Sinus- en cosinusregel
4p
Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 19 \text{,}\) \(Q\kern{-.8pt}R = 20\) en \(P\kern{-.8pt}R = 21 \text{.}\)
Bereken \(\angle \text{Q} \text{.}\)
Rond indien nodig af op één decimaal.
○
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 \(21^{2} = 19^{2} + 20^{2} - 2 ⋅ 19 ⋅ 20 ⋅ \cos(\angle Q)\)
dus \(441 = 761 - 760 ⋅ \cos(\angle Q) \text{.}\)
1p
○
Balansmethode geeft \(\cos(\angle Q) = {441 - 761 \over -760} = 0{,}421...\)
1p
○
Hieruit volgt \(\angle Q = \cos^{-1}(0{,}421...) ≈ 65{,}1\degree \text{.}\)
1p
4p
Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 24 \text{,}\) \(A\kern{-.8pt}B = 22\) en \(B\kern{-.8pt}C = 37 \text{.}\)
Bereken \(\angle \text{A} \text{.}\)
Rond indien nodig af op één decimaal.
○
De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(B\kern{-.8pt}C^{2} = A\kern{-.8pt}C^{2} + A\kern{-.8pt}B^{2} - 2 ⋅ A\kern{-.8pt}C ⋅ A\kern{-.8pt}B ⋅ \cos(\angle A) \text{.}\)
1p
○
Invullen geeft \(37^{2} = 24^{2} + 22^{2} - 2 ⋅ 24 ⋅ 22 ⋅ \cos(\angle A)\)
dus \(1\,369 = 1\,060 - 1\,056 ⋅ \cos(\angle A) \text{.}\)
1p
○
Balansmethode geeft \(\cos(\angle A) = {1\,369 - 1\,060 \over -1\,056} = -0{,}292...\)
1p
○
Hieruit volgt \(\angle A = \cos^{-1}(-0{,}292...) ≈ 107{,}0\degree \text{.}\)
1p
3p
Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 20 \text{,}\) \(A\kern{-.8pt}C = 21\) en \(\angle C = 57\degree \text{.}\)
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 \(A\kern{-.8pt}B^{2} = 20^{2} + 21^{2} - 2 ⋅ 20 ⋅ 21 ⋅ \cos(57\degree) = 383{,}503... \text{.}\)
1p
○
\(A\kern{-.8pt}B = \sqrt{383{,}503...} ≈ 19{,}6 \text{.}\)
1p
3p
Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 27 \text{,}\) \(A\kern{-.8pt}B = 19\) en \(\angle A = 106\degree \text{.}\)
Bereken de lengte van zijde \(B\kern{-.8pt}C \text{.}\)
Rond indien nodig af op één decimaal.
○
De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(B\kern{-.8pt}C^{2} = A\kern{-.8pt}C^{2} + A\kern{-.8pt}B^{2} - 2 ⋅ A\kern{-.8pt}C ⋅ A\kern{-.8pt}B ⋅ \cos(\angle A) \text{.}\)
1p
○
Dus \(B\kern{-.8pt}C^{2} = 27^{2} + 19^{2} - 2 ⋅ 27 ⋅ 19 ⋅ \cos(106\degree) = 1372{,}803... \text{.}\)
1p
○
\(B\kern{-.8pt}C = \sqrt{1372{,}803...} ≈ 37{,}1 \text{.}\)
1p
3p
Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 11 \text{,}\) \(P\kern{-.8pt}R = 16\) en \(\angle P = 31\degree \text{.}\)
Bereken \(\angle \text{Q} \text{.}\)
Rond indien nodig af op één decimaal.
