Sinus- en cosinusregel

De cosinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(B\kern{-.8pt}C^2=A\kern{-.8pt}C^2+A\kern{-.8pt}B^2-2⋅A\kern{-.8pt}C⋅A\kern{-.8pt}B⋅\cos(\angle A)\text{.}\)

Invullen geeft \(39^2=30^2+29^2-2⋅30⋅29⋅\cos(\angle A)\)
dus \(1\,521=1\,741-1\,740⋅\cos(\angle A)\text{.}\)

Balansmethode geeft \(\cos(\angle A)={1\,521-1\,741 \over -1\,740}=0{,}126...\)

Hieruit volgt \(\angle A=\cos^{-1}(0{,}126...)≈82{,}7\degree\text{.}\)

De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(L\kern{-.8pt}M^2=K\kern{-.8pt}M^2+K\kern{-.8pt}L^2-2⋅K\kern{-.8pt}M⋅K\kern{-.8pt}L⋅\cos(\angle K)\text{.}\)

Invullen geeft \(22^2=11^2+15^2-2⋅11⋅15⋅\cos(\angle K)\)
dus \(484=346-330⋅\cos(\angle K)\text{.}\)

Balansmethode geeft \(\cos(\angle K)={484-346 \over -330}=-0{,}418...\)

Hieruit volgt \(\angle K=\cos^{-1}(-0{,}418...)≈114{,}7\degree\text{.}\)

De cosinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \(L\kern{-.8pt}M^2=K\kern{-.8pt}M^2+K\kern{-.8pt}L^2-2⋅K\kern{-.8pt}M⋅K\kern{-.8pt}L⋅\cos(\angle K)\text{.}\)

Dus \(L\kern{-.8pt}M^2=12^2+20^2-2⋅12⋅20⋅\cos(89\degree)=535{,}622...\text{.}\)

\(L\kern{-.8pt}M=\sqrt{535{,}622...}≈23{,}1\text{.}\)

De cosinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(Q\kern{-.8pt}R^2=P\kern{-.8pt}R^2+P\kern{-.8pt}Q^2-2⋅P\kern{-.8pt}R⋅P\kern{-.8pt}Q⋅\cos(\angle P)\text{.}\)

Dus \(Q\kern{-.8pt}R^2=18^2+29^2-2⋅18⋅29⋅\cos(101\degree)=1364{,}204...\text{.}\)

\(Q\kern{-.8pt}R=\sqrt{1364{,}204...}≈36{,}9\text{.}\)

De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({Q\kern{-.8pt}R \over \sin(\angle P)}={P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}\text{.}\)

Daaruit volgt \(\sin(\angle Q)={P\kern{-.8pt}R⋅\sin(\angle P) \over Q\kern{-.8pt}R}={28⋅\sin(26\degree) \over 18}=0{,}681...\text{.}\)

Dit geeft \(\angle Q≈43{,}0\degree\) of \(\angle Q≈137{,}0\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle Q\) een scherpe hoek is, dus \(\angle Q≈43{,}0\degree\text{.}\)

De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({L\kern{-.8pt}M \over \sin(\angle K)}={K\kern{-.8pt}M \over \sin(\angle L)}={K\kern{-.8pt}L \over \sin(\angle M)}\text{.}\)

Daaruit volgt \(\sin(\angle L)={K\kern{-.8pt}M⋅\sin(\angle K) \over L\kern{-.8pt}M}={16⋅\sin(42\degree) \over 11}=0{,}973...\text{.}\)

Dit geeft \(\angle L≈76{,}7\degree\) of \(\angle L≈103{,}3\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle L\) een stompe hoek is, dus \(\angle L≈103{,}3\degree\text{.}\)

De sinusregel in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \({P\kern{-.8pt}R \over \sin(\angle Q)}={P\kern{-.8pt}Q \over \sin(\angle R)}={Q\kern{-.8pt}R \over \sin(\angle P)}\text{.}\)

Dus \(P\kern{-.8pt}Q={P\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle Q)}={23⋅\sin(88\degree) \over \sin(54\degree)}\text{.}\)

\(P\kern{-.8pt}Q≈28{,}4\text{.}\)

De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({A\kern{-.8pt}B \over \sin(\angle C)}={B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}\text{.}\)

Dus \(B\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle A) \over \sin(\angle C)}={10⋅\sin(119\degree) \over \sin(31\degree)}\text{.}\)

\(B\kern{-.8pt}C≈17{,}0\text{.}\)

Uit \(\angle A+\angle B+\angle C=180\degree\) volgt \(\angle B=180\degree-\angle A-\angle C=180\degree-54\degree-55\degree=71\degree\text{.}\)

De sinusregel in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \({B\kern{-.8pt}C \over \sin(\angle A)}={A\kern{-.8pt}C \over \sin(\angle B)}={A\kern{-.8pt}B \over \sin(\angle C)}\text{.}\)

Dus \(B\kern{-.8pt}C={A\kern{-.8pt}C⋅\sin(\angle A) \over \sin(\angle B)}={14⋅\sin(54\degree) \over \sin(71\degree)}\text{.}\)

\(B\kern{-.8pt}C≈12{,}0\text{.}\)

Uit \(\angle M+\angle K+\angle L=180\degree\) volgt \(\angle K=180\degree-\angle M-\angle L=180\degree-32\degree-28\degree=120\degree\text{.}\)

De sinusregel in \(\triangle K\kern{-.8pt}L\kern{-.8pt}M\) geeft \({K\kern{-.8pt}L \over \sin(\angle M)}={L\kern{-.8pt}M \over \sin(\angle K)}={K\kern{-.8pt}M \over \sin(\angle L)}\text{.}\)

Dus \(K\kern{-.8pt}L={L\kern{-.8pt}M⋅\sin(\angle M) \over \sin(\angle K)}={25⋅\sin(32\degree) \over \sin(120\degree)}\text{.}\)

\(K\kern{-.8pt}L≈15{,}3\text{.}\)