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 \(K\kern{-.8pt}L=27\text{,}\) \(\angle M=42\degree\) en \(\angle K=89\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis a 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 ○ Dus \(L\kern{-.8pt}M={K\kern{-.8pt}L⋅\sin(\angle K) \over \sin(\angle M)}={27⋅\sin(89\degree) \over \sin(42\degree)}\text{.}\) 1p ○ \(L\kern{-.8pt}M≈40{,}3\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=35\text{,}\) \(\angle A=44\degree\) en \(\angle B=103\degree\text{.}\) SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis b 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 \(A\kern{-.8pt}C={B\kern{-.8pt}C⋅\sin(\angle B) \over \sin(\angle A)}={35⋅\sin(103\degree) \over \sin(44\degree)}\text{.}\) 1p ○ \(A\kern{-.8pt}C≈49{,}1\text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=13\text{,}\) \(A\kern{-.8pt}C=22\) en \(\angle A=30\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis c 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 \(\sin(\angle B)={A\kern{-.8pt}C⋅\sin(\angle A) \over B\kern{-.8pt}C}={22⋅\sin(30\degree) \over 13}=0{,}846...\text{.}\) 1p ○ Dit geeft \(\angle B≈57{,}8\degree\) of \(\angle B≈122{,}2\degree\text{.}\) 1p 3p d Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M=6\text{,}\) \(K\kern{-.8pt}M=10\) en \(\angle K=29\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis d De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({L\kern{-.8pt}M \over \sin(\angle K)}={K\kern{-.8pt}M \over \sin(\angle L)}={K\kern{-.8pt}L \over \sin(\angle M)}\text{.}\) 1p ○ Daaruit volgt \(\sin(\angle L)={K\kern{-.8pt}M⋅\sin(\angle K) \over L\kern{-.8pt}M}={10⋅\sin(29\degree) \over 6}=0{,}808...\text{.}\) 1p ○ Dit geeft \(\angle L≈53{,}9\degree\) of \(\angle L≈126{,}1\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=40\text{,}\) \(\angle R=38\degree\) en \(\angle Q=56\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis a Uit \(\angle R+\angle P+\angle Q=180\degree\) volgt \(\angle P=180\degree-\angle R-\angle Q=180\degree-38\degree-56\degree=86\degree\text{.}\) 1p ○ De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}Q \over \sin(\angle R)}={Q\kern{-.8pt}R \over \sin(\angle P)}={P\kern{-.8pt}R \over \sin(\angle Q)}\text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q={Q\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle P)}={40⋅\sin(38\degree) \over \sin(86\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}Q≈24{,}7\text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=54\text{,}\) \(\angle A=42\degree\) en \(\angle C=29\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis b Uit \(\angle A+\angle B+\angle C=180\degree\) volgt \(\angle B=180\degree-\angle A-\angle C=180\degree-42\degree-29\degree=109\degree\text{.}\) 1p ○ 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 \(B\kern{-.8pt}C={A\kern{-.8pt}C⋅\sin(\angle A) \over \sin(\angle B)}={54⋅\sin(42\degree) \over \sin(109\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈38{,}2\text{.}\) 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 \(A\kern{-.8pt}C=16\text{,}\) \(A\kern{-.8pt}B=25\) en \(\angle A=85\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis 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 \(B\kern{-.8pt}C^2=16^2+25^2-2⋅16⋅25⋅\cos(85\degree)=811{,}275...\text{.}\) 1p ○ \(B\kern{-.8pt}C=\sqrt{811{,}275...}≈28{,}5\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C=30\text{,}\) \(A\kern{-.8pt}C=35\) en \(\angle C=116\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis b De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(A\kern{-.8pt}B^2=B\kern{-.8pt}C^2+A\kern{-.8pt}C^2-2⋅B\kern{-.8pt}C⋅A\kern{-.8pt}C⋅\cos(\angle C)\text{.}\) 1p ○ Dus \(A\kern{-.8pt}B^2=30^2+35^2-2⋅30⋅35⋅\cos(116\degree)=3045{,}579...\text{.}\) 1p ○ \(A\kern{-.8pt}B=\sqrt{3045{,}579...}≈55{,}2\text{.}\) 1p 4p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=28\text{,}\) \(A\kern{-.8pt}B=38\) en \(B\kern{-.8pt}C=45\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis c 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 ○ Invullen geeft \(45^2=28^2+38^2-2⋅28⋅38⋅\cos(\angle A)\) 1p ○ Balansmethode geeft \(\cos(\angle A)={2\,025-2\,228 \over -2\,128}=0{,}095...\) 1p ○ Hieruit volgt \(\angle A=\cos^{-1}(0{,}095...)≈84{,}5\degree\text{.}\) 1p 4p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q=26\text{,}\) \(Q\kern{-.8pt}R=16\) en \(P\kern{-.8pt}R=35\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis d 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 ○ Invullen geeft \(35^2=26^2+16^2-2⋅26⋅16⋅\cos(\angle Q)\) 1p ○ Balansmethode geeft \(\cos(\angle Q)={1\,225-932 \over -832}=-0{,}352...\) 1p ○ Hieruit volgt \(\angle Q=\cos^{-1}(-0{,}352...)≈110{,}6\degree\text{.}\) 1p |