Sinus- en cosinusregel

a

De cosinusregel in \(\triangle ABC\) geeft \(AC^2=AB^2+BC^2-2⋅AB⋅BC⋅\cos(\angle B)\text{.}\)

1p

Invullen geeft \(13^2=10^2+11^2-2⋅10⋅11⋅\cos(\angle B)\)
dus \(169=221-220⋅\cos(\angle B)\text{.}\)

1p

Balansmethode geeft \(\cos(\angle B)={169-221 \over -220}=0{,}236...\)

1p

Hieruit volgt \(\angle B=\cos^{-1}(0{,}236...)≈76{,}3\degree\text{.}\)

1p

a

De cosinusregel in \(\triangle KLM\) geeft \(LM^2=KM^2+KL^2-2⋅KM⋅KL⋅\cos(\angle K)\text{.}\)

1p

Invullen geeft \(30^2=18^2+22^2-2⋅18⋅22⋅\cos(\angle K)\)
dus \(900=808-792⋅\cos(\angle K)\text{.}\)

1p

Balansmethode geeft \(\cos(\angle K)={900-808 \over -792}=-0{,}116...\)

1p

Hieruit volgt \(\angle K=\cos^{-1}(-0{,}116...)≈96{,}7\degree\text{.}\)

1p

a

De cosinusregel in \(\triangle KLM\) geeft \(KM^2=KL^2+LM^2-2⋅KL⋅LM⋅\cos(\angle L)\text{.}\)

1p

Dus \(KM^2=33^2+27^2-2⋅33⋅27⋅\cos(85\degree)=1662{,}688...\text{.}\)

1p

\(KM=\sqrt{1662{,}688...}≈40{,}8\text{.}\)

1p

a

De cosinusregel in \(\triangle KLM\) geeft \(LM^2=KM^2+KL^2-2⋅KM⋅KL⋅\cos(\angle K)\text{.}\)

1p

Dus \(LM^2=31^2+25^2-2⋅31⋅25⋅\cos(96\degree)=1748{,}019...\text{.}\)

1p

\(LM=\sqrt{1748{,}019...}≈41{,}8\text{.}\)

1p

a

De sinusregel in \(\triangle ABC\) geeft \({BC \over \sin(\angle A)}={AC \over \sin(\angle B)}={AB \over \sin(\angle C)}\text{.}\)

1p

Daaruit volgt \(\sin(\angle B)={AC⋅\sin(\angle A) \over BC}={15⋅\sin(44\degree) \over 11}=0{,}947...\text{.}\)

1p

Dit geeft \(\angle B≈71{,}3\degree\) of \(\angle B≈108{,}7\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle B\) een scherpe hoek is, dus \(\angle B≈71{,}3\degree\text{.}\)

1p

a

De sinusregel in \(\triangle PQR\) geeft \({PQ \over \sin(\angle R)}={QR \over \sin(\angle P)}={PR \over \sin(\angle Q)}\text{.}\)

1p

Daaruit volgt \(\sin(\angle P)={QR⋅\sin(\angle R) \over PQ}={10⋅\sin(31\degree) \over 6}=0{,}858...\text{.}\)

1p

Dit geeft \(\angle P≈59{,}1\degree\) of \(\angle P≈120{,}9\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle P\) een stompe hoek is, dus \(\angle P≈120{,}9\degree\text{.}\)

1p

a

De sinusregel in \(\triangle PQR\) geeft \({PQ \over \sin(\angle R)}={QR \over \sin(\angle P)}={PR \over \sin(\angle Q)}\text{.}\)

1p

Dus \(QR={PQ⋅\sin(\angle P) \over \sin(\angle R)}={22⋅\sin(68\degree) \over \sin(64\degree)}\text{.}\)

1p

\(QR≈22{,}7\text{.}\)

1p

a

De sinusregel in \(\triangle ABC\) geeft \({AC \over \sin(\angle B)}={AB \over \sin(\angle C)}={BC \over \sin(\angle A)}\text{.}\)

1p

Dus \(AB={AC⋅\sin(\angle C) \over \sin(\angle B)}={26⋅\sin(106\degree) \over \sin(29\degree)}\text{.}\)

1p

\(AB≈51{,}6\text{.}\)

1p

a

Uit \(\angle L+\angle M+\angle K=180\degree\) volgt \(\angle M=180\degree-\angle L-\angle K=180\degree-50\degree-65\degree=65\degree\text{.}\)

1p

De sinusregel in \(\triangle KLM\) geeft \({KM \over \sin(\angle L)}={KL \over \sin(\angle M)}={LM \over \sin(\angle K)}\text{.}\)

1p

Dus \(KM={KL⋅\sin(\angle L) \over \sin(\angle M)}={15⋅\sin(50\degree) \over \sin(65\degree)}\text{.}\)

1p

\(KM≈12{,}7\text{.}\)

1p

a

Uit \(\angle K+\angle L+\angle M=180\degree\) volgt \(\angle L=180\degree-\angle K-\angle M=180\degree-49\degree-27\degree=104\degree\text{.}\)

1p

De sinusregel in \(\triangle KLM\) geeft \({LM \over \sin(\angle K)}={KM \over \sin(\angle L)}={KL \over \sin(\angle M)}\text{.}\)

1p

Dus \(LM={KM⋅\sin(\angle K) \over \sin(\angle L)}={38⋅\sin(49\degree) \over \sin(104\degree)}\text{.}\)

1p

\(LM≈29{,}6\text{.}\)

1p