MATHHX A

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 11.2.1 (Pythagoras’ læresætning)
    Betragt den retvinklede trekant:

    (-tikz- diagram)

    Der gælder:

    \[a^2+b^2=c^2\]

  • Sætning 11.2.2

    Betragt den retvinklede trekant:

    (-tikz- diagram)

    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):

    (image)

    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):

    (image)

    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.

  • a) Regn \(\sin (60\degree )\) i GeoGebra

  • b) Bestem \(v\) når \(\sin (v)=0{,}7\) i GeoGebra.

Løsning 11.2.1

  • a) \(\sin (60\degree )=0{,}87\)

  • b) \(v=44{,}4\degree \)

  • Eksempel 11.2.3
    Betragt trekanten

    (-tikz- diagram)

    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:

(-tikz- diagram)

  • a) Beregn sidelængden \(a\)

  • b) Beregn resten af siderne og vinklerne

Løsning 11.2.2

  • a) \(a=6{,}76\)

  • b) Den modstående katete er \(1{,}81\) og den sidste vinklen er \(75\degree \)

  • Eksempel 11.2.4
    Betragt trekanten:

    (-tikz- diagram)

    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:

(-tikz- diagram)

  • a) Bestem vinklen \(v\)

  • b) Bestem resten af vinklerne og siderne i trekanten.

Løsning 11.2.3

  • a) \(v=53{,}13\degree \)

  • b) Den sidste vinkel er \(36{,}87\degree \) og hypotenusen er \(5\)

Øvelse 11.2.4

Vi vender nu tilbage til figuren fra øvelse 11.1.4:

(-tikz- diagram)

  • 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:

(-tikz- diagram)

  • a) Gør rede for at \(F_1=F\cdot \sin (\theta )\)

  • b) Gør rede for at \(F_2=F\cdot \cos (\theta )\)

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:

(-tikz- diagram)

Den røde linje er parallel med y-aksen og går igennem \((1,0)\).

Øvelse 11.2.6

Du skal bevise ovenstående påstand.

  • a) Vis at tangens kan findes som det markerede stykke på tegningen oven over.

Løsning 11.2.6

  • a) Betragt den grønne trekant:
    (-tikz- diagram)

    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:

    (-tikz- diagram)

    Tag den røde trekant og spejl den i den lodrette side, så fås en en trekant hvor alle vinkler er \(60\degree \):

    (-tikz- diagram)

    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 \(\sin (60\degree )\)

    \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:

    \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 \(\tan (60\degree )=\sqrt {3}\)

Øvelse 11.2.7

Din tur.

  • a) Argumenter for resten af værdierne i tabellen.

Løsning 11.2.7

  • a) Spørg mig, hvis du er i tvivl.