MATHHX A
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11.2 Retvinklede trekanter
En trekant som har en ret vinkel kaldes en retvinklet trekant. Følgende sætninger burde være kendt fra folkeskolen (måske formuleret lidt anderledes).
Sætning 11.2.2
Betragt den retvinklede trekant:
Der gælder:
\[\cos (v)=\frac {\text {hosliggende katete}}{\text {hypotenusen}}\]
\[\sin (v)=\frac {\text {modstående katete}}{\text {hypotenusen}}\]
\[\tan (v)=\frac {\text {modstående katete}}{\text {hosliggende katete}}\]
Skal ovenstående sætning anvendes i praksis er det nødvendigt at kunne regne med cosinus, sinus og tangens:
Eksempel 11.2.1
Vi vil regne \(\cos (60\degree )\). vi åbner et algebravindue i GeoGebra og skriver cos(60) (læg mærke til at vi ikke behøver at skrive grader (\(\degree
\)) — det klarer GeoGebra for os):
Vi ser at \(\cos (60\degree )=0{,}5\)
Nogle gange kender vi f.eks. cosinusværdien og vil gerne regne tilbage til vinklen. Vi har derfor brug for den omvendte funktion til cosinus og den kaldes \(\cos ^{-1}\) eller arcus cosinus . Man kan regne den i GeoGebra
med kommandoen acos(x) .
Eksempel 11.2.2
Lad os sige at vi ved at \(\cos (v)=0{,}5\) og vi gerne vil regne tilbage til vinklen. Vi taster acosd(0.5) i et algebravindue i GeoGebra (d’et i ”acosd” står for
”degrees”, så vi får vinken i grader):
Vi ser at \(\cos ^{-1}(0{,}5)=60\degree \). Easy money. Læg mærke til, at det er GeoGebra, som skriver ”\(\cos ^{-1}(0.5)\). Vi har indtastet acosd(0.5)
Der findes selvfølgelig også omvendte funktioner til de andre trigonometriske funktioner og de hedder det tilsvarende i GeoGebra. F.eks. hedder \(\sin ^{-1}\) asind (når man vil have vinklen i
grader) i GeoGebra.
Øvelse 11.2.1
Du skal nu selv prøve.
Eksempel 11.2.3
Betragt trekanten
Vi vi beregne længden af hypotenusen \(c\). Vi finder en formel fra sætning 11.2.2 , som indeholder de to størrelser vi kender (vinkel og modstående katete)
\[\sin (v)=\frac {\text {modstående katete}}{\text {hypotenusen}}\]
Vi indsætter vores oplysninger:
\[\sin (30\degree )=\frac {2}{\text {c}}.\]
Vi isolerer \(c\):
\[\text {c}=\frac {2}{\sin (30\degree )},\]
og regner resultatet i GeoGebra/lommeregner:
\[\text {c}=4.\]
Længden af hypotenusen er altså \(4\).
Øvelse 11.2.2
Betragt den retvinklede trekant:
Eksempel 11.2.4
Betragt trekanten:
Vi vil bestemme \(v\). Vi finder den formel som indeholder de størrelser vi har:
\[\cos (v)=\frac {\text {hosliggende katete}}{\text {hypotenusen}}\]
Vi indsætter
\[\cos (v)=\frac {2}{4},\]
og regner
\[\cos (v)=0{,}5,\]
og isolerer \(v\):
\[v=\cos ^{-1}(0{,}5),\]
og får:
\[v=60\degree ,\]
og det passer da meget godt med tegningen.
Øvelse 11.2.3
Betragt trekanten:
Øvelse 11.2.4
Vi vender nu tilbage til figuren fra øvelse 11.1.4 :
a) Gør rede for at \(BD=AD\cdot \sin (i)\). Skrivemåden \(BD\) betyder længden af siden fra \(B\) til \(D\), hvis du ikke allerede kunne regne det ud:-)
b) Gør rede for at \(AC=AD\cdot \sin (b)\)
c) Gør rede for at \(\frac {BD}{AC}=\frac {\sin (i)}{\sin (b)}\)
Løsning 11.2.4
a) Trekant \(ABD\) er retvinklet så vi kan regne \(\sin (i)=\frac {BD}{AD}\). Ved at gange med \(AD\) på begge sider fås det ønskede resultat.
b) Fås på tilsvarende måde som a) men bare ved at kigge på trekant \(ACD\).
c) Vi bruger resultaterne i a) og b):
\[\frac {BD}{AC}=\frac {AD\cdot \sin (i)}{AD\cdot \sin (b)}=\frac {\sin (i)}{\sin (b)}\]
Øvelse 11.2.5
Vi vender tilbage til figuren fra øvelse 11.1.4 :
Løsning 11.2.5
a) I den farvede retvinklede trekant gælder \(\sin (\theta )=\frac {F_1}{F}\). Ved at gange med \(F\) på begge sider fås det ønskede resultat.
b) I den farvede retvinklede trekant gælder \(\cos (\theta )=\frac {F_2}{F}\). Ved at gange med \(F\) på begge sider fås det ønskede resultat.
