Sinus- en cosinusregel

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 \(26^{2} = 22^{2} + 16^{2} - 2 ⋅ 22 ⋅ 16 ⋅ \cos(\angle A)\)
dus \(676 = 740 - 704 ⋅ \cos(\angle A) \text{.}\)

Balansmethode geeft \(\cos(\angle A) = {676 - 740 \over -704} = 0{,}090...\)

Hieruit volgt \(\angle A = \cos^{-1}(0{,}090...) ≈ 84{,}8\degree \text{.}\)

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 \(46^{2} = 36^{2} + 26^{2} - 2 ⋅ 36 ⋅ 26 ⋅ \cos(\angle A)\)
dus \(2\,116 = 1\,972 - 1\,872 ⋅ \cos(\angle A) \text{.}\)

Balansmethode geeft \(\cos(\angle A) = {2\,116 - 1\,972 \over -1\,872} = -0{,}076...\)

Hieruit volgt \(\angle A = \cos^{-1}(-0{,}076...) ≈ 94{,}4\degree \text{.}\)

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

Dus \(Q\kern{-.8pt}R^{2} = 24^{2} + 18^{2} - 2 ⋅ 24 ⋅ 18 ⋅ \cos(79\degree) = 735{,}141... \text{.}\)

\(Q\kern{-.8pt}R = \sqrt{735{,}141...} ≈ 27{,}1 \text{.}\)

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

Dus \(P\kern{-.8pt}Q^{2} = 27^{2} + 27^{2} - 2 ⋅ 27 ⋅ 27 ⋅ \cos(105\degree) = 1835{,}358... \text{.}\)

\(P\kern{-.8pt}Q = \sqrt{1835{,}358...} ≈ 42{,}8 \text{.}\)

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 \(\sin(\angle Q) = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over Q\kern{-.8pt}R} = {21 ⋅ \sin(39\degree) \over 14} = 0{,}943... \text{.}\)

Dit geeft \(\angle Q ≈ 70{,}7\degree\) of \(\angle Q ≈ 109{,}3\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle Q\) een scherpe hoek is, dus \(\angle Q ≈ 70{,}7\degree \text{.}\)

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

Daaruit volgt \(\sin(\angle L) = {K\kern{-.8pt}M ⋅ \sin(\angle K) \over L\kern{-.8pt}M} = {18 ⋅ \sin(41\degree) \over 12} = 0{,}984... \text{.}\)

Dit geeft \(\angle L ≈ 79{,}8\degree\) of \(\angle L ≈ 100{,}2\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle L\) een stompe hoek is, dus \(\angle L ≈ 100{,}2\degree \text{.}\)

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 \(P\kern{-.8pt}Q = {P\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle Q)} = {30 ⋅ \sin(88\degree) \over \sin(33\degree)} \text{.}\)

\(P\kern{-.8pt}Q ≈ 55{,}0 \text{.}\)

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 \(A\kern{-.8pt}B = {A\kern{-.8pt}C ⋅ \sin(\angle C) \over \sin(\angle B)} = {53 ⋅ \sin(100\degree) \over \sin(54\degree)} \text{.}\)

\(A\kern{-.8pt}B ≈ 64{,}5 \text{.}\)

Uit \(\angle M + \angle K + \angle L = 180\degree\) volgt \(\angle K = 180\degree - \angle M - \angle L = 180\degree - 54\degree - 38\degree = 88\degree \text{.}\)

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 \(K\kern{-.8pt}L = {L\kern{-.8pt}M ⋅ \sin(\angle M) \over \sin(\angle K)} = {40 ⋅ \sin(54\degree) \over \sin(88\degree)} \text{.}\)

\(K\kern{-.8pt}L ≈ 32{,}4 \text{.}\)

Uit \(\angle B + \angle C + \angle A = 180\degree\) volgt \(\angle C = 180\degree - \angle B - \angle A = 180\degree - 40\degree - 33\degree = 107\degree \text{.}\)

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 \(A\kern{-.8pt}C = {A\kern{-.8pt}B ⋅ \sin(\angle B) \over \sin(\angle C)} = {42 ⋅ \sin(40\degree) \over \sin(107\degree)} \text{.}\)

\(A\kern{-.8pt}C ≈ 28{,}2 \text{.}\)