Lade Daten...
randRange(20, 140) randRange(20, 160 - ANGLE1) randRange(5, 25) shuffle([0, 1, 2]) [ KNOWN === 0 ? SIDELEN : null, KNOWN === 1 ? SIDELEN : null, KNOWN === 2 ? SIDELEN : null ] !!rand(2) [ KNOWN === 0 ? ANGLE1 : (UNKNOWN === 0 ^ GIVE_OPPOSITE ? null : ANGLE2), KNOWN === 1 ? ANGLE1 : (UNKNOWN === 1 ^ GIVE_OPPOSITE ? null : ANGLE2), KNOWN === 2 ? ANGLE1 : (UNKNOWN === 2 ^ GIVE_OPPOSITE ? null : ANGLE2) ] solveTriangle({ sides: SIDES.slice(), angles: ANGLES.slice(), sideLabels: SIDES.slice(), angleLabels: _.map(ANGLES, function(a) { return a == null ? a : a + "^\\circ"; }), vertexLabels: ["A", "B", "C"] }) ["BC", "AC", "AB"][UNKNOWN] roundTo(1,TRIANGLE.sides[UNKNOWN]) rand(2) ? randRange(-20, 20) : randRange(160, 200)

Bestimme UNKNOWN_MEASURE.

Runde die Antwort auf das nächstliegende Grad.

TRIANGLE = addTriangle(_.extend(TRIANGLE, { xPos: 1, yPos: 1, width: 10, height: 6, rot: ROTATION })); init({ range: [[0, TRIANGLE.width + 2], [0, TRIANGLE.height + 2]] }); TRIANGLE.draw();
SOLUTION

Benutze den Sinussatz:

\qquad \dfrac{BC}{\sin(m\angle A)} \quad = \quad \dfrac{AC}{\sin(m\angle B)} \quad = \quad \dfrac{AB}{\sin(m\angle C)}

Ergänze den fehlenden Winkel. (Die Summe aller Winkel in einem Dreieck ist immer 180^\circ.)

\qquad 180^\circ - ANGLE1^\circ - ANGLE2^\circ \quad = \quad \pink{TRIANGLE.angles[UNKNOWN]^\circ}

TRIANGLE.angleLabels[UNKNOWN] = "\\pink{" + TRIANGLE.angles[UNKNOWN] + "^\\circ}"; TRIANGLE.draw();

Aus dem Sinussatz können wir eine sinnvolle Beziehung zwischen den Winkeln und Seiten herstellen:

\qquad \dfrac{\pink{UNKNOWN_MEASURE}}{\sin( \pink{TRIANGLE.angles[UNKNOWN]^\circ})} \quad = \quad\dfrac{\blue{ TRIANGLE.sides[KNOWN]}}{\sin(\blue{ TRIANGLE.angles[KNOWN]^\circ})}

TRIANGLE.sideLabels[KNOWN] = "\\blue{" + TRIANGLE.sides[KNOWN] + "}"; TRIANGLE.angleLabels[KNOWN] = "\\blue{" + TRIANGLE.angles[KNOWN] + "^\\circ}"; TRIANGLE.sideLabels[UNKNOWN] = "\\pink{" + UNKNOWN_MEASURE + "}"; TRIANGLE.angleLabels[UNKNOWN] = "\\pink{" + TRIANGLE.angles[UNKNOWN] + "^\\circ}"; TRIANGLE.color = GRAY; TRIANGLE.draw();

Wir lösen nach der unbekannten Seite auf:

\qquad \pink{UNKNOWN_MEASURE} \quad = \quad \dfrac{\blue{TRIANGLE.sides[KNOWN]} \cdot \sin(\pink{TRIANGLE.angles[UNKNOWN]^\circ}) }{ \sin(\blue{TRIANGLE.angles[KNOWN]^\circ})}

Gerundet ist das Ergebnis:

\qquad \pink{UNKNOWN_MEASURE} \quad \approx \quad localeToFixed(SOLUTION,-1)

TRIANGLE.sideLabels[UNKNOWN] = "\\pink{" + localeToFixed(SOLUTION,-1) + "}"; TRIANGLE.draw();
randRange(5, 15, 3) shuffle([0, 1, 2]) solveTriangle({ sides: SIDES.slice(), angles: [null, null, null], sideLabels: SIDES.slice(), vertexLabels: ["A", "B", "C"] }) _.map(TRIANGLE.angles, round) "ABC"[UNKNOWN] round(Math.asin((TRIANGLE.sides[UNKNOWN] * sin(ANGLES[KNOWN] * Math.PI / 180)) / TRIANGLE.sides[KNOWN]) / Math.PI * 180) rand(2) ? randRange(-20, 20) : randRange(160, 200)

Bestimme m\angle UNKNOWN_MEASURE.

