Sinus- en cosinusregel

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

Invullen geeft \(19^2=12^2+15^2-2⋅12⋅15⋅\cos(\angle C)\)
dus \(361=369-360⋅\cos(\angle C)\text{.}\)

Balansmethode geeft \(\cos(\angle C)={361-369 \over -360}=0{,}022...\)

Hieruit volgt \(\angle C=\cos^{-1}(0{,}022...)≈88{,}7\degree\text{.}\)

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

Invullen geeft \(34^2=28^2+16^2-2⋅28⋅16⋅\cos(\angle L)\)
dus \(1\,156=1\,040-896⋅\cos(\angle L)\text{.}\)

Balansmethode geeft \(\cos(\angle L)={1\,156-1\,040 \over -896}=-0{,}129...\)

Hieruit volgt \(\angle L=\cos^{-1}(-0{,}129...)≈97{,}4\degree\text{.}\)

De cosinusregel in \(\triangle PQR\) geeft \(QR^2=PR^2+PQ^2-2⋅PR⋅PQ⋅\cos(\angle P)\text{.}\)

Dus \(QR^2=25^2+25^2-2⋅25⋅25⋅\cos(56\degree)=551{,}008...\text{.}\)

\(QR=\sqrt{551{,}008...}≈23{,}5\text{.}\)

De cosinusregel in \(\triangle PQR\) geeft \(PR^2=PQ^2+QR^2-2⋅PQ⋅QR⋅\cos(\angle Q)\text{.}\)

Dus \(PR^2=29^2+29^2-2⋅29⋅29⋅\cos(125\degree)=2646{,}755...\text{.}\)

\(PR=\sqrt{2646{,}755...}≈51{,}4\text{.}\)

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

Daaruit volgt \(\sin(\angle Q)={PR⋅\sin(\angle P) \over QR}={26⋅\sin(31\degree) \over 14}=0{,}956...\text{.}\)

Dit geeft \(\angle Q≈73{,}0\degree\) of \(\angle Q≈107{,}0\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle Q\) een scherpe hoek is, dus \(\angle Q≈73{,}0\degree\text{.}\)

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

Daaruit volgt \(\sin(\angle B)={AC⋅\sin(\angle A) \over BC}={19⋅\sin(28\degree) \over 12}=0{,}743...\text{.}\)

Dit geeft \(\angle B≈48{,}0\degree\) of \(\angle B≈132{,}0\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle B\) een stompe hoek is, dus \(\angle B≈132{,}0\degree\text{.}\)

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

Dus \(KM={LM⋅\sin(\angle L) \over \sin(\angle K)}={29⋅\sin(87\degree) \over \sin(37\degree)}\text{.}\)

\(KM≈48{,}1\text{.}\)

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

Dus \(AC={BC⋅\sin(\angle B) \over \sin(\angle A)}={38⋅\sin(97\degree) \over \sin(52\degree)}\text{.}\)

\(AC≈47{,}9\text{.}\)

Uit \(\angle Q+\angle R+\angle P=180\degree\) volgt \(\angle R=180\degree-\angle Q-\angle P=180\degree-47\degree-55\degree=78\degree\text{.}\)

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

Dus \(PR={PQ⋅\sin(\angle Q) \over \sin(\angle R)}={29⋅\sin(47\degree) \over \sin(78\degree)}\text{.}\)

\(PR≈21{,}7\text{.}\)

Uit \(\angle L+\angle M+\angle K=180\degree\) volgt \(\angle M=180\degree-\angle L-\angle K=180\degree-25\degree-52\degree=103\degree\text{.}\)

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

Dus \(KM={KL⋅\sin(\angle L) \over \sin(\angle M)}={43⋅\sin(25\degree) \over \sin(103\degree)}\text{.}\)

\(KM≈18{,}7\text{.}\)