Getal & Ruimte (13e editie) - vwo wiskunde B

'Sinus- en cosinusregel'.

vwo wiskunde B	3.4 De sinusregel en de cosinusregel
Sinus- en cosinusregel (10)	opgave 1 3p a Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 22 \text{,}\) \(\angle L = 53\degree\) en \(\angle M = 78\degree \text{.}\) Bereken de lengte van zijde \(K\kern{-.8pt}L \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeInScherp 007p - Sinus- en cosinusregel - basis - 0ms a De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}M \over \sin(\angle L)} = {K\kern{-.8pt}L \over \sin(\angle M)} = {L\kern{-.8pt}M \over \sin(\angle K)} \text{.}\) 1p ○ Dus \(K\kern{-.8pt}L = {K\kern{-.8pt}M ⋅ \sin(\angle M) \over \sin(\angle L)} = {22 ⋅ \sin(78\degree) \over \sin(53\degree)} \text{.}\) 1p ○ \(K\kern{-.8pt}L ≈ 26{,}9 \text{.}\) 1p 3p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 21 \text{,}\) \(\angle Q = 27\degree\) en \(\angle R = 115\degree \text{.}\) Bereken de lengte van zijde \(P\kern{-.8pt}Q \text{.}\) 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 \({P\kern{-.8pt}R \over \sin(\angle Q)} = {P\kern{-.8pt}Q \over \sin(\angle R)} = {Q\kern{-.8pt}R \over \sin(\angle P)} \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q = {P\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle Q)} = {21 ⋅ \sin(115\degree) \over \sin(27\degree)} \text{.}\) 1p ○ \(P\kern{-.8pt}Q ≈ 41{,}9 \text{.}\) 1p 3p c Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 20 \text{,}\) \(L\kern{-.8pt}M = 26\) en \(\angle M = 50\degree \text{.}\) Bereken \(\angle \text{K} \text{.}\) Rond indien nodig af op één decimaal. SinusregelHoekInScherp 007r - Sinus- en cosinusregel - basis - 5ms c De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}L \over \sin(\angle M)} = {L\kern{-.8pt}M \over \sin(\angle K)} = {K\kern{-.8pt}M \over \sin(\angle L)} \text{.}\) 1p ○ Daaruit volgt \(\sin(\angle K) = {L\kern{-.8pt}M ⋅ \sin(\angle M) \over K\kern{-.8pt}L} = {26 ⋅ \sin(50\degree) \over 20} = 0{,}995... \text{.}\) 1p ○ Dit geeft \(\angle K ≈ 84{,}8\degree\) of \(\angle K ≈ 95{,}2\degree \text{.}\) Uit de afbeelding volgt dat \(\angle K\) een scherpe hoek is, dus \(\angle K ≈ 84{,}8\degree \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 15 \text{,}\) \(P\kern{-.8pt}Q = 24\) en \(\angle Q = 31\degree \text{.}\) Bereken \(\angle \text{R} \text{.}\) 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 \({P\kern{-.8pt}R \over \sin(\angle Q)} = {P\kern{-.8pt}Q \over \sin(\angle R)} = {Q\kern{-.8pt}R \over \sin(\angle P)} \text{.}\) 1p ○ Daaruit volgt \(\sin(\angle R) = {P\kern{-.8pt}Q ⋅ \sin(\angle Q) \over P\kern{-.8pt}R} = {24 ⋅ \sin(31\degree) \over 15} = 0{,}824... \text{.}\) 1p ○ Dit geeft \(\angle R ≈ 55{,}5\degree\) of \(\angle R ≈ 124{,}5\degree \text{.}\) Uit de afbeelding volgt dat \(\angle R\) een stompe hoek is, dus \(\angle R ≈ 124{,}5\degree \text{.}\) 1p opgave 2 4p a Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 28 \text{,}\) \(\angle P = 41\degree\) en \(\angle R = 63\degree \text{.}\) Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeNaHoekInScherp 007t - Sinus- en cosinusregel - basis - 0ms a Uit \(\angle P + \angle Q + \angle R = 180\degree\) volgt \(\angle Q = 180\degree - \angle P - \angle R = 180\degree - 41\degree - 63\degree = 76\degree \text{.}\) 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 \(Q\kern{-.8pt}R = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over \sin(\angle Q)} = {28 ⋅ \sin(41\degree) \over \sin(76\degree)} \text{.