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 \(K\kern{-.8pt}L = {K\kern{-.8pt}M ⋅ \sin(\angle M) \over \sin(\angle L)} = {26 ⋅ \sin(88\degree) \over \sin(31\degree)} \text{.}\)

\(K\kern{-.8pt}L ≈ 50{,}5 \text{.}\)

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 \(Q\kern{-.8pt}R = {P\kern{-.8pt}Q ⋅ \sin(\angle P) \over \sin(\angle R)} = {29 ⋅ \sin(101\degree) \over \sin(52\degree)} \text{.}\)

\(Q\kern{-.8pt}R ≈ 36{,}1 \text{.}\)

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 \(\sin(\angle L) = {K\kern{-.8pt}M ⋅ \sin(\angle K) \over L\kern{-.8pt}M} = {18 ⋅ \sin(37\degree) \over 12} = 0{,}902... \text{.}\)

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

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

Daaruit volgt \(\sin(\angle Q) = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over Q\kern{-.8pt}R} = {21 ⋅ \sin(31\degree) \over 12} = 0{,}901... \text{.}\)

Dit geeft \(\angle Q ≈ 64{,}3\degree\) of \(\angle Q ≈ 115{,}7\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle Q\) een stompe hoek is, dus \(\angle Q ≈ 115{,}7\degree \text{.}\)

Uit \(\angle B + \angle C + \angle A = 180\degree\) volgt \(\angle C = 180\degree - \angle B - \angle A = 180\degree - 31\degree - 60\degree = 89\degree \text{.}\)

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 \(A\kern{-.8pt}C = {A\kern{-.8pt}B ⋅ \sin(\angle B) \over \sin(\angle C)} = {40 ⋅ \sin(31\degree) \over \sin(89\degree)} \text{.}\)

\(A\kern{-.8pt}C ≈ 20{,}6 \text{.}\)

Uit \(\angle C + \angle A + \angle B = 180\degree\) volgt \(\angle A = 180\degree - \angle C - \angle B = 180\degree - 27\degree - 36\degree = 117\degree \text{.}\)

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

Dus \(A\kern{-.8pt}B = {B\kern{-.8pt}C ⋅ \sin(\angle C) \over \sin(\angle A)} = {58 ⋅ \sin(27\degree) \over \sin(117\degree)} \text{.}\)

\(A\kern{-.8pt}B ≈ 29{,}6 \text{.}\)

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 \(A\kern{-.8pt}C^{2} = 33^{2} + 25^{2} - 2 ⋅ 33 ⋅ 25 ⋅ \cos(83\degree) = 1512{,}915... \text{.}\)

\(A\kern{-.8pt}C = \sqrt{1512{,}915...} ≈ 38{,}9 \text{.}\)

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 \(K\kern{-.8pt}L^{2} = 28^{2} + 21^{2} - 2 ⋅ 28 ⋅ 21 ⋅ \cos(110\degree) = 1627{,}215... \text{.}\)

\(K\kern{-.8pt}L = \sqrt{1627{,}215...} ≈ 40{,}3 \text{.}\)

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 \(33^{2} = 24^{2} + 24^{2} - 2 ⋅ 24 ⋅ 24 ⋅ \cos(\angle C)\)
dus \(1\,089 = 1\,152 - 1\,152 ⋅ \cos(\angle C) \text{.}\)

Balansmethode geeft \(\cos(\angle C) = {1\,089 - 1\,152 \over -1\,152} = 0{,}054...\)

Hieruit volgt \(\angle C = \cos^{-1}(0{,}054...) ≈ 86{,}9\degree \text{.}\)

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 \(49^{2} = 27^{2} + 36^{2} - 2 ⋅ 27 ⋅ 36 ⋅ \cos(\angle M)\)
dus \(2\,401 = 2\,025 - 1\,944 ⋅ \cos(\angle M) \text{.}\)

Balansmethode geeft \(\cos(\angle M) = {2\,401 - 2\,025 \over -1\,944} = -0{,}193...\)

Hieruit volgt \(\angle M = \cos^{-1}(-0{,}193...) ≈ 101{,}2\degree \text{.}\)