Getal & Ruimte (12e editie) - vwo wiskunde B

'Sinus- en cosinusregel'.

vwo wiskunde B	3.5 De sinusregel en de cosinusregel
Sinus- en cosinusregel (10)	opgave 1 3p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 30 \text{,}\) \(\angle A = 54\degree\) en \(\angle B = 65\degree \text{.}\) Bereken de lengte van zijde \(A\kern{-.8pt}C \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms a De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({B\kern{-.8pt}C \over \sin(\angle A)} = {A\kern{-.8pt}C \over \sin(\angle B)} = {A\kern{-.8pt}B \over \sin(\angle C)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C = {B\kern{-.8pt}C ⋅ \sin(\angle B) \over \sin(\angle A)} = {30 ⋅ \sin(65\degree) \over \sin(54\degree)} \text{.}\) 1p ○ \(A\kern{-.8pt}C ≈ 33{,}6 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 15 \text{,}\) \(\angle R = 33\degree\) en \(\angle P = 120\degree \text{.}\) Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeInStomp 007q - Sinus- en cosinusregel - basis - 0ms b De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}Q \over \sin(\angle R)} = {Q\kern{-.8pt}R \over \sin(\angle P)} = {P\kern{-.8pt}R \over \sin(\angle Q)} \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R = {P\kern{-.8pt}Q ⋅ \sin(\angle P) \over \sin(\angle R)} = {15 ⋅ \sin(120\degree) \over \sin(33\degree)} \text{.}\) 1p ○ \(Q\kern{-.8pt}R ≈ 23{,}9 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 19 \text{,}\) \(P\kern{-.8pt}Q = 28\) en \(\angle Q = 25\degree \text{.}\) Bereken \(\angle \text{R} \text{.}\) Rond indien nodig af op één decimaal. SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 4ms c De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}R \over \sin(\angle Q)} = {P\kern{-.8pt}Q \over \sin(\angle R)} = {Q\kern{-.8pt}R \over \sin(\angle P)} \text{.}\) 1p ○ Daaruit volgt \(\sin(\angle R) = {P\kern{-.8pt}Q ⋅ \sin(\angle Q) \over P\kern{-.8pt}R} = {28 ⋅ \sin(25\degree) \over 19} = 0{,}622... \text{.}\) 1p ○ Dit geeft \(\angle R ≈ 38{,}5\degree\) of \(\angle R ≈ 141{,}5\degree \text{.}\) Uit de afbeelding volgt dat \(\angle R\) een scherpe hoek is, dus \(\angle R ≈ 38{,}5\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 17 \text{,}\) \(Q\kern{-.8pt}R = 23\) en \(\angle R = 40\degree \text{.}\) Bereken \(\angle \text{P} \text{.}\) Rond indien nodig af op één decimaal. SinusregelHoekInStomp 007s - Sinus- en cosinusregel - basis - 0ms d De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}Q \over \sin(\angle R)} = {Q\kern{-.8pt}R \over \sin(\angle P)} = {P\kern{-.8pt}R \over \sin(\angle Q)} \text{.}\) 1p ○ Daaruit volgt \(\sin(\angle P) = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over P\kern{-.8pt}Q} = {23 ⋅ \sin(40\degree) \over 17} = 0{,}869... \text{.}\) 1p ○ Dit geeft \(\angle P ≈ 60{,}4\degree\) of \(\angle P ≈ 119{,}6\degree \text{.}\) Uit de afbeelding volgt dat \(\angle P\) een stompe hoek is, dus \(\angle P ≈ 119{,}6\degree \text{.}\) 1p opgave 2 4p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 27 \text{,}\) \(\angle B = 62\degree\) en \(\angle A = 57\degree \text{.}\) Bereken de lengte van zijde \(A\kern{-.8pt}C \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle B + \angle C + \angle A = 180\degree\) volgt \(\angle C = 180\degree - \angle B - \angle A = 180\degree - 62\degree - 57\degree = 61\degree \text{.}\) 1p ○ De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}C \over \sin(\angle B)} = {A\kern{-.8pt}B \over \sin(\angle C)} = {B\kern{-.8pt}C \over \sin(\angle A)} \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C = {A\kern{-.8pt}B ⋅ \sin(\angle B) \over \sin(\angle C)} = {27 ⋅ \sin(62\degree) \over \sin(61\degree)} \text{.