Sinus- en cosinusregel

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

Invullen geeft \(32^2=29^2+28^2-2⋅29⋅28⋅\cos(\angle B)\)
dus \(1\,024=1\,625-1\,624⋅\cos(\angle B)\text{.}\)

Balansmethode geeft \(\cos(\angle B)={1\,024-1\,625 \over -1\,624}=0{,}370...\)

Hieruit volgt \(\angle B=\cos^{-1}(0{,}370...)≈68{,}3\degree\text{.}\)

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

Invullen geeft \(56^2=29^2+43^2-2⋅29⋅43⋅\cos(\angle M)\)
dus \(3\,136=2\,690-2\,494⋅\cos(\angle M)\text{.}\)

Balansmethode geeft \(\cos(\angle M)={3\,136-2\,690 \over -2\,494}=-0{,}178...\)

Hieruit volgt \(\angle M=\cos^{-1}(-0{,}178...)≈100{,}3\degree\text{.}\)

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

Dus \(LM^2=10^2+13^2-2⋅10⋅13⋅\cos(76\degree)=206{,}100...\text{.}\)

\(LM=\sqrt{206{,}100...}≈14{,}4\text{.}\)

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

Dus \(AB^2=36^2+21^2-2⋅36⋅21⋅\cos(107\degree)=2179{,}066...\text{.}\)

\(AB=\sqrt{2179{,}066...}≈46{,}7\text{.}\)

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

Daaruit volgt \(\sin(\angle L)={KM⋅\sin(\angle K) \over LM}={18⋅\sin(40\degree) \over 12}=0{,}964...\text{.}\)

Dit geeft \(\angle L≈74{,}6\degree\) of \(\angle L≈105{,}4\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle L\) een scherpe hoek is, dus \(\angle L≈74{,}6\degree\text{.}\)

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

Daaruit volgt \(\sin(\angle R)={PQ⋅\sin(\angle Q) \over PR}={24⋅\sin(30\degree) \over 13}=0{,}923...\text{.}\)

Dit geeft \(\angle R≈67{,}4\degree\) of \(\angle R≈112{,}6\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle R\) een stompe hoek is, dus \(\angle R≈112{,}6\degree\text{.}\)

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

Dus \(AC={BC⋅\sin(\angle B) \over \sin(\angle A)}={41⋅\sin(88\degree) \over \sin(65\degree)}\text{.}\)

\(AC≈45{,}2\text{.}\)

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

Dus \(QR={PQ⋅\sin(\angle P) \over \sin(\angle R)}={15⋅\sin(105\degree) \over \sin(29\degree)}\text{.}\)

\(QR≈29{,}9\text{.}\)

Uit \(\angle A+\angle B+\angle C=180\degree\) volgt \(\angle B=180\degree-\angle A-\angle C=180\degree-46\degree-61\degree=73\degree\text{.}\)

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

Dus \(BC={AC⋅\sin(\angle A) \over \sin(\angle B)}={37⋅\sin(46\degree) \over \sin(73\degree)}\text{.}\)

\(BC≈27{,}8\text{.}\)

Uit \(\angle K+\angle L+\angle M=180\degree\) volgt \(\angle L=180\degree-\angle K-\angle M=180\degree-40\degree-25\degree=115\degree\text{.}\)

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

Dus \(LM={KM⋅\sin(\angle K) \over \sin(\angle L)}={25⋅\sin(40\degree) \over \sin(115\degree)}\text{.}\)

\(LM≈17{,}7\text{.}\)