4p
Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 15 \text{,}\) \(A\kern{-.8pt}C = 13\) en \(A\kern{-.8pt}B = 18 \text{.}\)
Bereken \(\angle \text{C} \text{.}\)
Rond indien nodig af op één decimaal.
○
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
○
Invullen geeft \(18^{2} = 15^{2} + 13^{2} - 2 ⋅ 15 ⋅ 13 ⋅ \cos(\angle C)\)
dus \(324 = 394 - 390 ⋅ \cos(\angle C) \text{.}\)
1p
○
Balansmethode geeft \(\cos(\angle C) = {324 - 394 \over -390} = 0{,}179...\)
1p
○
Hieruit volgt \(\angle C = \cos^{-1}(0{,}179...) ≈ 79{,}7\degree \text{.}\)
1p
4p
Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 23 \text{,}\) \(B\kern{-.8pt}C = 24\) en \(A\kern{-.8pt}C = 35 \text{.}\)
Bereken \(\angle \text{B} \text{.}\)
Rond indien nodig af op één decimaal.
○
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 \(35^{2} = 23^{2} + 24^{2} - 2 ⋅ 23 ⋅ 24 ⋅ \cos(\angle B)\)
dus \(1\,225 = 1\,105 - 1\,104 ⋅ \cos(\angle B) \text{.}\)
1p
○
Balansmethode geeft \(\cos(\angle B) = {1\,225 - 1\,105 \over -1\,104} = -0{,}108...\)
1p
○
Hieruit volgt \(\angle B = \cos^{-1}(-0{,}108...) ≈ 96{,}2\degree \text{.}\)
1p
3p
Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 16 \text{,}\) \(A\kern{-.8pt}C = 16\) en \(\angle C = 58\degree \text{.}\)
Bereken de lengte van zijde \(A\kern{-.8pt}B \text{.}\)
Rond indien nodig af op één decimaal.
○
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 \(A\kern{-.8pt}B^{2} = 16^{2} + 16^{2} - 2 ⋅ 16 ⋅ 16 ⋅ \cos(58\degree) = 240{,}681... \text{.}\)
1p
○
\(A\kern{-.8pt}B = \sqrt{240{,}681...} ≈ 15{,}5 \text{.}\)
1p
3p
Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 13 \text{,}\) \(P\kern{-.8pt}Q = 23\) en \(\angle P = 98\degree \text{.}\)
Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\)
Rond indien nodig af op één decimaal.
○
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 \(Q\kern{-.8pt}R^{2} = 13^{2} + 23^{2} - 2 ⋅ 13 ⋅ 23 ⋅ \cos(98\degree) = 781{,}225... \text{.}\)
1p
○
\(Q\kern{-.8pt}R = \sqrt{781{,}225...} ≈ 28{,}0 \text{.}\)
1p
3p
Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 19 \text{,}\) \(A\kern{-.8pt}C = 27\) en \(\angle A = 35\degree \text{.}\)
Bereken \(\angle \text{B} \text{.}\)
Rond indien nodig af op één decimaal.
○
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
○
Daaruit volgt \(\sin(\angle B) = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over B\kern{-.8pt}C} = {27 ⋅ \sin(35\degree) \over 19} = 0{,}815... \text{.}\)
1p
○
Dit geeft \(\angle B ≈ 54{,}6\degree\) of \(\angle B ≈ 125{,}4\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle B\) een scherpe hoek is, dus \(\angle B ≈ 54{,}6\degree \text{.}\)
1p
3p
Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(L\kern{-.8pt}M = 13 \text{,}\) \(K\kern{-.8pt}M = 18\) en \(\angle K = 41\degree \text{.}\)
Bereken \(\angle \text{L} \text{.}\)
Rond indien nodig af op één decimaal.
○
De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({L\kern{-.8pt}M \over \sin(\angle K)} = {K\kern{-.8pt}M \over \sin(\angle L)} = {K\kern{-.8pt}L \over \sin(\angle M)} \text{.}\)
1p
○
Daaruit volgt \(\sin(\angle L) = {K\kern{-.8pt}M ⋅ \sin(\angle K) \over L\kern{-.8pt}M} = {18 ⋅ \sin(41\degree) \over 13} = 0{,}908... \text{.}\)
1p
○
Dit geeft \(\angle L ≈ 65{,}3\degree\) of \(\angle L ≈ 114{,}7\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle L\) een stompe hoek is, dus \(\angle L ≈ 114{,}7\degree \text{.}\)
1p
3p
Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}L = 27 \text{,}\) \(\angle M = 44\degree\) en \(\angle K = 88\degree \text{.}\)
Bereken de lengte van zijde \(L\kern{-.8pt}M \text{.}\)
Rond indien nodig af op één decimaal.
○
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
○
Dus \(L\kern{-.8pt}M = {K\kern{-.8pt}L ⋅ \sin(\angle K) \over \sin(\angle M)} = {27 ⋅ \sin(88\degree) \over \sin(44\degree)} \text{.}\)
1p
○
\(L\kern{-.8pt}M ≈ 38{,}8 \text{.}\)
1p
3p
Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}R = 12 \text{,}\) \(\angle Q = 33\degree\) en \(\angle R = 119\degree \text{.}\)
Bereken de lengte van zijde \(P\kern{-.8pt}Q \text{.}\)
Rond indien nodig af op één decimaal.
○
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 \(P\kern{-.8pt}Q = {P\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle Q)} = {12 ⋅ \sin(119\degree) \over \sin(33\degree)} \text{.}\)
1p
○
\(P\kern{-.8pt}Q ≈ 19{,}3 \text{.}\)
1p
4p
Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 17 \text{,}\) \(\angle R = 65\degree\) en \(\angle Q = 45\degree \text{.}\)
Bereken de lengte van zijde \(P\kern{-.8pt}Q \text{.}\)
Rond indien nodig af op één decimaal.
○
Uit \(\angle R + \angle P + \angle Q = 180\degree\) volgt \(\angle P = 180\degree - \angle R - \angle Q = 180\degree - 65\degree - 45\degree = 70\degree \text{.}\)
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 \(P\kern{-.8pt}Q = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle P)} = {17 ⋅ \sin(65\degree) \over \sin(70\degree)} \text{.}\)
1p
○
\(P\kern{-.8pt}Q ≈ 16{,}4 \text{.}\)
1p
4p
Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 63 \text{,}\) \(\angle A = 63\degree\) en \(\angle C = 26\degree \text{.}\)
Bereken de lengte van zijde \(B\kern{-.8pt}C \text{.}\)
Rond indien nodig af op één decimaal.
○
Uit \(\angle A + \angle B + \angle C = 180\degree\) volgt \(\angle B = 180\degree - \angle A - \angle C = 180\degree - 63\degree - 26\degree = 91\degree \text{.}\)
1p
○
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 \(B\kern{-.8pt}C = {A\kern{-.8pt}C ⋅ \sin(\angle A) \over \sin(\angle B)} = {63 ⋅ \sin(63\degree) \over \sin(91\degree)} \text{.}\)
1p
○
\(B\kern{-.8pt}C ≈ 56{,}1 \text{.}\)
1p