○
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
○
Daaruit volgt \(\sin(\angle Q) = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over Q\kern{-.8pt}R} = {16 ⋅ \sin(31\degree) \over 11} = 0{,}749... \text{.}\)
1p
○
Dit geeft \(\angle Q ≈ 48{,}5\degree\) of \(\angle Q ≈ 131{,}5\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle Q\) een scherpe hoek is, dus \(\angle Q ≈ 48{,}5\degree \text{.}\)
1p
3p
Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 15 \text{,}\) \(P\kern{-.8pt}R = 27\) en \(\angle P = 28\degree \text{.}\)
Bereken \(\angle \text{Q} \text{.}\)
Rond indien nodig af op één decimaal.
○
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
○
Daaruit volgt \(\sin(\angle Q) = {P\kern{-.8pt}R ⋅ \sin(\angle P) \over Q\kern{-.8pt}R} = {27 ⋅ \sin(28\degree) \over 15} = 0{,}845... \text{.}\)
1p
○
Dit geeft \(\angle Q ≈ 57{,}7\degree\) of \(\angle Q ≈ 122{,}3\degree \text{.}\)
Uit de afbeelding volgt dat \(\angle Q\) een stompe hoek is, dus \(\angle Q ≈ 122{,}3\degree \text{.}\)
1p
3p
Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}B = 22 \text{,}\) \(\angle C = 59\degree\) en \(\angle A = 81\degree \text{.}\)
Bereken de lengte van zijde \(B\kern{-.8pt}C \text{.}\)
Rond indien nodig af op één decimaal.
○
De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}B \over \sin(\angle C)} = {B\kern{-.8pt}C \over \sin(\angle A)} = {A\kern{-.8pt}C \over \sin(\angle B)} \text{.}\)
1p
○
Dus \(B\kern{-.8pt}C = {A\kern{-.8pt}B ⋅ \sin(\angle A) \over \sin(\angle C)} = {22 ⋅ \sin(81\degree) \over \sin(59\degree)} \text{.}\)
1p
○
\(B\kern{-.8pt}C ≈ 25{,}3 \text{.}\)
1p
3p
Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(A\kern{-.8pt}C = 25 \text{,}\) \(\angle B = 42\degree\) en \(\angle C = 93\degree \text{.}\)
Bereken de lengte van zijde \(A\kern{-.8pt}B \text{.}\)
Rond indien nodig af op één decimaal.
○
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 \(A\kern{-.8pt}B = {A\kern{-.8pt}C ⋅ \sin(\angle C) \over \sin(\angle B)} = {25 ⋅ \sin(93\degree) \over \sin(42\degree)} \text{.}\)
1p
○
\(A\kern{-.8pt}B ≈ 37{,}3 \text{.}\)
1p
4p
Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(Q\kern{-.8pt}R = 31 \text{,}\) \(\angle R = 48\degree\) en \(\angle Q = 55\degree \text{.}\)
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 \(\angle P = 180\degree - \angle R - \angle Q = 180\degree - 48\degree - 55\degree = 77\degree \text{.}\)
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 \(P\kern{-.8pt}Q = {Q\kern{-.8pt}R ⋅ \sin(\angle R) \over \sin(\angle P)} = {31 ⋅ \sin(48\degree) \over \sin(77\degree)} \text{.}\)
1p
○
\(P\kern{-.8pt}Q ≈ 23{,}6 \text{.}\)
1p
4p
Gegeven is \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) met \(K\kern{-.8pt}M = 33 \text{,}\) \(\angle K = 25\degree\) en \(\angle M = 47\degree \text{.}\)
Bereken de lengte van zijde \(L\kern{-.8pt}M \text{.}\)
Rond indien nodig af op één decimaal.
○
Uit \(\angle K + \angle L + \angle M = 180\degree\) volgt \(\angle L = 180\degree - \angle K - \angle M = 180\degree - 25\degree - 47\degree = 108\degree \text{.}\)
1p
○
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
○
Dus \(L\kern{-.8pt}M = {K\kern{-.8pt}M ⋅ \sin(\angle K) \over \sin(\angle L)} = {33 ⋅ \sin(25\degree) \over \sin(108\degree)} \text{.}\)
1p
○
\(L\kern{-.8pt}M ≈ 14{,}7 \text{.}\)
1p