# Khan/khan-exercises

Fetching contributors…
Cannot retrieve contributors at this time
115 lines (106 sloc) 5.15 KB
 Vertical angles 2
randRange(6, 20) * randRangeNonZero(-1, 1) randRange(2, 8) randRangeNonZero(-200, 200) randRangeExclude(2, 9, [A, A - 1, A + 1])
A * SOLUTION + B - C * SOLUTION (A * D - B * C) / (A - C) shuffle(randFromArray([[2, 4], [3, 1]]))

Solve for x:

x= SOLUTION

var eq1 = A + "x + " + B + "^\\circ"; var eq2 = C + "x + " + D + "^\\circ"; init({ range: [[-4, 4], [-1, 1]], scale: 60 }); graph.pl = new ParallelLines(0, 0, 4, 0, 0); graph.pl.draw(); graph.pl.drawTransverse(KNOWN_INDEX % 2 === 0 ? ANCHOR : 180 - ANCHOR); graph.pl.drawAngle(KNOWN_INDEX, eq1); graph.pl.drawAngle(UNKNOWN_INDEX, eq2, GREEN);

We learned in Vertical angles 1 that vertical angles are equal. Watch this video to understand why.

Set the angle measures 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 \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^\circ) \pink{- Ax} = (Cx + D^\circ) \pink{- Ax}

B^\circ = C - Ax + D^\circ

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