}\) 1p ○ \(Q\kern{-.8pt}R ≈ 18{,}9 \text{.}\) 1p 4p b Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 43 \text{,}\) \(\angle R = 27\degree\) en \(\angle Q = 38\degree \text{.}\) Bereken de lengte van zijde \(P\kern{-.8pt}Q \text{.}\) Rond indien nodig af op één decimaal. SinusregelZijdeNaHoekInStomp 007u - Sinus- en cosinusregel - basis - 0ms b Uit \(\angle R + \angle P + \angle Q = 180\degree\) volgt \(\angle P = 180\degree - \angle R - \angle Q = 180\degree - 27\degree - 38\degree = 115\degree \text{.}\) 1p ○ 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 \(P\kern{-.8pt}Q = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle P)} = {43 ⋅ \sin(27\degree) \over \sin(115\degree)} \text{.}\) 1p ○ \(P\kern{-.8pt}Q ≈ 21{,}5 \text{.}\) 1p 3p c Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 16 \text{,}\) \(P\kern{-.8pt}R = 15\) en \(\angle R = 68\degree \text{.}\) Bereken de lengte van zijde \(P\kern{-.8pt}Q \text{.}\) Rond indien nodig af op één decimaal. CosinusregelZijdeInScherp 007v - Sinus- en cosinusregel - basis - 0ms c De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}Q^{2} = Q\kern{-.8pt}R^{2} + P\kern{-.8pt}R^{2} - 2 ⋅ Q\kern{-.8pt}R ⋅ P\kern{-.8pt}R ⋅ \cos(\angle R) \text{.}\) 1p ○ Dus \(P\kern{-.8pt}Q^{2} = 16^{2} + 15^{2} - 2 ⋅ 16 ⋅ 15 ⋅ \cos(68\degree) = 301{,}188... \text{.}\) 1p ○ \(P\kern{-.8pt}Q = \sqrt{301{,}188...} ≈ 17{,}4 \text{.}\) 1p 3p d Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 29 \text{,}\) \(P\kern{-.8pt}Q = 41\) en \(\angle P = 118\degree \text{.}\) Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\) Rond indien nodig af op één decimaal. CosinusregelZijdeInStomp 007w - Sinus- en cosinusregel - basis - 0ms d De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(Q\kern{-.8pt}R^{2} = P\kern{-.8pt}R^{2} + P\kern{-.8pt}Q^{2} - 2 ⋅ P\kern{-.8pt}R ⋅ P\kern{-.8pt}Q ⋅ \cos(\angle P) \text{.}\) 1p ○ Dus \(Q\kern{-.8pt}R^{2} = 29^{2} + 41^{2} - 2 ⋅ 29 ⋅ 41 ⋅ \cos(118\degree) = 3638{,}403... \text{.}\) 1p ○ \(Q\kern{-.8pt}R = \sqrt{3638{,}403...} ≈ 60{,}3 \text{.}\) 1p opgave 3 4p a Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 13 \text{,}\) \(B\kern{-.8pt}C = 10\) en \(A\kern{-.8pt}C = 13 \text{.}\) Bereken \(\angle \text{B} \text{.}\) Rond indien nodig af op één decimaal. CosinusregelHoekInScherp 007x - Sinus- en cosinusregel - basis - 5ms a 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 ○ Invullen geeft \(13^{2} = 13^{2} + 10^{2} - 2 ⋅ 13 ⋅ 10 ⋅ \cos(\angle B)\) dus \(169 = 269 - 260 ⋅ \cos(\angle B) \text{.}\) 1p ○ Balansmethode geeft \(\cos(\angle B) = {169 - 269 \over -260} = 0{,}384...\) 1p ○ Hieruit volgt \(\angle B = \cos^{-1}(0{,}384...) ≈ 67{,}4\degree \text{.}\) 1p 4p b Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 16 \text{,}\) \(B\kern{-.8pt}C = 34\) en \(A\kern{-.8pt}C = 37 \text{.}\) Bereken \(\angle \text{B} \text{.}\) Rond indien nodig af op één decimaal. CosinusregelHoekInStomp 007y - Sinus- en cosinusregel - basis - 0ms b 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 ○ Invullen geeft \(37^{2} = 16^{2} + 34^{2} - 2 ⋅ 16 ⋅ 34 ⋅ \cos(\angle B)\) dus \(1\,369 = 1\,412 - 1\,088 ⋅ \cos(\angle B) \text{.}\) 1p ○ Balansmethode geeft \(\cos(\angle B) = {1\,369 - 1\,412 \over -1\,088} = 0{,}039...\) 1p ○ Hieruit volgt \(\angle B = \cos^{-1}(0{,}039...) ≈ 87{,}7\degree \text{.}\) 1p

