degree) = {L\kern{-.8pt}M \over 72} \text{.}\) 1p ○ Hieruit volgt \(L\kern{-.8pt}M = 72 ⋅ \sin(40\degree) \text{.}\) 1p ○ Dus \(L\kern{-.8pt}M ≈ 46{,}3 \text{.}\) 1p 3p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 21 \text{,}\) \(\angle K = 39\degree\) en \(\angle L = 90\degree \text{.}\) Sinus (2) 007h - Sinus, cosinus en tangens - basis - 0ms b Sinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\sin(\angle K) = {L\kern{-.8pt}M \over K\kern{-.8pt}M}\) ofwel \(\sin(39\degree) = {21 \over K\kern{-.8pt}M} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}M = {21 \over \sin(39\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}M ≈ 33{,}4 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 43 \text{,}\) \(Q\kern{-.8pt}R = 52\) en \(\angle P = 90\degree \text{.}\) Sinus (3) 007i - Sinus, cosinus en tangens - basis - 0ms c Sinus in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(\sin(\angle R) = {P\kern{-.8pt}Q \over Q\kern{-.8pt}R}\) ofwel \(\sin(\angle R) = {43 \over 52} \text{.}\) 1p ○ Hieruit volgt \(\angle R = \sin^{-1}({43 \over 52}) \text{.}\) 1p ○ Dus \(\angle R ≈ 55{,}8\degree \text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 75 \text{,}\) \(\angle B = 31\degree\) en \(\angle C = 90\degree \text{.}\) Cosinus (1) 007j - Sinus, cosinus en tangens - basis - 0ms d Cosinus in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(\cos(\angle B) = {B\kern{-.8pt}C \over A\kern{-.8pt}B}\) ofwel \(\cos(31\degree) = {B\kern{-.8pt}C \over 75} \text{.}\) 1p ○ Hieruit volgt \(B\kern{-.8pt}C = 75 ⋅ \cos(31\degree) \text{.}\) 1p ○ Dus \(B\kern{-.8pt}C ≈ 64{,}3 \text{.}\) 1p opgave 23p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 52 \text{,}\) \(\angle L = 48\degree\) en \(\angle M = 90\degree \text{.}\) Cosinus (2) 007k - Sinus, cosinus en tangens - basis - 0ms a Cosinus in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(\cos(\angle L) = {L\kern{-.8pt}M \over K\kern{-.8pt}L}\) ofwel \(\cos(48\degree) = {52 \over K\kern{-.8pt}L} \text{.}\) 1p ○ Hieruit volgt \(K\kern{-.8pt}L = {52 \over \cos(48\degree)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L ≈ 77{,}7 \text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 39 \text{,}\) \(A\kern{-.8pt}C = 52\) 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 \(\cos(\angle A) = {39 \over 52} \text{.}\) 1p ○ Hieruit volgt \(\angle A = \cos^{-1}({39 \over 52}) \text{.}\) 1p ○ Dus \(\angle A ≈ 41{,}4\degree \text{.}\) 1p |