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 \(P\kern{-.8pt}Q={P\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle Q)}={32⋅\sin(76\degree) \over \sin(56\degree)}\text{.}\)

\(P\kern{-.8pt}Q≈37{,}5\text{.}\)

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

Dus \(L\kern{-.8pt}M={K\kern{-.8pt}L⋅\sin(\angle K) \over \sin(\angle M)}={30⋅\sin(117\degree) \over \sin(38\degree)}\text{.}\)

\(L\kern{-.8pt}M≈43{,}4\text{.}\)

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 \(\sin(\angle L)={K\kern{-.8pt}M⋅\sin(\angle K) \over L\kern{-.8pt}M}={13⋅\sin(35\degree) \over 9}=0{,}828...\text{.}\)

Dit geeft \(\angle L≈55{,}9\degree\) of \(\angle L≈124{,}1\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle L\) een scherpe hoek is, dus \(\angle L≈55{,}9\degree\text{.}\)

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

Daaruit volgt \(\sin(\angle C)={A\kern{-.8pt}B⋅\sin(\angle B) \over A\kern{-.8pt}C}={16⋅\sin(35\degree) \over 11}=0{,}834...\text{.}\)

Dit geeft \(\angle C≈56{,}5\degree\) of \(\angle C≈123{,}5\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle C\) een stompe hoek is, dus \(\angle C≈123{,}5\degree\text{.}\)

Uit \(\angle M+\angle K+\angle L=180\degree\) volgt \(\angle K=180\degree-\angle M-\angle L=180\degree-37\degree-59\degree=84\degree\text{.}\)

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

Dus \(K\kern{-.8pt}L={L\kern{-.8pt}M⋅\sin(\angle M) \over \sin(\angle K)}={19⋅\sin(37\degree) \over \sin(84\degree)}\text{.}\)

\(K\kern{-.8pt}L≈11{,}5\text{.}\)

Uit \(\angle L+\angle M+\angle K=180\degree\) volgt \(\angle M=180\degree-\angle L-\angle K=180\degree-33\degree-39\degree=108\degree\text{.}\)

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

Dus \(K\kern{-.8pt}M={K\kern{-.8pt}L⋅\sin(\angle L) \over \sin(\angle M)}={45⋅\sin(33\degree) \over \sin(108\degree)}\text{.}\)

\(K\kern{-.8pt}M≈25{,}8\text{.}\)

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

Dus \(B\kern{-.8pt}C^2=15^2+21^2-2⋅15⋅21⋅\cos(76\degree)=513{,}589...\text{.}\)

\(B\kern{-.8pt}C=\sqrt{513{,}589...}≈22{,}7\text{.}\)

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

Dus \(P\kern{-.8pt}Q^2=32^2+21^2-2⋅32⋅21⋅\cos(96\degree)=1605{,}486...\text{.}\)

\(P\kern{-.8pt}Q=\sqrt{1605{,}486...}≈40{,}1\text{.}\)

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

Invullen geeft \(19^2=14^2+14^2-2⋅14⋅14⋅\cos(\angle C)\)
dus \(361=392-392⋅\cos(\angle C)\text{.}\)

Balansmethode geeft \(\cos(\angle C)={361-392 \over -392}=0{,}079...\)

Hieruit volgt \(\angle C=\cos^{-1}(0{,}079...)≈85{,}5\degree\text{.}\)

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

Invullen geeft \(29^2=18^2+20^2-2⋅18⋅20⋅\cos(\angle L)\)
dus \(841=724-720⋅\cos(\angle L)\text{.}\)

Balansmethode geeft \(\cos(\angle L)={841-724 \over -720}=-0{,}162...\)

Hieruit volgt \(\angle L=\cos^{-1}(-0{,}162...)≈99{,}4\degree\text{.}\)