3.4 De sinusregel en de cosinusregel

Sinus- en cosinusregel (10)

opgave 1

Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 22 \text{,}\) \(\angle L = 53\degree\) en \(\angle M = 78\degree \text{.}\)
Bereken de lengte van zijde \(K\kern{-.8pt}L \text{.}\)
Rond indien nodig af op één decimaal.

SinusregelZijdeInScherp

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

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

○

Dus \(K\kern{-.8pt}L = {K\kern{-.8pt}M ⋅ \sin(\angle M) \over \sin(\angle L)} = {22 ⋅ \sin(78\degree) \over \sin(53\degree)} \text{.}\)

○

\(K\kern{-.8pt}L ≈ 26{,}9 \text{.}\)

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 21 \text{,}\) \(\angle Q = 27\degree\) en \(\angle R = 115\degree \text{.}\)
Bereken de lengte van zijde \(P\kern{-.8pt}Q \text{.}\)
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 \({P\kern{-.8pt}R \over \sin(\angle Q)} = {P\kern{-.8pt}Q \over \sin(\angle R)} = {Q\kern{-.8pt}R \over \sin(\angle P)} \text{.}\)

○

Dus \(P\kern{-.8pt}Q = {P\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle Q)} = {21 ⋅ \sin(115\degree) \over \sin(27\degree)} \text{.}\)

○

\(P\kern{-.8pt}Q ≈ 41{,}9 \text{.}\)

Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 20 \text{,}\) \(L\kern{-.8pt}M = 26\) en \(\angle M = 50\degree \text{.}\)
Bereken \(\angle \text{K} \text{.}\)
Rond indien nodig af op één decimaal.

SinusregelHoekInScherp

007r - Sinus- en cosinusregel - basis - 5ms

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

○

Daaruit volgt \(\sin(\angle K) = {L\kern{-.8pt}M ⋅ \sin(\angle M) \over K\kern{-.8pt}L} = {26 ⋅ \sin(50\degree) \over 20} = 0{,}995... \text{.}\)

○

Dit geeft \(\angle K ≈ 84{,}8\degree\) of \(\angle K ≈ 95{,}2\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle K\) een scherpe hoek is, dus \(\angle K ≈ 84{,}8\degree \text{.}\)

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 15 \text{,}\) \(P\kern{-.8pt}Q = 24\) en \(\angle Q = 31\degree \text{.}\)
Bereken \(\angle \text{R} \text{.}\)
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 \({P\kern{-.8pt}R \over \sin(\angle Q)} = {P\kern{-.8pt}Q \over \sin(\angle R)} = {Q\kern{-.8pt}R \over \sin(\angle P)} \text{.}\)

○

Daaruit volgt \(\sin(\angle R) = {P\kern{-.8pt}Q ⋅ \sin(\angle Q) \over P\kern{-.8pt}R} = {24 ⋅ \sin(31\degree) \over 15} = 0{,}824... \text{.}\)

○

Dit geeft \(\angle R ≈ 55{,}5\degree\) of \(\angle R ≈ 124{,}5\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle R\) een stompe hoek is, dus \(\angle R ≈ 124{,}5\degree \text{.}\)

opgave 2

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 28 \text{,}\) \(\angle P = 41\degree\) en \(\angle R = 63\degree \text{.}\)
Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\)
Rond indien nodig af op één decimaal.

