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 \(P\kern{-.8pt}Q = {P\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle Q)} = {28 ⋅ \sin(73\degree) \over \sin(49\degree)} \text{.}\)

\(P\kern{-.8pt}Q ≈ 35{,}5 \text{.}\)

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 \(P\kern{-.8pt}Q = {P\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle Q)} = {14 ⋅ \sin(91\degree) \over \sin(26\degree)} \text{.}\)

\(P\kern{-.8pt}Q ≈ 31{,}9 \text{.}\)

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 \(\sin(\angle C) = {A\kern{-.8pt}B ⋅ \sin(\angle B) \over A\kern{-.8pt}C} = {24 ⋅ \sin(28\degree) \over 12} = 0{,}938... \text{.}\)

Dit geeft \(\angle C ≈ 69{,}9\degree\) of \(\angle C ≈ 110{,}1\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle C\) een scherpe hoek is, dus \(\angle C ≈ 69{,}9\degree \text{.}\)

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 \(\sin(\angle M) = {K\kern{-.8pt}L ⋅ \sin(\angle L) \over K\kern{-.8pt}M} = {16 ⋅ \sin(30\degree) \over 9} = 0{,}888... \text{.}\)

Dit geeft \(\angle M ≈ 62{,}7\degree\) of \(\angle M ≈ 117{,}3\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle M\) een stompe hoek is, dus \(\angle M ≈ 117{,}3\degree \text{.}\)

Uit \(\angle R + \angle P + \angle Q = 180\degree\) volgt \(\angle P = 180\degree - \angle R - \angle Q = 180\degree - 56\degree - 40\degree = 84\degree \text{.}\)

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 \(P\kern{-.8pt}Q = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle P)} = {38 ⋅ \sin(56\degree) \over \sin(84\degree)} \text{.}\)

\(P\kern{-.8pt}Q ≈ 31{,}7 \text{.}\)

Uit \(\angle M + \angle K + \angle L = 180\degree\) volgt \(\angle K = 180\degree - \angle M - \angle L = 180\degree - 54\degree - 25\degree = 101\degree \text{.}\)

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 \(K\kern{-.8pt}L = {L\kern{-.8pt}M ⋅ \sin(\angle M) \over \sin(\angle K)} = {65 ⋅ \sin(54\degree) \over \sin(101\degree)} \text{.}\)

\(K\kern{-.8pt}L ≈ 53{,}6 \text{.}\)

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 \(B\kern{-.8pt}C^{2} = 25^{2} + 33^{2} - 2 ⋅ 25 ⋅ 33 ⋅ \cos(78\degree) = 1370{,}945... \text{.}\)

\(B\kern{-.8pt}C = \sqrt{1370{,}945...} ≈ 37{,}0 \text{.}\)

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 \(K\kern{-.8pt}M^{2} = 18^{2} + 20^{2} - 2 ⋅ 18 ⋅ 20 ⋅ \cos(101\degree) = 861{,}382... \text{.}\)

\(K\kern{-.8pt}M = \sqrt{861{,}382...} ≈ 29{,}3 \text{.}\)

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

Invullen geeft \(17^{2} = 14^{2} + 15^{2} - 2 ⋅ 14 ⋅ 15 ⋅ \cos(\angle R)\)
dus \(289 = 421 - 420 ⋅ \cos(\angle R) \text{.}\)

Balansmethode geeft \(\cos(\angle R) = {289 - 421 \over -420} = 0{,}314...\)

Hieruit volgt \(\angle R = \cos^{-1}(0{,}314...) ≈ 71{,}7\degree \text{.}\)

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

Invullen geeft \(41^{2} = 25^{2} + 25^{2} - 2 ⋅ 25 ⋅ 25 ⋅ \cos(\angle A)\)
dus \(1\,681 = 1\,250 - 1\,250 ⋅ \cos(\angle A) \text{.}\)

Balansmethode geeft \(\cos(\angle A) = {1\,681 - 1\,250 \over -1\,250} = -0{,}344...\)

Hieruit volgt \(\angle A = \cos^{-1}(-0{,}344...) ≈ 110{,}2\degree \text{.}\)