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 \(48^{2} = 41^{2} + 27^{2} - 2 ⋅ 41 ⋅ 27 ⋅ \cos(\angle K)\)
dus \(2\,304 = 2\,410 - 2\,214 ⋅ \cos(\angle K) \text{.}\)

Balansmethode geeft \(\cos(\angle K) = {2\,304 - 2\,410 \over -2\,214} = 0{,}047...\)

Hieruit volgt \(\angle K = \cos^{-1}(0{,}047...) ≈ 87{,}3\degree \text{.}\)

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

Invullen geeft \(22^{2} = 15^{2} + 15^{2} - 2 ⋅ 15 ⋅ 15 ⋅ \cos(\angle C)\)
dus \(484 = 450 - 450 ⋅ \cos(\angle C) \text{.}\)

Balansmethode geeft \(\cos(\angle C) = {484 - 450 \over -450} = -0{,}075...\)

Hieruit volgt \(\angle C = \cos^{-1}(-0{,}075...) ≈ 94{,}3\degree \text{.}\)

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 \(L\kern{-.8pt}M^{2} = 20^{2} + 18^{2} - 2 ⋅ 20 ⋅ 18 ⋅ \cos(78\degree) = 574{,}303... \text{.}\)

\(L\kern{-.8pt}M = \sqrt{574{,}303...} ≈ 24{,}0 \text{.}\)

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 \(A\kern{-.8pt}C^{2} = 30^{2} + 45^{2} - 2 ⋅ 30 ⋅ 45 ⋅ \cos(91\degree) = 2972{,}121... \text{.}\)

\(A\kern{-.8pt}C = \sqrt{2972{,}121...} ≈ 54{,}5 \text{.}\)

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

Daaruit volgt \(\sin(\angle C) = {A\kern{-.8pt}B ⋅ \sin(\angle B) \over A\kern{-.8pt}C} = {19 ⋅ \sin(43\degree) \over 14} = 0{,}925... \text{.}\)

Dit geeft \(\angle C ≈ 67{,}8\degree\) of \(\angle C ≈ 112{,}2\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle C\) een scherpe hoek is, dus \(\angle C ≈ 67{,}8\degree \text{.}\)

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

Daaruit volgt \(\sin(\angle R) = {P\kern{-.8pt}Q ⋅ \sin(\angle Q) \over P\kern{-.8pt}R} = {24 ⋅ \sin(30\degree) \over 16} = 0{,}75 \text{.}\)

Dit geeft \(\angle R ≈ 48{,}6\degree\) of \(\angle R ≈ 131{,}4\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle R\) een stompe hoek is, dus \(\angle R ≈ 131{,}4\degree \text{.}\)

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 \(K\kern{-.8pt}M = {L\kern{-.8pt}M ⋅ \sin(\angle L) \over \sin(\angle K)} = {30 ⋅ \sin(75\degree) \over \sin(42\degree)} \text{.}\)

\(K\kern{-.8pt}M ≈ 43{,}3 \text{.}\)

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 \(P\kern{-.8pt}Q = {P\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle Q)} = {12 ⋅ \sin(92\degree) \over \sin(43\degree)} \text{.}\)

\(P\kern{-.8pt}Q ≈ 17{,}6 \text{.}\)

Uit \(\angle A + \angle B + \angle C = 180\degree\) volgt \(\angle B = 180\degree - \angle A - \angle C = 180\degree - 61\degree - 65\degree = 54\degree \text{.}\)

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 \(B\kern{-.8pt}C = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over \sin(\angle B)} = {28 ⋅ \sin(61\degree) \over \sin(54\degree)} \text{.}\)

\(B\kern{-.8pt}C ≈ 30{,}3 \text{.}\)

Uit \(\angle R + \angle P + \angle Q = 180\degree\) volgt \(\angle P = 180\degree - \angle R - \angle Q = 180\degree - 26\degree - 57\degree = 97\degree \text{.}\)

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

Dus \(P\kern{-.8pt}Q = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle P)} = {27 ⋅ \sin(26\degree) \over \sin(97\degree)} \text{.}\)

\(P\kern{-.8pt}Q ≈ 11{,}9 \text{.}\)