Question

examine the figure below. if m∠2=(5x + 2)° and m∠7=(7x - 42)°, prove m∠6 = 68°. proof: 2 column given: 2 || lines crossed by a transversal, m∠2=(5x + 2)°, m∠7=(7x - 42)° prove: m∠6 = 68°

Explanation:

Step1: Use corresponding - angles property

Since the two lines are parallel and crossed by a transversal, $\angle2$ and $\angle7$ are corresponding angles, so $\angle2=\angle7$, then $m\angle2 = m\angle7$.

Step2: Set up the equation

Given $m\angle2=(5x + 2)^{\circ}$ and $m\angle7=(7x-42)^{\circ}$, we have the equation $5x + 2=7x-42$.

Step3: Solve the equation for $x$

Subtract $5x$ from both sides: $5x-5x + 2=7x-5x-42$, which simplifies to $2 = 2x-42$.
Add 42 to both sides: $2 + 42=2x-42 + 42$, getting $44 = 2x$.
Divide both sides by 2: $\frac{44}{2}=\frac{2x}{2}$, so $x = 22$.

Step4: Find $m\angle7$

Substitute $x = 22$ into the expression for $m\angle7$: $m\angle7=7\times22-42=154 - 42=112^{\circ}$.

Step5: Use the linear - pair property

$\angle6$ and $\angle7$ form a linear - pair, so $m\angle6+m\angle7 = 180^{\circ}$.
Substitute $m\angle7 = 112^{\circ}$ into the equation: $m\angle6+112=180$.
Subtract 112 from both sides: $m\angle6=180 - 112=68^{\circ}$.

Answer:

$m\angle6 = 68^{\circ}$