SinusregelZijdeNaHoekInScherp

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

Uit \(\angle P + \angle Q + \angle R = 180\degree\) volgt \(\angle Q = 180\degree - \angle P - \angle R = 180\degree - 41\degree - 63\degree = 76\degree \text{.}\)

○

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 \(Q\kern{-.8pt}R = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over \sin(\angle Q)} = {28 ⋅ \sin(41\degree) \over \sin(76\degree)} \text{.}\)

○

\(Q\kern{-.8pt}R ≈ 18{,}9 \text{.}\)

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 43 \text{,}\) \(\angle R = 27\degree\) en \(\angle Q = 38\degree \text{.}\)
Bereken de lengte van zijde \(P\kern{-.8pt}Q \text{.}\)
Rond indien nodig af op één decimaal.

SinusregelZijdeNaHoekInStomp

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

Uit \(\angle R + \angle P + \angle Q = 180\degree\) volgt \(\angle P = 180\degree - \angle R - \angle Q = 180\degree - 27\degree - 38\degree = 115\degree \text{.}\)

○

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 \(P\kern{-.8pt}Q = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle P)} = {43 ⋅ \sin(27\degree) \over \sin(115\degree)} \text{.}\)

○

\(P\kern{-.8pt}Q ≈ 21{,}5 \text{.}\)

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 16 \text{,}\) \(P\kern{-.8pt}R = 15\) en \(\angle R = 68\degree \text{.}\)
Bereken de lengte van zijde \(P\kern{-.8pt}Q \text{.}\)
Rond indien nodig af op één decimaal.

CosinusregelZijdeInScherp

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

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 \(P\kern{-.8pt}Q^{2} = 16^{2} + 15^{2} - 2 ⋅ 16 ⋅ 15 ⋅ \cos(68\degree) = 301{,}188... \text{.}\)

○

\(P\kern{-.8pt}Q = \sqrt{301{,}188...} ≈ 17{,}4 \text{.}\)

Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 29 \text{,}\) \(P\kern{-.8pt}Q = 41\) en \(\angle P = 118\degree \text{.}\)
Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\)
Rond indien nodig af op één decimaal.

CosinusregelZijdeInStomp

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

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 \(Q\kern{-.8pt}R^{2} = 29^{2} + 41^{2} - 2 ⋅ 29 ⋅ 41 ⋅ \cos(118\degree) = 3638{,}403... \text{.}\)

○

\(Q\kern{-.8pt}R = \sqrt{3638{,}403...} ≈ 60{,}3 \text{.}\)

opgave 3

Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 13 \text{,}\) \(B\kern{-.8pt}C = 10\) en \(A\kern{-.8pt}C = 13 \text{.}\)
Bereken \(\angle \text{B} \text{.}\)
Rond indien nodig af op één decimaal.

CosinusregelHoekInScherp

007x - Sinus- en cosinusregel - basis - 5ms

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

○

Invullen geeft \(13^{2} = 13^{2} + 10^{2} - 2 ⋅ 13 ⋅ 10 ⋅ \cos(\angle B)\)
dus \(169 = 269 - 260 ⋅ \cos(\angle B) \text{.}\)

○

Balansmethode geeft \(\cos(\angle B) = {169 - 269 \over -260} = 0{,}384...\)

○

Hieruit volgt \(\angle B = \cos^{-1}(0{,}384...) ≈ 67{,}4\degree \text{.}\)

Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 16 \text{,}\) \(B\kern{-.8pt}C = 34\) en \(A\kern{-.8pt}C = 37 \text{.}\)
Bereken \(\angle \text{B} \text{.}\)
Rond indien nodig af op één decimaal.

CosinusregelHoekInStomp

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

○

Invullen geeft \(37^{2} = 16^{2} + 34^{2} - 2 ⋅ 16 ⋅ 34 ⋅ \cos(\angle B)\)
dus \(1\,369 = 1\,412 - 1\,088 ⋅ \cos(\angle B) \text{.}\)

○

Balansmethode geeft \(\cos(\angle B) = {1\,369 - 1\,412 \over -1\,088} = 0{,}039...\)

○

Hieruit volgt \(\angle B = \cos^{-1}(0{,}039...) ≈ 87{,}7\degree \text{.}\)