}\) 1p ○ \(A\kern{-.8pt}C ≈ 27{,}3 \text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 51 \text{,}\) \(\angle P = 26\degree\) en \(\angle R = 53\degree \text{.}\) Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle P + \angle Q + \angle R = 180\degree\) volgt \(\angle Q = 180\degree - \angle P - \angle R = 180\degree - 26\degree - 53\degree = 101\degree \text{.}\) 1p ○ De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({Q\kern{-.8pt}R \over \sin(\angle P)} = {P\kern{-.8pt}R \over \sin(\angle Q)} = {P\kern{-.8pt}Q \over \sin(\angle R)} \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over \sin(\angle Q)} = {51 ⋅ \sin(26\degree) \over \sin(101\degree)} \text{.}\) 1p ○ \(Q\kern{-.8pt}R ≈ 22{,}8 \text{.}\) 1p 3p c Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 18 \text{,}\) \(B\kern{-.8pt}C = 19\) en \(\angle B = 83\degree \text{.}\) Bereken de lengte van zijde \(A\kern{-.8pt}C \text{.}\) Rond indien nodig af op één decimaal. CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 0ms c De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(A\kern{-.8pt}C^{2} = A\kern{-.8pt}B^{2} + B\kern{-.8pt}C^{2} - 2 ⋅ A\kern{-.8pt}B ⋅ B\kern{-.8pt}C ⋅ \cos(\angle B) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}C^{2} = 18^{2} + 19^{2} - 2 ⋅ 18 ⋅ 19 ⋅ \cos(83\degree) = 601{,}641... \text{.}\) 1p ○ \(A\kern{-.8pt}C = \sqrt{601{,}641...} ≈ 24{,}5 \text{.}\) 1p 3p d Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 27 \text{,}\) \(A\kern{-.8pt}C = 19\) en \(\angle C = 109\degree \text{.}\) Bereken de lengte van zijde \(A\kern{-.8pt}B \text{.}\) Rond indien nodig af op één decimaal. CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms d De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(A\kern{-.8pt}B^{2} = B\kern{-.8pt}C^{2} + A\kern{-.8pt}C^{2} - 2 ⋅ B\kern{-.8pt}C ⋅ A\kern{-.8pt}C ⋅ \cos(\angle C) \text{.}\) 1p ○ Dus \(A\kern{-.8pt}B^{2} = 27^{2} + 19^{2} - 2 ⋅ 27 ⋅ 19 ⋅ \cos(109\degree) = 1424{,}032... \text{.}\) 1p ○ \(A\kern{-.8pt}B = \sqrt{1424{,}032...} ≈ 37{,}7 \text{.}\) 1p opgave 3 4p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 15 \text{,}\) \(Q\kern{-.8pt}R = 16\) en \(P\kern{-.8pt}R = 15 \text{.}\) Bereken \(\angle \text{Q} \text{.}\) Rond indien nodig af op één decimaal. CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 4ms a De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}R^{2} = P\kern{-.8pt}Q^{2} + Q\kern{-.8pt}R^{2} - 2 ⋅ P\kern{-.8pt}Q ⋅ Q\kern{-.8pt}R ⋅ \cos(\angle Q) \text{.}\) 1p ○ Invullen geeft \(15^{2} = 15^{2} + 16^{2} - 2 ⋅ 15 ⋅ 16 ⋅ \cos(\angle Q)\) dus \(225 = 481 - 480 ⋅ \cos(\angle Q) \text{.}\) 1p ○ Balansmethode geeft \(\cos(\angle Q) = {225 - 481 \over -480} = 0{,}533...\) 1p ○ Hieruit volgt \(\angle Q = \cos^{-1}(0{,}533...) ≈ 57{,}8\degree \text{.}\) 1p 4p b Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 16 \text{,}\) \(L\kern{-.8pt}M = 13\) en \(K\kern{-.8pt}M = 22 \text{.}\) Bereken \(\angle \text{L} \text{.}\) Rond indien nodig af op één decimaal. CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms b De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(K\kern{-.8pt}M^{2} = K\kern{-.8pt}L^{2} + L\kern{-.8pt}M^{2} - 2 ⋅ K\kern{-.8pt}L ⋅ L\kern{-.8pt}M ⋅ \cos(\angle L) \text{.}\) 1p ○ Invullen geeft \(22^{2} = 16^{2} + 13^{2} - 2 ⋅ 16 ⋅ 13 ⋅ \cos(\angle L)\) dus \(484 = 425 - 416 ⋅ \cos(\angle L) \text{.}\) 1p ○ Balansmethode geeft \(\cos(\angle L) = {484 - 425 \over -416} = -0{,}141...\) 1p ○ Hieruit volgt \(\angle L = \cos^{-1}(-0{,}141...) ≈ 98{,}2\degree \text{.}\) 1p

