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 \(39^2=26^2+33^2-2⋅26⋅33⋅\cos(\angle A)\)
dus \(1\,521=1\,765-1\,716⋅\cos(\angle A)\text{.}\)

Balansmethode geeft \(\cos(\angle A)={1\,521-1\,765 \over -1\,716}=0{,}142...\)

Hieruit volgt \(\angle A=\cos^{-1}(0{,}142...)≈81{,}8\degree\text{.}\)

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

Invullen geeft \(51^2=41^2+29^2-2⋅41⋅29⋅\cos(\angle B)\)
dus \(2\,601=2\,522-2\,378⋅\cos(\angle B)\text{.}\)

Balansmethode geeft \(\cos(\angle B)={2\,601-2\,522 \over -2\,378}=-0{,}033...\)

Hieruit volgt \(\angle B=\cos^{-1}(-0{,}033...)≈91{,}9\degree\text{.}\)

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 \(K\kern{-.8pt}M^2=28^2+24^2-2⋅28⋅24⋅\cos(88\degree)=1313{,}095...\text{.}\)

\(K\kern{-.8pt}M=\sqrt{1313{,}095...}≈36{,}2\text{.}\)

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

Dus \(B\kern{-.8pt}C^2=22^2+30^2-2⋅22⋅30⋅\cos(111\degree)=1857{,}045...\text{.}\)

\(B\kern{-.8pt}C=\sqrt{1857{,}045...}≈43{,}1\text{.}\)

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

Daaruit volgt \(\sin(\angle R)={P\kern{-.8pt}Q⋅\sin(\angle Q) \over P\kern{-.8pt}R}={11⋅\sin(34\degree) \over 8}=0{,}768...\text{.}\)

Dit geeft \(\angle R≈50{,}3\degree\) of \(\angle R≈129{,}7\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle R\) een scherpe hoek is, dus \(\angle R≈50{,}3\degree\text{.}\)

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 \(\sin(\angle L)={K\kern{-.8pt}M⋅\sin(\angle K) \over L\kern{-.8pt}M}={26⋅\sin(34\degree) \over 16}=0{,}908...\text{.}\)

Dit geeft \(\angle L≈65{,}3\degree\) of \(\angle L≈114{,}7\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle L\) een stompe hoek is, dus \(\angle L≈114{,}7\degree\text{.}\)

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

Dus \(A\kern{-.8pt}B={A\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle B)}={35⋅\sin(79\degree) \over \sin(57\degree)}\text{.}\)

\(A\kern{-.8pt}B≈41{,}0\text{.}\)

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 \(K\kern{-.8pt}L={K\kern{-.8pt}M⋅\sin(\angle M) \over \sin(\angle L)}={34⋅\sin(95\degree) \over \sin(53\degree)}\text{.}\)

\(K\kern{-.8pt}L≈42{,}4\text{.}\)

Uit \(\angle M+\angle K+\angle L=180\degree\) volgt \(\angle K=180\degree-\angle M-\angle L=180\degree-33\degree-58\degree=89\degree\text{.}\)

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 \(K\kern{-.8pt}L={L\kern{-.8pt}M⋅\sin(\angle M) \over \sin(\angle K)}={20⋅\sin(33\degree) \over \sin(89\degree)}\text{.}\)

\(K\kern{-.8pt}L≈10{,}9\text{.}\)

Uit \(\angle L+\angle M+\angle K=180\degree\) volgt \(\angle M=180\degree-\angle L-\angle K=180\degree-27\degree-31\degree=122\degree\text{.}\)

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 \(K\kern{-.8pt}M={K\kern{-.8pt}L⋅\sin(\angle L) \over \sin(\angle M)}={44⋅\sin(27\degree) \over \sin(122\degree)}\text{.}\)

\(K\kern{-.8pt}M≈23{,}6\text{.}\)