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

Dus \(A\kern{-.8pt}C = {B\kern{-.8pt}C ⋅ \sin(\angle B) \over \sin(\angle A)} = {12 ⋅ \sin(83\degree) \over \sin(32\degree)} \text{.}\)

\(A\kern{-.8pt}C ≈ 22{,}5 \text{.}\)

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 \(B\kern{-.8pt}C = {A\kern{-.8pt}B ⋅ \sin(\angle A) \over \sin(\angle C)} = {39 ⋅ \sin(102\degree) \over \sin(50\degree)} \text{.}\)

\(B\kern{-.8pt}C ≈ 49{,}8 \text{.}\)

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 \(\sin(\angle L) = {K\kern{-.8pt}M ⋅ \sin(\angle K) \over L\kern{-.8pt}M} = {22 ⋅ \sin(35\degree) \over 16} = 0{,}788... \text{.}\)

Dit geeft \(\angle L ≈ 52{,}1\degree\) of \(\angle L ≈ 127{,}9\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle L\) een scherpe hoek is, dus \(\angle L ≈ 52{,}1\degree \text{.}\)

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

Daaruit volgt \(\sin(\angle A) = {B\kern{-.8pt}C ⋅ \sin(\angle C) \over A\kern{-.8pt}B} = {26 ⋅ \sin(39\degree) \over 19} = 0{,}861... \text{.}\)

Dit geeft \(\angle A ≈ 59{,}4\degree\) of \(\angle A ≈ 120{,}6\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle A\) een stompe hoek is, dus \(\angle A ≈ 120{,}6\degree \text{.}\)

Uit \(\angle Q + \angle R + \angle P = 180\degree\) volgt \(\angle R = 180\degree - \angle Q - \angle P = 180\degree - 59\degree - 38\degree = 83\degree \text{.}\)

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 \(P\kern{-.8pt}R = {P\kern{-.8pt}Q ⋅ \sin(\angle Q) \over \sin(\angle R)} = {29 ⋅ \sin(59\degree) \over \sin(83\degree)} \text{.}\)

\(P\kern{-.8pt}R ≈ 25{,}0 \text{.}\)

Uit \(\angle K + \angle L + \angle M = 180\degree\) volgt \(\angle L = 180\degree - \angle K - \angle M = 180\degree - 28\degree - 47\degree = 105\degree \text{.}\)

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 \(L\kern{-.8pt}M = {K\kern{-.8pt}M ⋅ \sin(\angle K) \over \sin(\angle L)} = {57 ⋅ \sin(28\degree) \over \sin(105\degree)} \text{.}\)

\(L\kern{-.8pt}M ≈ 27{,}7 \text{.}\)

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 \(A\kern{-.8pt}C^{2} = 22^{2} + 37^{2} - 2 ⋅ 22 ⋅ 37 ⋅ \cos(83\degree) = 1654{,}596... \text{.}\)

\(A\kern{-.8pt}C = \sqrt{1654{,}596...} ≈ 40{,}7 \text{.}\)

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

Dus \(K\kern{-.8pt}M^{2} = 23^{2} + 26^{2} - 2 ⋅ 23 ⋅ 26 ⋅ \cos(101\degree) = 1433{,}207... \text{.}\)

\(K\kern{-.8pt}M = \sqrt{1433{,}207...} ≈ 37{,}9 \text{.}\)

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 \(58^{2} = 50^{2} + 30^{2} - 2 ⋅ 50 ⋅ 30 ⋅ \cos(\angle Q)\)
dus \(3\,364 = 3\,400 - 3\,000 ⋅ \cos(\angle Q) \text{.}\)

Balansmethode geeft \(\cos(\angle Q) = {3\,364 - 3\,400 \over -3\,000} = 0{,}012\)

Hieruit volgt \(\angle Q = \cos^{-1}(0{,}012) ≈ 89{,}3\degree \text{.}\)

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 \(39^{2} = 30^{2} + 24^{2} - 2 ⋅ 30 ⋅ 24 ⋅ \cos(\angle C)\)
dus \(1\,521 = 1\,476 - 1\,440 ⋅ \cos(\angle C) \text{.}\)

Balansmethode geeft \(\cos(\angle C) = {1\,521 - 1\,476 \over -1\,440} = -0{,}031...\)

Hieruit volgt \(\angle C = \cos^{-1}(-0{,}031...) ≈ 91{,}8\degree \text{.}\)