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 \(A\kern{-.8pt}B=21\text{,}\) \(\angle C=54\degree\) en \(\angle A=73\degree\text{.}\) SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms a 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 ○ Dus \(B\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle A) \over \sin(\angle C)}={21⋅\sin(73\degree) \over \sin(54\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈24{,}8\text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R=36\text{,}\) \(\angle Q=38\degree\) en \(\angle R=115\degree\text{.}\) 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 \(P\kern{-.8pt}Q={P\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle Q)}={36⋅\sin(115\degree) \over \sin(38\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}Q≈53{,}0\text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L=17\text{,}\) \(L\kern{-.8pt}M=25\) en \(\angle M=36\degree\text{.}\) SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 8ms 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 \(\sin(\angle K)={L\kern{-.8pt}M⋅\sin(\angle M) \over K\kern{-.8pt}L}={25⋅\sin(36\degree) \over 17}=0{,}864...\text{.}\) 1p ○ Dit geeft \(\angle K≈59{,}8\degree\) of \(\angle K≈120{,}2\degree\text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=17\text{,}\) \(P\kern{-.8pt}R=26\) en \(\angle P=30\degree\text{.}\) SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d 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 ○ Daaruit volgt \(\sin(\angle Q)={P\kern{-.8pt}R⋅\sin(\angle P) \over Q\kern{-.8pt}R}={26⋅\sin(30\degree) \over 17}=0{,}764...\text{.}\) 1p ○ Dit geeft \(\angle Q≈49{,}9\degree\) of \(\angle Q≈130{,}1\degree\text{.}\) 1p opgave 24p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=12\text{,}\) \(\angle A=53\degree\) en \(\angle C=59\degree\text{.}\) SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle A+\angle B+\angle C=180\degree\) volgt \(\angle B=180\degree-\angle A-\angle C=180\degree-53\degree-59\degree=68\degree\text{.}\) 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 \(B\kern{-.8pt}C={A\kern{-.8pt}C⋅\sin(\angle A) \over \sin(\angle B)}={12⋅\sin(53\degree) \over \sin(68\degree)}\text{.}\) 1p ○ \(B\kern{-.8pt}C≈10{,}3\text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=41\text{,}\) \(\angle R=49\degree\) en \(\angle Q=28\degree\text{.}\) SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle R+\angle P+\angle Q=180\degree\) volgt \(\angle P=180\degree-\angle R-\angle Q=180\degree-49\degree-28\degree=103\degree\text{.}\) 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 \(P\kern{-.8pt}Q={Q\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle P)}={41⋅\sin(49\degree) \over \sin(103\degree)}\text{.}\) 1p ○ \(P\kern{-.8pt}Q≈31{,}8\text{.}\) 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 \(A\kern{-.8pt}C=17\text{,}\) \(A\kern{-.8pt}B=15\) en \(\angle A=67\degree\text{.}\) CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 1ms 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 \(B\kern{-.8pt}C^2=17^2+15^2-2⋅17⋅15⋅\cos(67\degree)=314{,}727...\text{.}\) 1p ○ \(B\kern{-.8pt}C=\sqrt{314{,}727...}≈17{,}7\text{.}\) 1p 3p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=48\text{,}\) \(A\kern{-.8pt}B=29\) en \(\angle A=107\degree\text{.}\) CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms b 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 \(B\kern{-.8pt}C^2=48^2+29^2-2⋅48⋅29⋅\cos(107\degree)=3958{,}962...\text{.}\) 1p ○ \(B\kern{-.8pt}C=\sqrt{3958{,}962...}≈62{,}9\text{.}\) 1p 4p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C=27\text{,}\) \(A\kern{-.8pt}B=27\) en \(B\kern{-.8pt}C=33\text{.}\) CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 5ms 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 \(33^2=27^2+27^2-2⋅27⋅27⋅\cos(\angle A)\) 1p ○ Balansmethode geeft \(\cos(\angle A)={1\,089-1\,458 \over -1\,458}=0{,}253...\) 1p ○ Hieruit volgt \(\angle A=\cos^{-1}(0{,}253...)≈75{,}3\degree\text{.}\) 1p 4p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R=24\text{,}\) \(P\kern{-.8pt}R=23\) en \(P\kern{-.8pt}Q=37\text{.}\) CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms d 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 ○ Invullen geeft \(37^2=24^2+23^2-2⋅24⋅23⋅\cos(\angle R)\) 1p ○ Balansmethode geeft \(\cos(\angle R)={1\,369-1\,105 \over -1\,104}=-0{,}239...\) 1p ○ Hieruit volgt \(\angle R=\cos^{-1}(-0{,}239...)≈103{,}8\degree\text{.}\) 1p |