Sinus- en cosinusregel

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 \(19^{2} = 17^{2} + 15^{2} - 2 ⋅ 17 ⋅ 15 ⋅ \cos(\angle R)\)
dus \(361 = 514 - 510 ⋅ \cos(\angle R) \text{.}\)

Balansmethode geeft \(\cos(\angle R) = {361 - 514 \over -510} = 0{,}3\)

Hieruit volgt \(\angle R = \cos^{-1}(0{,}3) ≈ 72{,}5\degree \text{.}\)

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 \(34^{2} = 20^{2} + 20^{2} - 2 ⋅ 20 ⋅ 20 ⋅ \cos(\angle M)\)
dus \(1\,156 = 800 - 800 ⋅ \cos(\angle M) \text{.}\)

Balansmethode geeft \(\cos(\angle M) = {1\,156 - 800 \over -800} = -0{,}445\)

Hieruit volgt \(\angle M = \cos^{-1}(-0{,}445) ≈ 116{,}4\degree \text{.}\)

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

Dus \(K\kern{-.8pt}M^{2} = 10^{2} + 10^{2} - 2 ⋅ 10 ⋅ 10 ⋅ \cos(70\degree) = 131{,}595... \text{.}\)

\(K\kern{-.8pt}M = \sqrt{131{,}595...} ≈ 11{,}5 \text{.}\)

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 \(K\kern{-.8pt}L^{2} = 16^{2} + 22^{2} - 2 ⋅ 16 ⋅ 22 ⋅ \cos(92\degree) = 764{,}569... \text{.}\)

\(K\kern{-.8pt}L = \sqrt{764{,}569...} ≈ 27{,}7 \text{.}\)

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

Daaruit volgt \(\sin(\angle P) = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over P\kern{-.8pt}Q} = {29 ⋅ \sin(28\degree) \over 14} = 0{,}972... \text{.}\)

Dit geeft \(\angle P ≈ 76{,}5\degree\) of \(\angle P ≈ 103{,}5\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle P\) een scherpe hoek is, dus \(\angle P ≈ 76{,}5\degree \text{.}\)

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 \(\sin(\angle A) = {B\kern{-.8pt}C ⋅ \sin(\angle C) \over A\kern{-.8pt}B} = {23 ⋅ \sin(27\degree) \over 14} = 0{,}745... \text{.}\)

Dit geeft \(\angle A ≈ 48{,}2\degree\) of \(\angle A ≈ 131{,}8\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle A\) een stompe hoek is, dus \(\angle A ≈ 131{,}8\degree \text{.}\)

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 \(A\kern{-.8pt}B = {A\kern{-.8pt}C ⋅ \sin(\angle C) \over \sin(\angle B)} = {25 ⋅ \sin(77\degree) \over \sin(58\degree)} \text{.}\)

\(A\kern{-.8pt}B ≈ 28{,}7 \text{.}\)

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 \(P\kern{-.8pt}Q = {P\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle Q)} = {26 ⋅ \sin(111\degree) \over \sin(27\degree)} \text{.}\)

\(P\kern{-.8pt}Q ≈ 53{,}5 \text{.}\)

Uit \(\angle B + \angle C + \angle A = 180\degree\) volgt \(\angle C = 180\degree - \angle B - \angle A = 180\degree - 59\degree - 38\degree = 83\degree \text{.}\)

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 \(A\kern{-.8pt}C = {A\kern{-.8pt}B ⋅ \sin(\angle B) \over \sin(\angle C)} = {24 ⋅ \sin(59\degree) \over \sin(83\degree)} \text{.}\)

\(A\kern{-.8pt}C ≈ 20{,}7 \text{.}\)

Uit \(\angle L + \angle M + \angle K = 180\degree\) volgt \(\angle M = 180\degree - \angle L - \angle K = 180\degree - 37\degree - 39\degree = 104\degree \text{.}\)

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 \(K\kern{-.8pt}M = {K\kern{-.8pt}L ⋅ \sin(\angle L) \over \sin(\angle M)} = {25 ⋅ \sin(37\degree) \over \sin(104\degree)} \text{.}\)

\(K\kern{-.8pt}M ≈ 15{,}5 \text{.}\)