Sinus- en cosinusregel

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

Invullen geeft \(44^2=29^2+41^2-2⋅29⋅41⋅\cos(\angle M)\)
dus \(1\,936=2\,522-2\,378⋅\cos(\angle M)\text{.}\)

Balansmethode geeft \(\cos(\angle M)={1\,936-2\,522 \over -2\,378}=0{,}246...\)

Hieruit volgt \(\angle M=\cos^{-1}(0{,}246...)≈75{,}7\degree\text{.}\)

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 \(65^2=54^2+29^2-2⋅54⋅29⋅\cos(\angle P)\)
dus \(4\,225=3\,757-3\,132⋅\cos(\angle P)\text{.}\)

Balansmethode geeft \(\cos(\angle P)={4\,225-3\,757 \over -3\,132}=-0{,}149...\)

Hieruit volgt \(\angle P=\cos^{-1}(-0{,}149...)≈98{,}6\degree\text{.}\)

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

Dus \(A\kern{-.8pt}B^2=20^2+18^2-2⋅20⋅18⋅\cos(87\degree)=686{,}318...\text{.}\)

\(A\kern{-.8pt}B=\sqrt{686{,}318...}≈26{,}2\text{.}\)

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

Dus \(A\kern{-.8pt}B^2=24^2+17^2-2⋅24⋅17⋅\cos(96\degree)=950{,}295...\text{.}\)

\(A\kern{-.8pt}B=\sqrt{950{,}295...}≈30{,}8\text{.}\)

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 \(\sin(\angle A)={B\kern{-.8pt}C⋅\sin(\angle C) \over A\kern{-.8pt}B}={14⋅\sin(42\degree) \over 10}=0{,}936...\text{.}\)

Dit geeft \(\angle A≈69{,}5\degree\) of \(\angle A≈110{,}5\degree\text{.}\)
Uit de afbeelding volgt dat \(\angle A\) een scherpe hoek is, dus \(\angle A≈69{,}5\degree\text{.}\)

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

Daaruit volgt \(\sin(\angle Q)={P\kern{-.8pt}R⋅\sin(\angle P) \over Q\kern{-.8pt}R}={12⋅\sin(32\degree) \over 7}=0{,}908...\text{.}\)

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

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 \(A\kern{-.8pt}B={A\kern{-.8pt}C⋅\sin(\angle C) \over \sin(\angle B)}={28⋅\sin(83\degree) \over \sin(34\degree)}\text{.}\)

\(A\kern{-.8pt}B≈49{,}7\text{.}\)

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 \(P\kern{-.8pt}R={Q\kern{-.8pt}R⋅\sin(\angle Q) \over \sin(\angle P)}={20⋅\sin(96\degree) \over \sin(26\degree)}\text{.}\)

\(P\kern{-.8pt}R≈45{,}4\text{.}\)

Uit \(\angle A+\angle B+\angle C=180\degree\) volgt \(\angle B=180\degree-\angle A-\angle C=180\degree-54\degree-49\degree=77\degree\text{.}\)

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 \(B\kern{-.8pt}C={A\kern{-.8pt}C⋅\sin(\angle A) \over \sin(\angle B)}={19⋅\sin(54\degree) \over \sin(77\degree)}\text{.}\)

\(B\kern{-.8pt}C≈15{,}8\text{.}\)

Uit \(\angle R+\angle P+\angle Q=180\degree\) volgt \(\angle P=180\degree-\angle R-\angle Q=180\degree-25\degree-47\degree=108\degree\text{.}\)

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 \(P\kern{-.8pt}Q={Q\kern{-.8pt}R⋅\sin(\angle R) \over \sin(\angle P)}={38⋅\sin(25\degree) \over \sin(108\degree)}\text{.}\)

\(P\kern{-.8pt}Q≈16{,}9\text{.}\)