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 \(43^{2} = 39^{2} + 24^{2} - 2 ⋅ 39 ⋅ 24 ⋅ \cos(\angle R)\)
dus \(1\,849 = 2\,097 - 1\,872 ⋅ \cos(\angle R) \text{.}\)

Balansmethode geeft \(\cos(\angle R) = {1\,849 - 2\,097 \over -1\,872} = 0{,}132...\)

Hieruit volgt \(\angle R = \cos^{-1}(0{,}132...) ≈ 82{,}4\degree \text{.}\)

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 \(42^{2} = 21^{2} + 29^{2} - 2 ⋅ 21 ⋅ 29 ⋅ \cos(\angle A)\)
dus \(1\,764 = 1\,282 - 1\,218 ⋅ \cos(\angle A) \text{.}\)

Balansmethode geeft \(\cos(\angle A) = {1\,764 - 1\,282 \over -1\,218} = -0{,}395...\)

Hieruit volgt \(\angle A = \cos^{-1}(-0{,}395...) ≈ 113{,}3\degree \text{.}\)

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 \(K\kern{-.8pt}M^{2} = 27^{2} + 21^{2} - 2 ⋅ 27 ⋅ 21 ⋅ \cos(75\degree) = 876{,}499... \text{.}\)

\(K\kern{-.8pt}M = \sqrt{876{,}499...} ≈ 29{,}6 \text{.}\)

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 \(K\kern{-.8pt}L^{2} = 18^{2} + 16^{2} - 2 ⋅ 18 ⋅ 16 ⋅ \cos(120\degree) = 867 \text{.}\)

\(K\kern{-.8pt}L = \sqrt{867} ≈ 29{,}5 \text{.}\)

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 \(\sin(\angle Q) = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over Q\kern{-.8pt}R} = {27 ⋅ \sin(31\degree) \over 19} = 0{,}731... \text{.}\)

Dit geeft \(\angle Q ≈ 47{,}0\degree\) of \(\angle Q ≈ 133{,}0\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle Q\) een scherpe hoek is, dus \(\angle Q ≈ 47{,}0\degree \text{.}\)

De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}L \over \sin(\angle M)} = {L\kern{-.8pt}M \over \sin(\angle K)} = {K\kern{-.8pt}M \over \sin(\angle L)} \text{.}\)

Daaruit volgt \(\sin(\angle K) = {L\kern{-.8pt}M ⋅ \sin(\angle M) \over K\kern{-.8pt}L} = {23 ⋅ \sin(27\degree) \over 14} = 0{,}745... \text{.}\)

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

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 \(K\kern{-.8pt}M = {L\kern{-.8pt}M ⋅ \sin(\angle L) \over \sin(\angle K)} = {16 ⋅ \sin(83\degree) \over \sin(53\degree)} \text{.}\)

\(K\kern{-.8pt}M ≈ 19{,}9 \text{.}\)

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 \(L\kern{-.8pt}M = {K\kern{-.8pt}L ⋅ \sin(\angle K) \over \sin(\angle M)} = {13 ⋅ \sin(123\degree) \over \sin(26\degree)} \text{.}\)

\(L\kern{-.8pt}M ≈ 24{,}9 \text{.}\)

Uit \(\angle Q + \angle R + \angle P = 180\degree\) volgt \(\angle R = 180\degree - \angle Q - \angle P = 180\degree - 56\degree - 50\degree = 74\degree \text{.}\)

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 \(P\kern{-.8pt}R = {P\kern{-.8pt}Q ⋅ \sin(\angle Q) \over \sin(\angle R)} = {25 ⋅ \sin(56\degree) \over \sin(74\degree)} \text{.}\)

\(P\kern{-.8pt}R ≈ 21{,}6 \text{.}\)

Uit \(\angle C + \angle A + \angle B = 180\degree\) volgt \(\angle A = 180\degree - \angle C - \angle B = 180\degree - 29\degree - 41\degree = 110\degree \text{.}\)

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 \(A\kern{-.8pt}B = {B\kern{-.8pt}C ⋅ \sin(\angle C) \over \sin(\angle A)} = {40 ⋅ \sin(29\degree) \over \sin(110\degree)} \text{.}\)

\(A\kern{-.8pt}B ≈ 20{,}6 \text{.}\)