# Khan/khan-exercises

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 Alternate interior angles 2
randRangeNonZero( -20, 20 ) randRange( 2, 9 ) randRangeNonZero( -200, 200 ) randRange( 2, 9 ) A * SOLUTION + B - C * SOLUTION (A*D-B*C) / (A-C) shuffle( randFromArray( [ [ 0, 6 ], [ 1, 7 ] ] ) ) KNOWN_INDEX % 2 === 0 ? ANCHOR : 180 - ANCHOR

The two horizontal lines are parallel, and there is a third line that intersects them as shown below.

Solve for x:

SOLUTION

var eq1 = A + "x + " + B + "^\\circ"; var eq2 = C + "x + " + D + "^\\circ"; init({ range: [ [ -1, 11 ], [-1, 4] ] }); graph.pl = new ParallelLines( 0, 0, 10, 0, 3 ); graph.pl.draw(); graph.pl.drawTransverse( ANCHOR ); graph.pl.drawAngle( KNOWN_INDEX, eq1 ); graph.pl.drawAngle( UNKNOWN_INDEX, eq2, "#FFA500" );

We learned in Alternate interior angles 1 that alternate interior angles are equal. Watch this video to understand why.

Set the \color{BLUE}{\text{blue angle}} equal to the \color{ORANGE}{\text{orange angle}}.

\color{BLUE}{Ax + B} = \color{ORANGE}{Cx + D}

Subtract \color{PINK}{Cx} from both sides.

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

A - Cx + B = D

B > 0 ? "Subtract" : "Add" \color{PINK}{abs(B)} B > 0 ? "from" : "to" both sides.

(A - Cx + B) \color{PINK}{+ -B} = D \color{PINK}{+ -B}

A - Cx = D - B

Divide both sides by \color{PINK}{A - C}.

\dfrac{A - Cx}{\color{PINK}{A - C}} = \dfrac{D - B}{\color{PINK}{A - C}}

Simplify.

x = SOLUTION

Subtract \color{PINK}{Ax} from both sides.

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

B = C - Ax + D

D > 0 ? "Subtract" : "Add" \color{PINK}{abs(D)} D > 0 ? "from" : "to" both sides.

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

B - D = C - Ax

Divide both sides by \color{PINK}{C - A}.

\dfrac{B - D}{\color{PINK}{C - A}} = \dfrac{C - Ax}{\color{PINK}{C - A}}

Simplify.

SOLUTION = x

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