3.5 De sinusregel en de cosinusregel

Sinus- en cosinusregel (10)

opgave 1

Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 30 \text{,}\) \(\angle A = 54\degree\) en \(\angle B = 65\degree \text{.}\)
Bereken de lengte van zijde \(A\kern{-.8pt}C \text{.}\)
Rond indien nodig af op één decimaal.

SinusregelZijdeInScherp

007p - Sinus- en cosinusregel - basis - 0ms

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 \(A\kern{-.8pt}C = {B\kern{-.8pt}C ⋅ \sin(\angle B) \over \sin(\angle A)} = {30 ⋅ \sin(65\degree) \over \sin(54\degree)} \text{.}\)

○

\(A\kern{-.8pt}C ≈ 33{,}6 \text{.}\)

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 15 \text{,}\) \(\angle R = 33\degree\) en \(\angle P = 120\degree \text{.}\)
Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\)
Rond indien nodig af op één decimaal.

SinusregelZijdeInStomp

007q - Sinus- en cosinusregel - basis - 0ms

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 \(Q\kern{-.8pt}R = {P\kern{-.8pt}Q ⋅ \sin(\angle P) \over \sin(\angle R)} = {15 ⋅ \sin(120\degree) \over \sin(33\degree)} \text{.}\)

○

\(Q\kern{-.8pt}R ≈ 23{,}9 \text{.}\)

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 19 \text{,}\) \(P\kern{-.8pt}Q = 28\) en \(\angle Q = 25\degree \text{.}\)
Bereken \(\angle \text{R} \text{.}\)
Rond indien nodig af op één decimaal.

SinusregelHoekInScherp

007r - Sinus- en cosinusregel - basis - 4ms

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 \(\sin(\angle R) = {P\kern{-.8pt}Q ⋅ \sin(\angle Q) \over P\kern{-.8pt}R} = {28 ⋅ \sin(25\degree) \over 19} = 0{,}622... \text{.}\)

○

Dit geeft \(\angle R ≈ 38{,}5\degree\) of \(\angle R ≈ 141{,}5\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle R\) een scherpe hoek is, dus \(\angle R ≈ 38{,}5\degree \text{.}\)

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 17 \text{,}\) \(Q\kern{-.8pt}R = 23\) en \(\angle R = 40\degree \text{.}\)
Bereken \(\angle \text{P} \text{.}\)
Rond indien nodig af op één decimaal.

SinusregelHoekInStomp

007s - Sinus- en cosinusregel - basis - 0ms

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 \(\sin(\angle P) = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over P\kern{-.8pt}Q} = {23 ⋅ \sin(40\degree) \over 17} = 0{,}869... \text{.}\)

○

Dit geeft \(\angle P ≈ 60{,}4\degree\) of \(\angle P ≈ 119{,}6\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle P\) een stompe hoek is, dus \(\angle P ≈ 119{,}6\degree \text{.}\)

opgave 2

Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 27 \text{,}\) \(\angle B = 62\degree\) en \(\angle A = 57\degree \text{.}\)
Bereken de lengte van zijde \(A\kern{-.8pt}C \text{.}\)
Rond indien nodig af op één decimaal.