Runde die Antwort auf das nächstliegende Grad.

TRIANGLE.angleLabels = [ KNOWN === 0 ? ANGLES[0] + "^\\circ" : null, KNOWN === 1 ? ANGLES[1] + "^\\circ" : null, KNOWN === 2 ? ANGLES[2] + "^\\circ" : null ]; TRIANGLE = addTriangle(_.extend(TRIANGLE, { xPos: 1, yPos: 1, width: 10, height: 6, rot: ROTATION })); init({ range: [[0, TRIANGLE.width + 2], [0, TRIANGLE.height + 2]] }); TRIANGLE.draw();
SOLUTION \Large{^\circ}

Benutze den Sinussatz:

\qquad \dfrac{BC}{\sin(m\angle A)} \quad = \quad \dfrac{AC}{\sin(m\angle B)} \quad = \quad \dfrac{AB}{\sin(m\angle C)}

Aus dem Sinussatz ergibt sich:

\qquad \dfrac{\pink{SIDES[UNKNOWN]}}{\sin( \pink{m\angle UNKNOWN_MEASURE})} \quad = \quad\dfrac{\blue{ TRIANGLE.sides[KNOWN]}}{\sin(\blue{ ANGLES[KNOWN]^\circ})}

TRIANGLE.sideLabels[KNOWN] = "\\blue{" + TRIANGLE.sides[KNOWN] + "}"; TRIANGLE.angleLabels[KNOWN] = "\\blue{" + ANGLES[KNOWN] + "^\\circ}"; TRIANGLE.sideLabels[UNKNOWN] = "\\pink{" + TRIANGLE.sides[UNKNOWN] + "}"; TRIANGLE.angleLabels[UNKNOWN] = "\\pink{\\boldsymbol{\\unicode{x003F}}}"; TRIANGLE.color = GRAY; TRIANGLE.draw();

Löse nach dem Sinus des unbekannten Winkels auf:

\qquad \sin(\pink{m\angle UNKNOWN_MEASURE}) \quad = \quad \dfrac{\pink{TRIANGLE.sides[UNKNOWN]} \cdot \sin(\blue{ANGLES[KNOWN]^\circ}) }{\blue{TRIANGLE.sides[KNOWN]}}

Die rechte Seite lässt sich wie folgt berechnen:

\qquad \sin(\pink{m\angle UNKNOWN_MEASURE}) \quad \approx \quad localeToFixed(roundTo(9, (TRIANGLE.sides[UNKNOWN] * sin(ANGLES[KNOWN] * Math.PI / 180)) / TRIANGLE.sides[KNOWN]),-1)

Mit der Umkehrfunktion des Sinus (Arcussinus) können wir den Winkel m\angle UNKNOWN_MEASURE bestimmen (und runden):

\begin{aligned} \qquad \sin^{-1}\big(\sin(\pink{m\angle UNKNOWN_MEASURE})\big) \quad &\approx \quad \sin^{-1}( localeToFixed(roundTo(9, (TRIANGLE.sides[UNKNOWN] * sin(ANGLES[KNOWN] * Math.PI / 180)) / TRIANGLE.sides[KNOWN]),-1)) \\[1ex] \qquad =\; \pink{m\angle UNKNOWN_MEASURE} \quad &\approx \quad \sin^{-1}( localeToFixed(roundTo(9, (TRIANGLE.sides[UNKNOWN] * sin(ANGLES[KNOWN] * Math.PI / 180)) / TRIANGLE.sides[KNOWN]),-1)) \\[1ex] \qquad \Rightarrow\pink{m\angle UNKNOWN_MEASURE} \quad &\approx \quad SOLUTION^\circ \end{aligned}

TRIANGLE.angleLabels[UNKNOWN] = "\\pink{" + SOLUTION + "^\\circ}"; TRIANGLE.draw();