Getal & Ruimte (13e editie) - havo wiskunde B
'Sinus- en cosinusregel'.
| havo wiskunde B | 3.2 De sinusregel |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=25\text{,}\) \(\angle A=54\degree\) en \(\angle B=89\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms a De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}={A\kern{-.8pt}B \over \sin(\angle C)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C={B\kern{-.8pt}C⋅\sin(\angle B) \over \sin(\angle A)}={25⋅\sin(89\degree) \over \sin(54\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}C≈30{,}9\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=43\text{,}\) \(\angle Q=50\degree\) en \(\angle R=99\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}={Q\kern{-.8pt}R \over \sin(\angle P)}\text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q={P\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle Q)}={43⋅\sin(99\degree) \over \sin(50\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}Q≈55{,}4\text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=21\text{,}\) \(P\kern{-.8pt}Q=29\) en \(\angle Q=43\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 5ms c De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}={Q\kern{-.8pt}R \over \sin(\angle P)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle R)={P\kern{-.8pt}Q⋅\sin(\angle Q) \over P\kern{-.8pt}R}={29⋅\sin(43\degree) \over 21}=0{,}941...\text{.}\) 1p ○ Dit geeft \(\angle R≈70{,}4\degree\) of \(\angle R≈109{,}6\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=21\text{,}\) \(A\kern{-.8pt}C=30\) en \(\angle A=41\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}={A\kern{-.8pt}B \over \sin(\angle C)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle B)={A\kern{-.8pt}C⋅\sin(\angle A) \over B\kern{-.8pt}C}={30⋅\sin(41\degree) \over 21}=0{,}937...\text{.}\) 1p ○ Dit geeft \(\angle B≈69{,}6\degree\) of \(\angle B≈110{,}4\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=22\text{,}\) \(\angle P=39\degree\) en \(\angle R=54\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle P+\angle Q+\angle R=180\degree\) volgt \(\angle Q=180\degree-\angle P-\angle R=180\degree-39\degree-54\degree=87\degree\text{.}\) 1p ○ De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({Q\kern{-.8pt}R \over \sin(\angle P)}={P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}\text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R={P\kern{-.8pt}R⋅\sin(\angle P) \over \sin(\angle Q)}={22⋅\sin(39\degree) \over \sin(87\degree)}\text{.}\) 1p ○ \(Q\kern{-.8pt}R≈13{,}9\text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=44\text{,}\) \(\angle P=40\degree\) en \(\angle R=44\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle P+\angle Q+\angle R=180\degree\) volgt \(\angle Q=180\degree-\angle P-\angle R=180\degree-40\degree-44\degree=96\degree\text{.}\) 1p ○ De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({Q\kern{-.8pt}R \over \sin(\angle P)}={P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}\text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R={P\kern{-.8pt}R⋅\sin(\angle P) \over \sin(\angle Q)}={44⋅\sin(40\degree) \over \sin(96\degree)}\text{.}\) 1p ○ \(Q\kern{-.8pt}R≈28{,}4\text{.}\) 1p |
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| havo wiskunde B | 3.3 De cosinusregel |
opgave 13p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=29\text{,}\) \(A\kern{-.8pt}B=29\) en \(\angle A=61\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 0ms a De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(B\kern{-.8pt}C^2=A\kern{-.8pt}C^2+A\kern{-.8pt}B^2-2⋅A\kern{-.8pt}C⋅A\kern{-.8pt}B⋅\cos(\angle A)\text{.}\) 1p ○ Dus \(B\kern{-.8pt}C^2=29^2+29^2-2⋅29⋅29⋅\cos(61\degree)=866{,}550...\text{.}\) 1p ○ \(B\kern{-.8pt}C=\sqrt{866{,}550...}≈29{,}4\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=17\text{,}\) \(P\kern{-.8pt}R=24\) en \(\angle R=117\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms b De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}Q^2=Q\kern{-.8pt}R^2+P\kern{-.8pt}R^2-2⋅Q\kern{-.8pt}R⋅P\kern{-.8pt}R⋅\cos(\angle R)\text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q^2=17^2+24^2-2⋅17⋅24⋅\cos(117\degree)=1235{,}456...\text{.}\) 1p ○ \(P\kern{-.8pt}Q=\sqrt{1235{,}456...}≈35{,}1\text{.}\) 1p 4p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=21\text{,}\) \(L\kern{-.8pt}M=17\) en \(K\kern{-.8pt}M=24\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 5ms c De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(K\kern{-.8pt}M^2=K\kern{-.8pt}L^2+L\kern{-.8pt}M^2-2⋅K\kern{-.8pt}L⋅L\kern{-.8pt}M⋅\cos(\angle L)\text{.}\) 1p ○ Invullen geeft \(24^2=21^2+17^2-2⋅21⋅17⋅\cos(\angle L)\) 1p ○ Balansmethode geeft \(\cos(\angle L)={576-730 \over -714}=0{,}215...\) 1p ○ Hieruit volgt \(\angle L=\cos^{-1}(0{,}215...)≈77{,}5\degree\text{.}\) 1p 4p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=12\text{,}\) \(P\kern{-.8pt}Q=15\) en \(Q\kern{-.8pt}R=22\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms d De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(Q\kern{-.8pt}R^2=P\kern{-.8pt}R^2+P\kern{-.8pt}Q^2-2⋅P\kern{-.8pt}R⋅P\kern{-.8pt}Q⋅\cos(\angle P)\text{.}\) 1p ○ Invullen geeft \(22^2=12^2+15^2-2⋅12⋅15⋅\cos(\angle P)\) 1p ○ Balansmethode geeft \(\cos(\angle P)={484-369 \over -360}=-0{,}319...\) 1p ○ Hieruit volgt \(\angle P=\cos^{-1}(-0{,}319...)≈108{,}6\degree\text{.}\) 1p |