Getal & Ruimte (13e editie) - havo wiskunde B
'Sinus- en cosinusregel'.
| havo wiskunde B | 3.2 De sinusregel |
opgave 13p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M=14\text{,}\) \(\angle L=34\degree\) en \(\angle M=84\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 1ms a De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}M \over \sin(\angle L)}={K\kern{-.8pt}L \over \sin(\angle M)}={L\kern{-.8pt}M \over \sin(\angle K)}\text{.}\) 1p ○ Dus \(K\kern{-.8pt}L={K\kern{-.8pt}M⋅\sin(\angle M) \over \sin(\angle L)}={14⋅\sin(84\degree) \over \sin(34\degree)}\text{.}\) 1p ○ \(K\kern{-.8pt}L≈24{,}9\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=13\text{,}\) \(\angle B=42\degree\) en \(\angle C=97\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}C \over \sin(\angle B)}={A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}B={A\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle B)}={13⋅\sin(97\degree) \over \sin(42\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}B≈19{,}3\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=9\text{,}\) \(L\kern{-.8pt}M=13\) en \(\angle M=25\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 9ms c De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}L \over \sin(\angle M)}={L\kern{-.8pt}M \over \sin(\angle K)}={K\kern{-.8pt}M \over \sin(\angle L)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle K)={L\kern{-.8pt}M⋅\sin(\angle M) \over K\kern{-.8pt}L}={13⋅\sin(25\degree) \over 9}=0{,}610...\text{.}\) 1p ○ Dit geeft \(\angle K≈37{,}6\degree\) of \(\angle K≈142{,}4\degree\text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=16\text{,}\) \(B\kern{-.8pt}C=20\) en \(\angle C=52\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle A)={B\kern{-.8pt}C⋅\sin(\angle C) \over A\kern{-.8pt}B}={20⋅\sin(52\degree) \over 16}=0{,}985...\text{.}\) 1p ○ Dit geeft \(\angle A≈80{,}1\degree\) of \(\angle A≈99{,}9\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=28\text{,}\) \(\angle Q=58\degree\) en \(\angle P=46\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 1ms a Uit \(\angle Q+\angle R+\angle P=180\degree\) volgt \(\angle R=180\degree-\angle Q-\angle P=180\degree-58\degree-46\degree=76\degree\text{.}\) 1p ○ 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 \(P\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle Q) \over \sin(\angle R)}={28⋅\sin(58\degree) \over \sin(76\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}R≈24{,}5\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B=36\text{,}\) \(\angle B=38\degree\) en \(\angle A=44\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle B+\angle C+\angle A=180\degree\) volgt \(\angle C=180\degree-\angle B-\angle A=180\degree-38\degree-44\degree=98\degree\text{.}\) 1p ○ De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}C \over \sin(\angle B)}={A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle B) \over \sin(\angle C)}={36⋅\sin(38\degree) \over \sin(98\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}C≈22{,}4\text{.}\) 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 \(A\kern{-.8pt}B=22\text{,}\) \(B\kern{-.8pt}C=28\) en \(\angle B=82\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 1ms a De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(A\kern{-.8pt}C^2=A\kern{-.8pt}B^2+B\kern{-.8pt}C^2-2⋅A\kern{-.8pt}B⋅B\kern{-.8pt}C⋅\cos(\angle B)\text{.}\) 1p ○ Dus \(A\kern{-.8pt}C^2=22^2+28^2-2⋅22⋅28⋅\cos(82\degree)=1096{,}538...\text{.}\) 1p ○ \(A\kern{-.8pt}C=\sqrt{1096{,}538...}≈33{,}1\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=21\text{,}\) \(Q\kern{-.8pt}R=40\) en \(\angle Q=92\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms b De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}R^2=P\kern{-.8pt}Q^2+Q\kern{-.8pt}R^2-2⋅P\kern{-.8pt}Q⋅Q\kern{-.8pt}R⋅\cos(\angle Q)\text{.}\) 1p ○ Dus \(P\kern{-.8pt}R^2=21^2+40^2-2⋅21⋅40⋅\cos(92\degree)=2099{,}631...\text{.}\) 1p ○ \(P\kern{-.8pt}R=\sqrt{2099{,}631...}≈45{,}8\text{.}\) 1p 4p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=24\text{,}\) \(L\kern{-.8pt}M=23\) en \(K\kern{-.8pt}M=31\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 8ms 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 \(31^2=24^2+23^2-2⋅24⋅23⋅\cos(\angle L)\) 1p ○ Balansmethode geeft \(\cos(\angle L)={961-1\,105 \over -1\,104}=0{,}130...\) 1p ○ Hieruit volgt \(\angle L=\cos^{-1}(0{,}130...)≈82{,}5\degree\text{.}\) 1p 4p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=13\text{,}\) \(L\kern{-.8pt}M=15\) en \(K\kern{-.8pt}M=20\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms d 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 \(20^2=13^2+15^2-2⋅13⋅15⋅\cos(\angle L)\) 1p ○ Balansmethode geeft \(\cos(\angle L)={400-394 \over -390}=-0{,}015...\) 1p ○ Hieruit volgt \(\angle L=\cos^{-1}(-0{,}015...)≈90{,}9\degree\text{.}\) 1p |