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 \(25^{2} = 24^{2} + 22^{2} - 2 ⋅ 24 ⋅ 22 ⋅ \cos(\angle R)\)
dus \(625 = 1\,060 - 1\,056 ⋅ \cos(\angle R) \text{.}\)

Balansmethode geeft \(\cos(\angle R) = {625 - 1\,060 \over -1\,056} = 0{,}411...\)

Hieruit volgt \(\angle R = \cos^{-1}(0{,}411...) ≈ 65{,}7\degree \text{.}\)

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 \(48^{2} = 30^{2} + 33^{2} - 2 ⋅ 30 ⋅ 33 ⋅ \cos(\angle C)\)
dus \(2\,304 = 1\,989 - 1\,980 ⋅ \cos(\angle C) \text{.}\)

Balansmethode geeft \(\cos(\angle C) = {2\,304 - 1\,989 \over -1\,980} = -0{,}159...\)

Hieruit volgt \(\angle C = \cos^{-1}(-0{,}159...) ≈ 99{,}2\degree \text{.}\)

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 \(K\kern{-.8pt}M^{2} = 22^{2} + 22^{2} - 2 ⋅ 22 ⋅ 22 ⋅ \cos(64\degree) = 543{,}656... \text{.}\)

\(K\kern{-.8pt}M = \sqrt{543{,}656...} ≈ 23{,}3 \text{.}\)

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 \(K\kern{-.8pt}L^{2} = 17^{2} + 32^{2} - 2 ⋅ 17 ⋅ 32 ⋅ \cos(98\degree) = 1464{,}420... \text{.}\)

\(K\kern{-.8pt}L = \sqrt{1464{,}420...} ≈ 38{,}3 \text{.}\)

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 \(\sin(\angle P) = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over P\kern{-.8pt}Q} = {24 ⋅ \sin(31\degree) \over 13} = 0{,}950... \text{.}\)

Dit geeft \(\angle P ≈ 72{,}0\degree\) of \(\angle P ≈ 108{,}0\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle P\) een scherpe hoek is, dus \(\angle P ≈ 72{,}0\degree \text{.}\)

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 \(\sin(\angle B) = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over B\kern{-.8pt}C} = {12 ⋅ \sin(32\degree) \over 8} = 0{,}794... \text{.}\)

Dit geeft \(\angle B ≈ 52{,}6\degree\) of \(\angle B ≈ 127{,}4\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle B\) een stompe hoek is, dus \(\angle B ≈ 127{,}4\degree \text{.}\)

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 \(P\kern{-.8pt}R = {Q\kern{-.8pt}R ⋅ \sin(\angle Q) \over \sin(\angle P)} = {19 ⋅ \sin(89\degree) \over \sin(50\degree)} \text{.}\)

\(P\kern{-.8pt}R ≈ 24{,}8 \text{.}\)

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 \(A\kern{-.8pt}B = {A\kern{-.8pt}C ⋅ \sin(\angle C) \over \sin(\angle B)} = {29 ⋅ \sin(95\degree) \over \sin(33\degree)} \text{.}\)

\(A\kern{-.8pt}B ≈ 53{,}0 \text{.}\)

Uit \(\angle K + \angle L + \angle M = 180\degree\) volgt \(\angle L = 180\degree - \angle K - \angle M = 180\degree - 59\degree - 58\degree = 63\degree \text{.}\)

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 \(L\kern{-.8pt}M = {K\kern{-.8pt}M ⋅ \sin(\angle K) \over \sin(\angle L)} = {10 ⋅ \sin(59\degree) \over \sin(63\degree)} \text{.}\)

\(L\kern{-.8pt}M ≈ 9{,}6 \text{.}\)

Uit \(\angle A + \angle B + \angle C = 180\degree\) volgt \(\angle B = 180\degree - \angle A - \angle C = 180\degree - 36\degree - 44\degree = 100\degree \text{.}\)

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 \(B\kern{-.8pt}C = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over \sin(\angle B)} = {31 ⋅ \sin(36\degree) \over \sin(100\degree)} \text{.}\)

\(B\kern{-.8pt}C ≈ 18{,}5 \text{.}\)