# Khan/khan-exercises

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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 ]) [randRange(30, 170), 180, 90][PROBLEM] 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]])

Given

\qquad m ANGLE_ONE = COEF_1x + CONST_1^\circ
\qquad m ANGLE_TWO = COEF_2x + CONST_2^\circ

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: 0.1, fill: "black", stroke: "none" }; line( [0, 0], [5, 0], { arrows: "->" } ); line( [0, 0], polar(5, largeAngle), {arrows: "->"} ); if (ANGLE_ONE === ANGLE_TOP){ line( [0, 0], polar(5, subAngle2), {arrows: "->"} ); } else { 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" ); if (ANGLE_ONE === ANGLE_TOP){ circle( polar(4.5, subAngle2), dotAttr ); label( polar(4.5, subAngle2), B, (subAngle2 > 45) ? "right" : "above" ); } else { circle( polar(4.5, subAngle1), dotAttr ); label( polar(4.5, subAngle1), B, (subAngle1 > 45) ? "right" : "above" ); }

From the diagram, we see that together \blue{ANGLE_TOP} and \green{ANGLE_BOT} form \purple{LARGE_ANGLE}, so

\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(); if (PROBLEM === 0){ style({ stroke: "purple", strokeWidth: 3 }); arc( [ 0, 0 ], 1.2, 0, largeAngle ); $("#givenLarge").addClass("hint_purple"); } if (ANGLE_ONE === ANGLE_TOP){$("#given0").addClass("hint_blue"); $("#given1").addClass("hint_green"); } else {$("#given0").addClass("hint_green"); $("#given1").addClass("hint_blue"); }; Since LARGE_ANGLE is a straight angle, we know \purple{mLARGE_ANGLE = 180^\circ}. style({ stroke: "purple", strokeWidth: 3 }); arc( [ 0, 0 ], 1.2, 0, 180 );$("#givenLarge").addClass("hint_purple");

Since we are given that \overline{OA}\perp\overline{OC}, we know \purple{mLARGE_ANGLE = 90^\circ}.

style({ stroke: "purple", strokeWidth: 3 }); path( [[0, .5], [.5, .5], [.5, 0]] ); \$("#givenLarge").addClass("hint_purple");

Substitute in the expressions that were given for each measure:

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

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

Combine like terms:

\qquadCOEF_1 + COEF_2x + CONST_1 + CONST_2^\circ = largeAngle^\circ.

Subtract CONST_1 + CONST_2^\circ from both sides:

Add -(CONST_1 + CONST_2)^\circ to both sides:

\qquadCOEF_1 + COEF_2x = largeAngle - CONST_1 - CONST_2^\circ.

Divide both sides by COEF_1 + COEF_2 to find x:

Substitute \pink{X}^\circ for \pink{x} in the expression that was given for \green{mANSWER[0]}:

Substitute \pink{X^\circ} for \pink{x} in the expression that was given for \blue{mANSWER[0]}: