Sinus- en cosinusregel

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

Invullen geeft \(12^{2} = 11^{2} + 10^{2} - 2 ⋅ 11 ⋅ 10 ⋅ \cos(\angle K)\)
dus \(144 = 221 - 220 ⋅ \cos(\angle K) \text{.}\)

Balansmethode geeft \(\cos(\angle K) = {144 - 221 \over -220} = 0{,}35\)

Hieruit volgt \(\angle K = \cos^{-1}(0{,}35) ≈ 69{,}5\degree \text{.}\)

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 \(36^{2} = 25^{2} + 22^{2} - 2 ⋅ 25 ⋅ 22 ⋅ \cos(\angle R)\)
dus \(1\,296 = 1\,109 - 1\,100 ⋅ \cos(\angle R) \text{.}\)

Balansmethode geeft \(\cos(\angle R) = {1\,296 - 1\,109 \over -1\,100} = -0{,}17\)

Hieruit volgt \(\angle R = \cos^{-1}(-0{,}17) ≈ 99{,}8\degree \text{.}\)

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

Dus \(L\kern{-.8pt}M^{2} = 26^{2} + 36^{2} - 2 ⋅ 26 ⋅ 36 ⋅ \cos(76\degree) = 1519{,}122... \text{.}\)

\(L\kern{-.8pt}M = \sqrt{1519{,}122...} ≈ 39{,}0 \text{.}\)

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

Dus \(P\kern{-.8pt}R^{2} = 23^{2} + 16^{2} - 2 ⋅ 23 ⋅ 16 ⋅ \cos(112\degree) = 1060{,}710... \text{.}\)

\(P\kern{-.8pt}R = \sqrt{1060{,}710...} ≈ 32{,}6 \text{.}\)

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 \(\sin(\angle P) = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over P\kern{-.8pt}Q} = {21 ⋅ \sin(27\degree) \over 10} = 0{,}953... \text{.}\)

Dit geeft \(\angle P ≈ 72{,}4\degree\) of \(\angle P ≈ 107{,}6\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle P\) een scherpe hoek is, dus \(\angle P ≈ 72{,}4\degree \text{.}\)

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

Daaruit volgt \(\sin(\angle B) = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over B\kern{-.8pt}C} = {26 ⋅ \sin(37\degree) \over 17} = 0{,}920... \text{.}\)

Dit geeft \(\angle B ≈ 67{,}0\degree\) of \(\angle B ≈ 113{,}0\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle B\) een stompe hoek is, dus \(\angle B ≈ 113{,}0\degree \text{.}\)

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 \(B\kern{-.8pt}C = {A\kern{-.8pt}B ⋅ \sin(\angle A) \over \sin(\angle C)} = {28 ⋅ \sin(85\degree) \over \sin(33\degree)} \text{.}\)

\(B\kern{-.8pt}C ≈ 51{,}2 \text{.}\)

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 \(B\kern{-.8pt}C = {A\kern{-.8pt}B ⋅ \sin(\angle A) \over \sin(\angle C)} = {29 ⋅ \sin(114\degree) \over \sin(36\degree)} \text{.}\)

\(B\kern{-.8pt}C ≈ 45{,}1 \text{.}\)

Uit \(\angle P + \angle Q + \angle R = 180\degree\) volgt \(\angle Q = 180\degree - \angle P - \angle R = 180\degree - 53\degree - 41\degree = 86\degree \text{.}\)

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

Dus \(Q\kern{-.8pt}R = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over \sin(\angle Q)} = {31 ⋅ \sin(53\degree) \over \sin(86\degree)} \text{.}\)

\(Q\kern{-.8pt}R ≈ 24{,}8 \text{.}\)

Uit \(\angle M + \angle K + \angle L = 180\degree\) volgt \(\angle K = 180\degree - \angle M - \angle L = 180\degree - 61\degree - 28\degree = 91\degree \text{.}\)

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 \(K\kern{-.8pt}L = {L\kern{-.8pt}M ⋅ \sin(\angle M) \over \sin(\angle K)} = {40 ⋅ \sin(61\degree) \over \sin(91\degree)} \text{.}\)

\(K\kern{-.8pt}L ≈ 35{,}0 \text{.}\)