Because the two garden borders are parallel and the walking path crosses both of them, the matching acute angles formed are congruent. So the acute angle at the second border is also $68^{\circ }$. The question asks for the obtuse angle at the second intersection, which forms a linear pair with the $68^{\circ }$ angle. Linear pairs sum to $180^{\circ }$, so the obtuse angle is $180-68=112$. Therefore, the correct answer is $112^{\circ }$.

Because $m\parallel n$, alternate interior angles are congruent. So set the expressions equal:

Add 18 to both sides and subtract $3x$ from both sides:

Now find the angle measure:

Since $57^{\circ }$ is acute, the acute angle formed by line $t$ with line $m$ is $57^{\circ }$.

Because $p\parallel q$, corresponding angles formed by transversal $t$ are congruent. That means the angle measuring $68^{\circ }$ and the angle measuring $(2x+8{)}^{\circ }$ must be equal:

Subtract 8 from both sides:

Then divide by 2:

So the statement that must be true is $x=30$.

In any triangle, the angle measures add to $180^{\circ }$. One angle is $90^{\circ }$, so the other two angles must add to $180-90=90^{\circ }$. Since those two angles are in the ratio $2:3$, let them be $2x$ and $3x$. Then $2x+3x=90$, so $5x=90$ and $x=18$. The smaller angle is $2x=2(18)=36^{\circ }$. Therefore, the correct answer is $36$.

When a transversal cuts two parallel lines, all acute angles are congruent, all obtuse angles are congruent, and any acute angle and obtuse angle are supplementary. That means their measures add to $180$. So

Combine like terms:

Add $40$ to both sides:

Divide by $4$:

Since $x$ represents the measure of each acute angle, the equivalent expression for the measure of each acute angle is $55$.