Ekstra
Man kan aflæse tangens i enhedscirklen på følgende måde:
Den røde linje er parallel med y-aksen og går igennem \((1,0)\).
Øvelse 11.2.6
Du skal bevise ovenstående påstand.
Løsning 11.2.6
a) Betragt den grønne trekant:
Vi bruger formlen
\[\tan (v)=\frac {\text {modstående katete}}{\text {hosliggende katete}}\]
på vinklen \(v\) og den grønne trekant. Det giver:
\[\tan (v)=\frac {d}{1}\]
Dvs.
\[d=\tan (v)\]
Der findes ”pæne” værdier for cosinus, sinus og tangens for udvalgte vinkler.
.
Grader
Radianer
\(\sin (v))\)
\(\cos (v)\)
\(\tan (v)\)
\(0^\circ \)
\(0\)
\(0\)
\(1\)
\(0\)
\(30^\circ \)
\(\frac {\pi }{6}\)
\(\frac {1}{2}\)
\(\frac {\sqrt {3}}{2}\)
\(\frac {1}{\sqrt {3}}\)
\(45^\circ \)
\(\frac {\pi }{4}\)
\(\frac {1}{\sqrt {2}}\)
\(\frac {1}{\sqrt {2}}\)
\(1\)
\(60^\circ \)
\(\frac {\pi }{3}\)
\(\frac {\sqrt {3}}{2}\)
\(\frac {1}{2}\)
\(\sqrt {3}\)
\(90^\circ \)
\(\frac {\pi }{2}\)
\(1\)
\(0\)
Udefineret
\(180^\circ \)
\(\pi \)
\(0\)
\(-1\)
\(0\)
\(270^\circ \)
\(\frac {3\pi }{2}\)
\(-1\)
\(0\)
Udefineret
\(360^\circ \)
\(2\pi \)
\(0\)
\(1\)
\(0\)
Tabel 11.1: Udvalgte værdier for sinus cosinus og tangens
Eksempel 11.2.5
Vi vil regne os frem til en af rækkerne i tabellen:
.
Grader
Radianer
\(\sin (v)\)
\(\cos (v)\)
\(\tan (v)\)
\(60^\circ \)
\(\frac {\pi }{3}\)
\(\frac {\sqrt {3}}{2}\)
\(\frac {1}{2}\)
\(\sqrt {3}\)
Da \(60\degree \) er \(\frac {1}{6}\) af \(360\degree \), må vinklen være på \(\frac {2\pi }{6}=\frac {\pi }{3}\) radianter. Nemt. Vi tegner nu vinklen ind i enhedscirklen:
Tag den røde trekant og spejl den i den lodrette side, så fås en en trekant hvor alle vinkler er \(60\degree \):
Grundlinjen i denne trekant må være
\[2\cdot \cos (60\degree )\]
og da den store trekant er ligevinklet er den også ligesidet derfor må den sidste side også være \(1\), så
\[2\cdot \cos (60\degree )=1\]
dvs.
\[\cos (60\degree )=\frac {1}{2}\]
Vi kan regne sinusværdien ud fra den røde trekant nu, hvor vi ved at \(\cos (60\degree )=\frac {1}{2}\). Vi bruger Pythagoras:
\[\left (\frac {1}{2}\right )^2+\sin (60\degree )^2=1^2\]
Vi finder \(\sin (60\degree )\)
\(\seteqnumber{0}{11.}{0}\)
\begin{align*}
\sin (60\degree ) & = \sqrt {1-\left (\frac {1}{2}\right )^2}\\ & = \sqrt {\frac {3}{4}}\\ & = \frac {\sqrt {3}}{2}
\end{align*}
Altså \(\sin (60\degree )=\frac {\sqrt {3}}{2}\).
Vi kan til sidst regne tangens:
\(\seteqnumber{0}{11.}{0}\)
\begin{align*}
\tan (60\degree ) & = \frac {\sin (60\degree )}{\cos (60\degree )}\\ & =\frac {\ \frac {\sqrt {3}}{2}\ }{\frac {1}{2}}\\ & = \sqrt {3}
\end{align*}
Altså \(\tan (60\degree )=\sqrt {3}\)