randRange(0,2) randFromArray([["O","A","B","C"],["B","A","C","D"],["O","L","M","N"],["P","Q","R","S"]]) "\\angle " + A + O + B "\\angle " + B + O + C "\\angle " + A + O + C shuffle([ ANGLE_BOT, ANGLE_TOP ]) function(){return 0===PROBLEM?randRange(30,170):1===PROBLEM?180:90}() randRange(15, largeAngle-15) largeAngle - subAngle1 randRange( 1, 20 ) randRange( 2, 9 ) subAngle1 - COEF_1*X randRange( 2, 9 ) subAngle2 - COEF_2*X randFromArray([ [ANGLE_ONE, COEF_1, CONST_1], [ANGLE_TWO, COEF_2, CONST_2] ])

Dado

\qquad \overrightarrow{OA}\perp\overrightarrow{OC},
 \qquad m ANGLE_ONE = COEF_1x + CONST_1, y
\qquad m ANGLE_TWO = COEF_2x + CONST_2,
 , y
\qquad m LARGE_ANGLE = largeAngle,

encuentra mANSWER[0].

init({range:[[Math.min(polar(5.5,largeAngle)[0],-.5)-.5,5.5],[-1,Math.min(Math.max(polar(5.5,largeAngle)[1],polar(5.5,subAngle2)[1])+.5,5.5)]],scale:40}),dotAttr={r:.1,fill:"black",stroke:"none"},line([0,0],[5,0],{arrows:"->"}),line([0,0],polar(5,largeAngle),{arrows:"->"}),ANGLE_ONE===ANGLE_TOP?line([0,0],polar(5,subAngle2),{arrows:"->"}):line([0,0],polar(5,subAngle1),{arrows:"->"}),circle([0,0],dotAttr),label([0,0],O,"below left"),circle(polar(4.5,largeAngle),dotAttr),label(polar(4.5,largeAngle),A,largeAngle>120?"below left":"left"),circle([4.5,0],dotAttr),label([4.5,0],C,"below"),ANGLE_ONE===ANGLE_TOP?(circle(polar(4.5,subAngle2),dotAttr),label(polar(4.5,subAngle2),B,subAngle2>45?"right":"above")):(circle(polar(4.5,subAngle1),dotAttr),label(polar(4.5,subAngle1),B,subAngle1>45?"right":"above"))
ANSWER[1] * X + ANSWER[2] {}^{\circ}

Del diagrama vemos que \blue{ANGLE_TOP} y \green{ANGLE_BOT} forman \purple{LARGE_ANGLE}, por tanto

\qquad \blue{mANGLE_TOP} + \green{mANGLE_BOT} = \purple{mLARGE_ANGLE}.

style({stroke:GREEN,strokeWidth:3}),arc([0,0],1,0,ANGLE_ONE===ANGLE_TOP?subAngle2:subAngle1).toBack(),style({stroke:BLUE,strokeWidth:3}),arc([0,0],1,ANGLE_ONE===ANGLE_TOP?subAngle2:subAngle1,largeAngle).toBack(),0===PROBLEM&&(style({stroke:"purple",strokeWidth:3}),arc([0,0],1.2,0,largeAngle),jQuery("#givenLarge").css({color:"purple"})),ANGLE_ONE===ANGLE_TOP?(jQuery("#given0").css({color:BLUE}),jQuery("#given1").css({color:GREEN})):(jQuery("#given0").css({color:GREEN}),jQuery("#given1").css({color:BLUE}))

Puesto que LARGE_ANGLE es un ángulo recto, sabemos que \purple{mLARGE_ANGLE = 180}.

style({stroke:"purple",strokeWidth:3}),arc([0,0],1.2,0,180)

Puesto que se nos da \overrightarrow{OA}\perp\overrightarrow{OC}, sabemos que \purple{mLARGE_ANGLE = 90}.

style({stroke:"purple",strokeWidth:3}),path([[0,.5],[.5,.5],[.5,0]])

Sustituye en la expresión que fue dada para cada medida:

\qquad \blue{COEF_1x + CONST_1} + \green{COEF_2x + CONST_2} = \purple{largeAngle}.

\qquad \blue{COEF_2x + CONST_2} + \green{COEF_1x + CONST_1} = \purple{largeAngle}.

Combina términos semejantes:

\qquadCOEF_1 + COEF_2x + CONST_1 + CONST_2 = largeAngle.

Resta CONST_1 + CONST_2 de ambos lados:

Suma -(CONST_1 + CONST_2) en ambos lados:

\qquadCOEF_1 + COEF_2x = largeAngle - CONST_1 - CONST_2.

Divide ambos lados entre COEF_1 + COEF_2 para encontrar x:

\qquad x = X.

Sustituye x por X en la expresión que fue dada para mANSWER[0]:

\qquad mANSWER[0] = ANSWER[1](\pink{X}) + ANSWER[2].

Sustituye X por x en la expresión dada para mANSWER[0]:

\qquad mANSWER[0] = ANSWER[1](\pink{X}) + ANSWER[2].

Simplifica:

\qquad \green{mANSWER[0] = ANSWER[1]*X + ANSWER[2]}.

Simplifica:

\qquad \blue{mANSWER[0] = ANSWER[1]*X + ANSWER[2]}.

Por tanto \green{mANSWER[0] = ANSWER[1] * X + ANSWER[2]}.

Por tanto \blue{mANSWER[0] = ANSWER[1] * X + ANSWER[2]}.