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.

Add `\pink{abs(B)^\circ}` to 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.

Add `\pink{abs(B)^\circ}` to 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.

Add `\pink{abs(B)^\circ}` to 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.

Add `\pink{abs(D)}` to 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.

graph.pl.drawAdjacentAngles( KNOWN_INDEX, "C", PINK );

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.

graph.pl.drawAdjacentAngles(KNOWN_INDEX, "C", PINK);

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`