SinusregelZijdeNaHoekInScherp

007t - Sinus- en cosinusregel - basis - 0ms

Uit \(\angle B + \angle C + \angle A = 180\degree\) volgt \(\angle C = 180\degree - \angle B - \angle A = 180\degree - 62\degree - 57\degree = 61\degree \text{.}\)

○

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 \(A\kern{-.8pt}C = {A\kern{-.8pt}B ⋅ \sin(\angle B) \over \sin(\angle C)} = {27 ⋅ \sin(62\degree) \over \sin(61\degree)} \text{.}\)

○

\(A\kern{-.8pt}C ≈ 27{,}3 \text{.}\)

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 51 \text{,}\) \(\angle P = 26\degree\) en \(\angle R = 53\degree \text{.}\)
Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\)
Rond indien nodig af op één decimaal.

SinusregelZijdeNaHoekInStomp

007u - Sinus- en cosinusregel - basis - 0ms

Uit \(\angle P + \angle Q + \angle R = 180\degree\) volgt \(\angle Q = 180\degree - \angle P - \angle R = 180\degree - 26\degree - 53\degree = 101\degree \text{.}\)

○

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 \(Q\kern{-.8pt}R = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over \sin(\angle Q)} = {51 ⋅ \sin(26\degree) \over \sin(101\degree)} \text{.}\)

○

\(Q\kern{-.8pt}R ≈ 22{,}8 \text{.}\)

Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 18 \text{,}\) \(B\kern{-.8pt}C = 19\) en \(\angle B = 83\degree \text{.}\)
Bereken de lengte van zijde \(A\kern{-.8pt}C \text{.}\)
Rond indien nodig af op één decimaal.

CosinusregelZijdeInScherp

007v - Sinus- en cosinusregel - basis - 0ms

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 \(A\kern{-.8pt}C^{2} = 18^{2} + 19^{2} - 2 ⋅ 18 ⋅ 19 ⋅ \cos(83\degree) = 601{,}641... \text{.}\)

○

\(A\kern{-.8pt}C = \sqrt{601{,}641...} ≈ 24{,}5 \text{.}\)

Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 27 \text{,}\) \(A\kern{-.8pt}C = 19\) en \(\angle C = 109\degree \text{.}\)
Bereken de lengte van zijde \(A\kern{-.8pt}B \text{.}\)
Rond indien nodig af op één decimaal.

CosinusregelZijdeInStomp

007w - Sinus- en cosinusregel - basis - 0ms

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

○

Dus \(A\kern{-.8pt}B^{2} = 27^{2} + 19^{2} - 2 ⋅ 27 ⋅ 19 ⋅ \cos(109\degree) = 1424{,}032... \text{.}\)

○

\(A\kern{-.8pt}B = \sqrt{1424{,}032...} ≈ 37{,}7 \text{.}\)

opgave 3

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 15 \text{,}\) \(Q\kern{-.8pt}R = 16\) en \(P\kern{-.8pt}R = 15 \text{.}\)
Bereken \(\angle \text{Q} \text{.}\)
Rond indien nodig af op één decimaal.

CosinusregelHoekInScherp

007x - Sinus- en cosinusregel - basis - 4ms

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

○

Invullen geeft \(15^{2} = 15^{2} + 16^{2} - 2 ⋅ 15 ⋅ 16 ⋅ \cos(\angle Q)\)
dus \(225 = 481 - 480 ⋅ \cos(\angle Q) \text{.}\)

○

Balansmethode geeft \(\cos(\angle Q) = {225 - 481 \over -480} = 0{,}533...\)

○

Hieruit volgt \(\angle Q = \cos^{-1}(0{,}533...) ≈ 57{,}8\degree \text{.}\)

Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 16 \text{,}\) \(L\kern{-.8pt}M = 13\) en \(K\kern{-.8pt}M = 22 \text{.}\)
Bereken \(\angle \text{L} \text{.}\)
Rond indien nodig af op één decimaal.

CosinusregelHoekInStomp

007y - Sinus- en cosinusregel - basis - 0ms

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

○

Invullen geeft \(22^{2} = 16^{2} + 13^{2} - 2 ⋅ 16 ⋅ 13 ⋅ \cos(\angle L)\)
dus \(484 = 425 - 416 ⋅ \cos(\angle L) \text{.}\)

○

Balansmethode geeft \(\cos(\angle L) = {484 - 425 \over -416} = -0{,}141...\)

○

Hieruit volgt \(\angle L = \cos^{-1}(-0{,}141...) ≈ 98{,}2\degree \text{.}\)