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 \(34^2=16^2+30^2-2⋅16⋅30⋅\cos(\angle R)\)
dus \(1\,156=1\,156-960⋅\cos(\angle R)\text{.}\)

Balansmethode geeft \(\cos(\angle R)={1\,156-1\,156 \over -960}=-0\)

Hieruit volgt \(\angle R=\cos^{-1}(-0)=90{,}0\degree\text{.}\)

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 \(26^2=20^2+12^2-2⋅20⋅12⋅\cos(\angle L)\)
dus \(676=544-480⋅\cos(\angle L)\text{.}\)

Balansmethode geeft \(\cos(\angle L)={676-544 \over -480}=-0{,}275\)

Hieruit volgt \(\angle L=\cos^{-1}(-0{,}275)≈106{,}0\degree\text{.}\)

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

Dus \(Q\kern{-.8pt}R^2=28^2+41^2-2⋅28⋅41⋅\cos(86\degree)=2304{,}839...\text{.}\)

\(Q\kern{-.8pt}R=\sqrt{2304{,}839...}≈48{,}0\text{.}\)

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 \(K\kern{-.8pt}L^2=16^2+26^2-2⋅16⋅26⋅\cos(99\degree)=1062{,}153...\text{.}\)

\(K\kern{-.8pt}L=\sqrt{1062{,}153...}≈32{,}6\text{.}\)

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

Daaruit volgt \(\sin(\angle P)={Q\kern{-.8pt}R⋅\sin(\angle R) \over P\kern{-.8pt}Q}={14⋅\sin(32\degree) \over 10}=0{,}741...\text{.}\)

Dit geeft \(\angle P≈47{,}9\degree\) of \(\angle P≈132{,}1\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle P\) een scherpe hoek is, dus \(\angle P≈47{,}9\degree\text{.}\)

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

Daaruit volgt \(\sin(\angle M)={K\kern{-.8pt}L⋅\sin(\angle L) \over K\kern{-.8pt}M}={24⋅\sin(25\degree) \over 16}=0{,}633...\text{.}\)

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

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 \(A\kern{-.8pt}B={A\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle B)}={21⋅\sin(61\degree) \over \sin(62\degree)}\text{.}\)

\(A\kern{-.8pt}B≈20{,}8\text{.}\)

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 \(A\kern{-.8pt}B={A\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle B)}={27⋅\sin(99\degree) \over \sin(40\degree)}\text{.}\)

\(A\kern{-.8pt}B≈41{,}5\text{.}\)

Uit \(\angle P+\angle Q+\angle R=180\degree\) volgt \(\angle Q=180\degree-\angle P-\angle R=180\degree-52\degree-64\degree=64\degree\text{.}\)

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

Dus \(Q\kern{-.8pt}R={P\kern{-.8pt}R⋅\sin(\angle P) \over \sin(\angle Q)}={19⋅\sin(52\degree) \over \sin(64\degree)}\text{.}\)

\(Q\kern{-.8pt}R≈16{,}7\text{.}\)

Uit \(\angle R+\angle P+\angle Q=180\degree\) volgt \(\angle P=180\degree-\angle R-\angle Q=180\degree-39\degree-25\degree=116\degree\text{.}\)

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 \(P\kern{-.8pt}Q={Q\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle P)}={57⋅\sin(39\degree) \over \sin(116\degree)}\text{.}\)

\(P\kern{-.8pt}Q≈39{,}9\text{.}\)