Sinus- en cosinusregel

De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(L\kern{-.8pt}M^2=K\kern{-.8pt}M^2+K\kern{-.8pt}L^2-2⋅K\kern{-.8pt}M⋅K\kern{-.8pt}L⋅\cos(\angle K)\text{.}\)

Invullen geeft \(38^2=35^2+23^2-2⋅35⋅23⋅\cos(\angle K)\)
dus \(1\,444=1\,754-1\,610⋅\cos(\angle K)\text{.}\)

Balansmethode geeft \(\cos(\angle K)={1\,444-1\,754 \over -1\,610}=0{,}192...\)

Hieruit volgt \(\angle K=\cos^{-1}(0{,}192...)≈78{,}9\degree\text{.}\)

De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(K\kern{-.8pt}L^2=L\kern{-.8pt}M^2+K\kern{-.8pt}M^2-2⋅L\kern{-.8pt}M⋅K\kern{-.8pt}M⋅\cos(\angle M)\text{.}\)

Invullen geeft \(50^2=42^2+27^2-2⋅42⋅27⋅\cos(\angle M)\)
dus \(2\,500=2\,493-2\,268⋅\cos(\angle M)\text{.}\)

Balansmethode geeft \(\cos(\angle M)={2\,500-2\,493 \over -2\,268}=-0{,}003...\)

Hieruit volgt \(\angle M=\cos^{-1}(-0{,}003...)≈90{,}2\degree\text{.}\)

De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(K\kern{-.8pt}L^2=L\kern{-.8pt}M^2+K\kern{-.8pt}M^2-2⋅L\kern{-.8pt}M⋅K\kern{-.8pt}M⋅\cos(\angle M)\text{.}\)

Dus \(K\kern{-.8pt}L^2=11^2+11^2-2⋅11⋅11⋅\cos(68\degree)=151{,}345...\text{.}\)

\(K\kern{-.8pt}L=\sqrt{151{,}345...}≈12{,}3\text{.}\)

De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(A\kern{-.8pt}C^2=A\kern{-.8pt}B^2+B\kern{-.8pt}C^2-2⋅A\kern{-.8pt}B⋅B\kern{-.8pt}C⋅\cos(\angle B)\text{.}\)

Dus \(A\kern{-.8pt}C^2=26^2+49^2-2⋅26⋅49⋅\cos(94\degree)=3254{,}739...\text{.}\)

\(A\kern{-.8pt}C=\sqrt{3254{,}739...}≈57{,}1\text{.}\)

De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}L \over \sin(\angle M)}={L\kern{-.8pt}M \over \sin(\angle K)}={K\kern{-.8pt}M \over \sin(\angle L)}\text{.}\)

Daaruit volgt \(\sin(\angle K)={L\kern{-.8pt}M⋅\sin(\angle M) \over K\kern{-.8pt}L}={27⋅\sin(37\degree) \over 18}=0{,}902...\text{.}\)

Dit geeft \(\angle K≈64{,}5\degree\) of \(\angle K≈115{,}5\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle K\) een scherpe hoek is, dus \(\angle K≈64{,}5\degree\text{.}\)

De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}M \over \sin(\angle L)}={K\kern{-.8pt}L \over \sin(\angle M)}={L\kern{-.8pt}M \over \sin(\angle K)}\text{.}\)

Daaruit volgt \(\sin(\angle M)={K\kern{-.8pt}L⋅\sin(\angle L) \over K\kern{-.8pt}M}={13⋅\sin(46\degree) \over 10}=0{,}935...\text{.}\)

Dit geeft \(\angle M≈69{,}3\degree\) of \(\angle M≈110{,}7\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle M\) een stompe hoek is, dus \(\angle M≈110{,}7\degree\text{.}\)

De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}Q \over \sin(\angle R)}={Q\kern{-.8pt}R \over \sin(\angle P)}={P\kern{-.8pt}R \over \sin(\angle Q)}\text{.}\)

Dus \(Q\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle P) \over \sin(\angle R)}={15⋅\sin(69\degree) \over \sin(57\degree)}\text{.}\)

\(Q\kern{-.8pt}R≈16{,}7\text{.}\)

De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}\text{.}\)

Dus \(B\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle A) \over \sin(\angle C)}={20⋅\sin(93\degree) \over \sin(33\degree)}\text{.}\)

\(B\kern{-.8pt}C≈36{,}7\text{.}\)

Uit \(\angle P+\angle Q+\angle R=180\degree\) volgt \(\angle Q=180\degree-\angle P-\angle R=180\degree-59\degree-45\degree=76\degree\text{.}\)

De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({Q\kern{-.8pt}R \over \sin(\angle P)}={P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}\text{.}\)

Dus \(Q\kern{-.8pt}R={P\kern{-.8pt}R⋅\sin(\angle P) \over \sin(\angle Q)}={26⋅\sin(59\degree) \over \sin(76\degree)}\text{.}\)

\(Q\kern{-.8pt}R≈23{,}0\text{.}\)

Uit \(\angle Q+\angle R+\angle P=180\degree\) volgt \(\angle R=180\degree-\angle Q-\angle P=180\degree-35\degree-45\degree=100\degree\text{.}\)

De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}={Q\kern{-.8pt}R \over \sin(\angle P)}\text{.}\)

Dus \(P\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle Q) \over \sin(\angle R)}={25⋅\sin(35\degree) \over \sin(100\degree)}\text{.}\)

\(P\kern{-.8pt}R≈14{,}6\text{.}\)