# Khan/khan-exercises

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randRange(0, 8) * Math.PI / 8
randRange(6, 20) * randRangeNonZero(-1, 1) randRange(2, 6) randRangeNonZero(2, 8 - DIFFERENCE) + DIFFERENCE DIFFERENCE * randRange(2, 15)
A - DIFFERENCE A * SOLUTION + B - C * SOLUTION max(30, min(150, A * SOLUTION + B)) randRange(0, 3) * 2 + ((ANGLE < 50) ? 1 : ((ANGLE > 130) ? 0 : randRange(0, 1))) (KNOWN_INDEX + 4) % 8

Corresponding angles are equal to one another. Watch this video to understand why.

\blue{\angle A} and \green{\angle B} are corresponding angles. Therefore, we can set them equal to one another.

\blue{Ax + B^\circ} = \green{Cx + D^\circ}

Subtract \pink{Cx} from both sides.

(Ax + B^\circ) \pink{- Cx} = (Cx + D^\circ) \pink{- Cx}

A - Cx + B^\circ = D^\circ

Subtract \pink{abs(B)^\circ} from both sides.

(A - Cx + B^\circ) \pink{+ -B^\circ} = D^\circ \pink{+ -B^\circ}

A - Cx = D - B^\circ

Divide both sides by \pink{A - C}.

\dfrac{A - Cx}{\pink{A - C}} = \dfrac{D - B^\circ}{\pink{A - C}}

Simplify.

x = SOLUTION^\circ

randRange(6, 20) * randRangeNonZero(-1, 1) randRange(2, 6) randRangeNonZero(2, 8 - DIFFERENCE) + DIFFERENCE DIFFERENCE * randRange(2, 15)
A - DIFFERENCE A * SOLUTION + B - C * SOLUTION max(30, min(150, A * SOLUTION + B)) shuffle(randFromArray([[0, 6], [1, 7]]))

Alternate interior angles are equal to one another. Watch this video to understand why.

\blue{\angle A} and \green{\angle B} are alternate interior angles. Therefore, we can set them equal to one another.

\blue{Ax + B^\circ} = \green{Cx + D^\circ}

Subtract \pink{Cx} from both sides.

(Ax + B^\circ) \pink{- Cx} = (Cx + D^\circ) \pink{- Cx}

A - Cx + B^\circ = D^\circ

Subtract \pink{abs(B)^\circ} from both sides.

(A - Cx + B^\circ) \pink{+ -B^\circ} = D^\circ \pink{+ -B^\circ}

A - Cx = D - B^\circ

Divide both sides by \pink{A - C}.

\dfrac{A - Cx}{\pink{A - C}} = \dfrac{D - B^\circ}{\pink{A - C}}

Simplify.

x = SOLUTION^\circ

randRange(6, 20) * randRangeNonZero(-1, 1) randRange(2, 6) randRangeNonZero(2, 8 - DIFFERENCE) + DIFFERENCE DIFFERENCE * randRange(2, 15)
A - DIFFERENCE A * SOLUTION + B - C * SOLUTION max(30, min(150, A * SOLUTION + B)) shuffle(randFromArray([[2, 4], [3, 5]]))

Alternate exterior angles are equal to one another. Watch this video to understand why.

\blue{\angle A} and \green{\angle B} are alternate exterior angles. Therefore, we can set them equal to one another.

\blue{Ax + B^\circ} = \green{Cx + D^\circ}

Subtract \pink{Cx} from both sides.

(Ax + B^\circ) \pink{- Cx} = (Cx + D^\circ) \pink{- Cx}

A - Cx + B^\circ = D^\circ

Subtract \pink{abs(B)^\circ} from both sides.

(A - Cx + B^\circ) \pink{+ -B^\circ} = D^\circ \pink{+ -B^\circ}

A - Cx = D - B^\circ

Divide both sides by \pink{A - C}.

\dfrac{A - Cx}{\pink{A - C}} = \dfrac{D - B^\circ}{\pink{A - C}}

Simplify.

x = SOLUTION^\circ

Subtract \pink{Ax} from both sides.

(Ax + B) \pink{- Ax} = (Cx + D) \pink{- Ax}

B = C - Ax + D

Subtract \pink{abs(D)} from both sides.

B \pink{+ -D} = (C - Ax + D) \pink{+ -D}

B - D = C - Ax

Divide both sides by \pink{C - A}.

