Sinus- en cosinusregel

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 \(13^2=12^2+10^2-2⋅12⋅10⋅\cos(\angle P)\)
dus \(169=244-240⋅\cos(\angle P)\text{.}\)

Balansmethode geeft \(\cos(\angle P)={169-244 \over -240}=0{,}312...\)

Hieruit volgt \(\angle P=\cos^{-1}(0{,}312...)≈71{,}8\degree\text{.}\)

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

Invullen geeft \(30^2=22^2+15^2-2⋅22⋅15⋅\cos(\angle L)\)
dus \(900=709-660⋅\cos(\angle L)\text{.}\)

Balansmethode geeft \(\cos(\angle L)={900-709 \over -660}=-0{,}289...\)

Hieruit volgt \(\angle L=\cos^{-1}(-0{,}289...)≈106{,}8\degree\text{.}\)

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

Dus \(P\kern{-.8pt}Q^2=30^2+30^2-2⋅30⋅30⋅\cos(87\degree)=1705{,}795...\text{.}\)

\(P\kern{-.8pt}Q=\sqrt{1705{,}795...}≈41{,}3\text{.}\)

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 \(K\kern{-.8pt}M^2=20^2+15^2-2⋅20⋅15⋅\cos(96\degree)=687{,}717...\text{.}\)

\(K\kern{-.8pt}M=\sqrt{687{,}717...}≈26{,}2\text{.}\)

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 \(\sin(\angle Q)={P\kern{-.8pt}R⋅\sin(\angle P) \over Q\kern{-.8pt}R}={24⋅\sin(33\degree) \over 14}=0{,}933...\text{.}\)

Dit geeft \(\angle Q≈69{,}0\degree\) of \(\angle Q≈111{,}0\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle Q\) een scherpe hoek is, dus \(\angle Q≈69{,}0\degree\text{.}\)

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

Daaruit volgt \(\sin(\angle C)={A\kern{-.8pt}B⋅\sin(\angle B) \over A\kern{-.8pt}C}={15⋅\sin(25\degree) \over 10}=0{,}633...\text{.}\)

Dit geeft \(\angle C≈39{,}3\degree\) of \(\angle C≈140{,}7\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle C\) een stompe hoek is, dus \(\angle C≈140{,}7\degree\text{.}\)

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

Dus \(Q\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle P) \over \sin(\angle R)}={28⋅\sin(59\degree) \over \sin(58\degree)}\text{.}\)

\(Q\kern{-.8pt}R≈28{,}3\text{.}\)

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

Dus \(Q\kern{-.8pt}R={P\kern{-.8pt}Q⋅\sin(\angle P) \over \sin(\angle R)}={24⋅\sin(91\degree) \over \sin(29\degree)}\text{.}\)

\(Q\kern{-.8pt}R≈49{,}5\text{.}\)

Uit \(\angle C+\angle A+\angle B=180\degree\) volgt \(\angle A=180\degree-\angle C-\angle B=180\degree-64\degree-58\degree=58\degree\text{.}\)

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 \(A\kern{-.8pt}B={B\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle A)}={25⋅\sin(64\degree) \over \sin(58\degree)}\text{.}\)

\(A\kern{-.8pt}B≈26{,}5\text{.}\)

Uit \(\angle B+\angle C+\angle A=180\degree\) volgt \(\angle C=180\degree-\angle B-\angle A=180\degree-42\degree-35\degree=103\degree\text{.}\)

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 \(A\kern{-.8pt}C={A\kern{-.8pt}B⋅\sin(\angle B) \over \sin(\angle C)}={30⋅\sin(42\degree) \over \sin(103\degree)}\text{.}\)

\(A\kern{-.8pt}C≈20{,}6\text{.}\)