Sinus- en cosinusregel

De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}R^2=P\kern{-.8pt}Q^2+Q\kern{-.8pt}R^2-2⋅P\kern{-.8pt}Q⋅Q\kern{-.8pt}R⋅\cos(\angle Q)\text{.}\)

Invullen geeft \(31^2=26^2+21^2-2⋅26⋅21⋅\cos(\angle Q)\)
dus \(961=1\,117-1\,092⋅\cos(\angle Q)\text{.}\)

Balansmethode geeft \(\cos(\angle Q)={961-1\,117 \over -1\,092}=0{,}142...\)

Hieruit volgt \(\angle Q=\cos^{-1}(0{,}142...)≈81{,}8\degree\text{.}\)

De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}R^2=P\kern{-.8pt}Q^2+Q\kern{-.8pt}R^2-2⋅P\kern{-.8pt}Q⋅Q\kern{-.8pt}R⋅\cos(\angle Q)\text{.}\)

Invullen geeft \(19^2=11^2+15^2-2⋅11⋅15⋅\cos(\angle Q)\)
dus \(361=346-330⋅\cos(\angle Q)\text{.}\)

Balansmethode geeft \(\cos(\angle Q)={361-346 \over -330}=-0{,}045...\)

Hieruit volgt \(\angle Q=\cos^{-1}(-0{,}045...)≈92{,}6\degree\text{.}\)

De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}R^2=P\kern{-.8pt}Q^2+Q\kern{-.8pt}R^2-2⋅P\kern{-.8pt}Q⋅Q\kern{-.8pt}R⋅\cos(\angle Q)\text{.}\)

Dus \(P\kern{-.8pt}R^2=22^2+23^2-2⋅22⋅23⋅\cos(87\degree)=960{,}036...\text{.}\)

\(P\kern{-.8pt}R=\sqrt{960{,}036...}≈31{,}0\text{.}\)

De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(Q\kern{-.8pt}R^2=P\kern{-.8pt}R^2+P\kern{-.8pt}Q^2-2⋅P\kern{-.8pt}R⋅P\kern{-.8pt}Q⋅\cos(\angle P)\text{.}\)

Dus \(Q\kern{-.8pt}R^2=13^2+21^2-2⋅13⋅21⋅\cos(99\degree)=695{,}413...\text{.}\)

\(Q\kern{-.8pt}R=\sqrt{695{,}413...}≈26{,}4\text{.}\)

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

Daaruit volgt \(\sin(\angle Q)={P\kern{-.8pt}R⋅\sin(\angle P) \over Q\kern{-.8pt}R}={20⋅\sin(34\degree) \over 12}=0{,}931...\text{.}\)

Dit geeft \(\angle Q≈68{,}7\degree\) of \(\angle Q≈111{,}3\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle Q\) een scherpe hoek is, dus \(\angle Q≈68{,}7\degree\text{.}\)

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

Daaruit volgt \(\sin(\angle L)={K\kern{-.8pt}M⋅\sin(\angle K) \over L\kern{-.8pt}M}={26⋅\sin(29\degree) \over 15}=0{,}840...\text{.}\)

Dit geeft \(\angle L≈57{,}2\degree\) of \(\angle L≈122{,}8\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle L\) een stompe hoek is, dus \(\angle L≈122{,}8\degree\text{.}\)

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

Dus \(A\kern{-.8pt}B={A\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle B)}={28⋅\sin(59\degree) \over \sin(65\degree)}\text{.}\)

\(A\kern{-.8pt}B≈26{,}5\text{.}\)

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

Dus \(K\kern{-.8pt}L={K\kern{-.8pt}M⋅\sin(\angle M) \over \sin(\angle L)}={19⋅\sin(110\degree) \over \sin(38\degree)}\text{.}\)

\(K\kern{-.8pt}L≈29{,}0\text{.}\)

Uit \(\angle K+\angle L+\angle M=180\degree\) volgt \(\angle L=180\degree-\angle K-\angle M=180\degree-57\degree-58\degree=65\degree\text{.}\)

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

Dus \(L\kern{-.8pt}M={K\kern{-.8pt}M⋅\sin(\angle K) \over \sin(\angle L)}={16⋅\sin(57\degree) \over \sin(65\degree)}\text{.}\)

\(L\kern{-.8pt}M≈14{,}8\text{.}\)

Uit \(\angle B+\angle C+\angle A=180\degree\) volgt \(\angle C=180\degree-\angle B-\angle A=180\degree-26\degree-49\degree=105\degree\text{.}\)

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

Dus \(A\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle B) \over \sin(\angle C)}={46⋅\sin(26\degree) \over \sin(105\degree)}\text{.}\)

\(A\kern{-.8pt}C≈20{,}9\text{.}\)