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

Dus \(P\kern{-.8pt}R = {Q\kern{-.8pt}R ⋅ \sin(\angle Q) \over \sin(\angle P)} = {32 ⋅ \sin(52\degree) \over \sin(63\degree)} \text{.}\)

\(P\kern{-.8pt}R ≈ 28{,}3 \text{.}\)

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 \(K\kern{-.8pt}L = {K\kern{-.8pt}M ⋅ \sin(\angle M) \over \sin(\angle L)} = {10 ⋅ \sin(127\degree) \over \sin(25\degree)} \text{.}\)

\(K\kern{-.8pt}L ≈ 18{,}9 \text{.}\)

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 \(\sin(\angle L) = {K\kern{-.8pt}M ⋅ \sin(\angle K) \over L\kern{-.8pt}M} = {23 ⋅ \sin(25\degree) \over 10} = 0{,}972... \text{.}\)

Dit geeft \(\angle L ≈ 76{,}4\degree\) of \(\angle L ≈ 103{,}6\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle L\) een scherpe hoek is, dus \(\angle L ≈ 76{,}4\degree \text{.}\)

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 \(\sin(\angle Q) = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over Q\kern{-.8pt}R} = {23 ⋅ \sin(31\degree) \over 12} = 0{,}987... \text{.}\)

Dit geeft \(\angle Q ≈ 80{,}8\degree\) of \(\angle Q ≈ 99{,}2\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle Q\) een stompe hoek is, dus \(\angle Q ≈ 99{,}2\degree \text{.}\)

Uit \(\angle M + \angle K + \angle L = 180\degree\) volgt \(\angle K = 180\degree - \angle M - \angle L = 180\degree - 55\degree - 45\degree = 80\degree \text{.}\)

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 \(K\kern{-.8pt}L = {L\kern{-.8pt}M ⋅ \sin(\angle M) \over \sin(\angle K)} = {36 ⋅ \sin(55\degree) \over \sin(80\degree)} \text{.}\)

\(K\kern{-.8pt}L ≈ 29{,}9 \text{.}\)

Uit \(\angle A + \angle B + \angle C = 180\degree\) volgt \(\angle B = 180\degree - \angle A - \angle C = 180\degree - 28\degree - 45\degree = 107\degree \text{.}\)

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 \(B\kern{-.8pt}C = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over \sin(\angle B)} = {30 ⋅ \sin(28\degree) \over \sin(107\degree)} \text{.}\)

\(B\kern{-.8pt}C ≈ 14{,}7 \text{.}\)

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 \(A\kern{-.8pt}C^{2} = 15^{2} + 13^{2} - 2 ⋅ 15 ⋅ 13 ⋅ \cos(69\degree) = 254{,}236... \text{.}\)

\(A\kern{-.8pt}C = \sqrt{254{,}236...} ≈ 15{,}9 \text{.}\)

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

Dus \(A\kern{-.8pt}B^{2} = 26^{2} + 20^{2} - 2 ⋅ 26 ⋅ 20 ⋅ \cos(91\degree) = 1094{,}150... \text{.}\)

\(A\kern{-.8pt}B = \sqrt{1094{,}150...} ≈ 33{,}1 \text{.}\)

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 \(58^{2} = 51^{2} + 30^{2} - 2 ⋅ 51 ⋅ 30 ⋅ \cos(\angle M)\)
dus \(3\,364 = 3\,501 - 3\,060 ⋅ \cos(\angle M) \text{.}\)

Balansmethode geeft \(\cos(\angle M) = {3\,364 - 3\,501 \over -3\,060} = 0{,}044...\)

Hieruit volgt \(\angle M = \cos^{-1}(0{,}044...) ≈ 87{,}4\degree \text{.}\)

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 \(23^{2} = 16^{2} + 11^{2} - 2 ⋅ 16 ⋅ 11 ⋅ \cos(\angle A)\)
dus \(529 = 377 - 352 ⋅ \cos(\angle A) \text{.}\)

Balansmethode geeft \(\cos(\angle A) = {529 - 377 \over -352} = -0{,}431...\)

Hieruit volgt \(\angle A = \cos^{-1}(-0{,}431...) ≈ 115{,}6\degree \text{.}\)