Sinus- en cosinusregel

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

Invullen geeft \(56^2=30^2+49^2-2⋅30⋅49⋅\cos(\angle R)\)
dus \(3\,136=3\,301-2\,940⋅\cos(\angle R)\text{.}\)

Balansmethode geeft \(\cos(\angle R)={3\,136-3\,301 \over -2\,940}=0{,}056...\)

Hieruit volgt \(\angle R=\cos^{-1}(0{,}056...)≈86{,}8\degree\text{.}\)

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

Invullen geeft \(27^2=17^2+17^2-2⋅17⋅17⋅\cos(\angle P)\)
dus \(729=578-578⋅\cos(\angle P)\text{.}\)

Balansmethode geeft \(\cos(\angle P)={729-578 \over -578}=-0{,}261...\)

Hieruit volgt \(\angle P=\cos^{-1}(-0{,}261...)≈105{,}1\degree\text{.}\)

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

Dus \(K\kern{-.8pt}M^2=23^2+23^2-2⋅23⋅23⋅\cos(62\degree)=561{,}299...\text{.}\)

\(K\kern{-.8pt}M=\sqrt{561{,}299...}≈23{,}7\text{.}\)

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

Dus \(K\kern{-.8pt}L^2=29^2+38^2-2⋅29⋅38⋅\cos(109\degree)=3002{,}552...\text{.}\)

\(K\kern{-.8pt}L=\sqrt{3002{,}552...}≈54{,}8\text{.}\)

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

Daaruit volgt \(\sin(\angle K)={L\kern{-.8pt}M⋅\sin(\angle M) \over K\kern{-.8pt}L}={13⋅\sin(47\degree) \over 10}=0{,}950...\text{.}\)

Dit geeft \(\angle K≈71{,}9\degree\) of \(\angle K≈108{,}1\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle K\) een scherpe hoek is, dus \(\angle K≈71{,}9\degree\text{.}\)

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

Daaruit volgt \(\sin(\angle A)={B\kern{-.8pt}C⋅\sin(\angle C) \over A\kern{-.8pt}B}={22⋅\sin(28\degree) \over 13}=0{,}794...\text{.}\)

Dit geeft \(\angle A≈52{,}6\degree\) of \(\angle A≈127{,}4\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle A\) een stompe hoek is, dus \(\angle A≈127{,}4\degree\text{.}\)

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

Dus \(K\kern{-.8pt}M={L\kern{-.8pt}M⋅\sin(\angle L) \over \sin(\angle K)}={17⋅\sin(89\degree) \over \sin(46\degree)}\text{.}\)

\(K\kern{-.8pt}M≈23{,}6\text{.}\)

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

Dus \(K\kern{-.8pt}M={L\kern{-.8pt}M⋅\sin(\angle L) \over \sin(\angle K)}={27⋅\sin(103\degree) \over \sin(32\degree)}\text{.}\)

\(K\kern{-.8pt}M≈49{,}6\text{.}\)

Uit \(\angle C+\angle A+\angle B=180\degree\) volgt \(\angle A=180\degree-\angle C-\angle B=180\degree-50\degree-59\degree=71\degree\text{.}\)

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 \(A\kern{-.8pt}B={B\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle A)}={13⋅\sin(50\degree) \over \sin(71\degree)}\text{.}\)

\(A\kern{-.8pt}B≈10{,}5\text{.}\)

Uit \(\angle L+\angle M+\angle K=180\degree\) volgt \(\angle M=180\degree-\angle L-\angle K=180\degree-39\degree-49\degree=92\degree\text{.}\)

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

Dus \(K\kern{-.8pt}M={K\kern{-.8pt}L⋅\sin(\angle L) \over \sin(\angle M)}={19⋅\sin(39\degree) \over \sin(92\degree)}\text{.}\)

\(K\kern{-.8pt}M≈12{,}0\text{.}\)