\dfrac{B - D}{\pink{C - A}} = \dfrac{C - Ax}{\pink{C - A}}

Simplify.

SOLUTION = x

randRange(6, 20) * randRangeNonZero(-1, 1) randRange(2, 8) randRangeNonZero(-200, 200) randRangeExclude(2, 9, [A, A - 1, A + 1])
180 - (A + C) * SOLUTION - B max(30, min(150, A * SOLUTION + B)) A * SOLUTION + B shuffle(randFromArray([[1, 6], [0, 7]]))

The pink angles are adjacent to \blue{\angle A} and form a straight line, so we know that:

\blue{Ax + B^\circ} + \pink{C} = 180^\circ

The pink angles equal each other because they are vertical angles.

One of the pink angles corresponds with \green{\angle B}, and the other pink angle forms an alternative interior angle. Therefore, \green{\angle B} equals the pink angle measure.

\pink{C} = \green{Cx + D^\circ}

Substitute \green{Cx + D^\circ} for \pink{C} in our first equation.

\blue{Ax + B^\circ} + \green{Cx + D^\circ} = 180^\circ

Combine like terms.

A + Cx + B + D^\circ = 180^\circ

Subtract \pink{abs(B + D)^\circ} from both sides.

Add \pink{abs(B + D)^\circ} to both sides.

(A + Cx + B + D^\circ) \pink{+ -(B + D)^\circ} = 180^\circ \pink{+ -(B + D)^\circ}

A + Cx = 180 - B - D^\circ

Divide by \pink{A + C}.

\dfrac{A + Cx}{\pink{A + C}} = \dfrac{180 - B - D^\circ}{\pink{A + C}}

Simplify.

x = (180 - B - D) / (A + C)^\circ

Note that the pinks angles are supplementary to \blue{\angle A}.

randRange(6, 20) * randRangeNonZero(-1, 1) randRange(2, 8) randRangeNonZero(-200, 200) randRangeExclude(2, 9, [A, A - 1, A + 1])
180 - (A + C) * SOLUTION - B max(30, min(150, A * SOLUTION + B)) A * SOLUTION + B shuffle(randFromArray([[3, 4], [2, 5]]))

The pink angles are adjacent to \blue{\angle A} and form a straight line, so we know that:

\blue{Ax + B^\circ} + \pink{C} = 180^\circ

The pink angles equal each other because they are vertical angles.

One of the pink angles corresponds with \green{\angle B}, and the other pink angle forms an alternative interior angle. Therefore, \green{\angle B} equals the pink angle measure.

\pink{C} = \green{Cx + D^\circ}

Substitute \green{Cx + D^\circ} for \pink{C} in our first equation.

\blue{Ax + B^\circ} + \green{Cx + D^\circ} = 180^\circ

Combine like terms.

A + Cx + B + D^\circ = 180^\circ

Subtract \pink{abs(B + D)^\circ} from both sides.

Add \pink{abs(B + D)^\circ} to both sides.

(A + Cx + B + D^\circ) \pink{+ -(B + D)^\circ} = 180^\circ \pink{+ -(B + D)^\circ}

A + Cx = 180 - B - D^\circ

Divide by \pink{A + C}.

\dfrac{A + Cx}{\pink{A + C}} = \dfrac{180 - B - D^\circ}{\pink{A + C}}

Simplify.

x = (180 - B - D) / (A + C)^\circ

Note that the pinks angles are supplementary to \blue{\angle A}.

Below are two parallel lines with a third line intersecting them.

\qquad \begin{eqnarray} \blue{\angle A} &=& \blue{Ax + B^\circ} \\ \green{\angle B} &=& \green{Cx + D^\circ} \end{eqnarray}

Solve for x:

init({ range: [[-4, 4], [-3, 3]], scale: 60 }); graph.pl = new ParallelLines(0, 0, 6, 2.5, ROTATION); graph.pl.draw(); var drawAnchor = KNOWN_INDEX % 2 === 0 ? ANGLE : 180 - ANGLE; graph.pl.drawMarkers(drawAnchor); graph.pl.drawTransverse(drawAnchor); graph.pl.drawAngle(KNOWN_INDEX, "A"); graph.pl.drawAngle(UNKNOWN_INDEX, "B", GREEN);

SOLUTION ^\circ