Finding Unknown Angles with Parallel Lines

$\angle APQ$ (below $AB$, left of $EF$) and $\angle PQD$ (above $CD$, right of $EF$) are alternate angles.

$\angle PQD = 72^{\circ}$ (alt. ∠s, AB // CD)

$\angle PQD$ and $x^{\circ}$ are side by side on the straight line $EF$ at $Q$.

$x = 180 - 72 = 108$ (adj. ∠s on a str. line)

$a = 180 - 63 = 117$ (adj. ∠s on a str. line)

$a^{\circ}$ (below $AB$, right of $EF$) and $b^{\circ}$ (above $CD$, left of $EF$) are alternate angles.

$b = 117$ (alt. ∠s, AB // CD)

$c = 180 - 117 = 63$ (adj. ∠s on a str. line)

Check by another route: $\angle PQD = 63^{\circ}$ (corr. ∠s, AB // CD), and $c^{\circ}$ is opposite $\angle PQD$, so $c = 63$ (vert. opp. ∠s), as found.

$p^{\circ}$ is in the same position at $Q$ as the $105^{\circ}$ angle at $P$ (above the road, right of the bypass).

$p = 105$ (corr. ∠s, AB // CD)

$p^{\circ}$ and $q^{\circ}$ are side by side on the straight road $CD$ at $Q$.

$q = 180 - 105 = 75$ (adj. ∠s on a str. line)

The two angles below $CD$ at $Q$ are opposite these, so they are $105^{\circ}$ and $75^{\circ}$ as well (vert. opp. ∠s).

$\angle APQ$ and $\angle APE$ are side by side on the straight line $EF$ at $P$.

$\angle APQ = 180^{\circ} - 118^{\circ} = 62^{\circ}$ (adj. ∠s on a str. line)

$\angle APQ$ (below $AB$, left of $EF$) and $\angle PQD$ (above $CD$, right of $EF$) are alternate angles.

$\angle PQD = 62^{\circ}$ (alt. ∠s, AB // CD)

At $Q$, the angle $\angle PQD$ and the reflex angle $PQD$ together make a complete turn.

$x = 360 - 62 = 298$ (∠s at a pt)

Check: $\angle PQC = 118^{\circ}$ (corr. ∠s, AB // CD), and the reflex angle is $\angle PQC$ and the straight angle $CQD$ together: $118^{\circ} + 180^{\circ} = 298^{\circ}$.

$\angle APQ = 180^{\circ} - 55^{\circ} = 125^{\circ}$ (adj. ∠s on a str. line)

$p^{\circ}$ is in the same position at $Q$ as $\angle APQ$ at $P$ (below the line, left of $GH$).

$p = 125$ (corr. ∠s, AB // CD)

$p^{\circ}$ (below $CD$, left of $GH$) and $\angle QRF$ (above $EF$, right of $GH$) are alternate angles.

$\angle QRF = 125^{\circ}$ (alt. ∠s, CD // EF)

$q = 180 - 125 = 55$ (adj. ∠s on a str. line)

$x^{\circ}$ is at $R$, on $GH$, so use the $70^{\circ}$ angle, which is also on $GH$.

$\angle RSD = 180^{\circ} - 70^{\circ} = 110^{\circ}$ (adj. ∠s on a str. line)

$x^{\circ}$ (below $AB$, left of $GH$) and $\angle RSD$ (above $CD$, right of $GH$) are alternate angles.

$x = 110$ (alt. ∠s, AB // CD)

$y^{\circ}$ is at $Q$, on $EF$, so use the $64^{\circ}$ angle.

$\angle PQD = 64^{\circ}$ (corr. ∠s, AB // CD)

$y = 180 - 64 = 116$ (adj. ∠s on a str. line)

Take the line $PQ$ as a transversal of $AB$ and $CD$: $\angle BPQ$ and $a^{\circ}$ are between $AB$ and $CD$, on the same side of $PQ$ (above it).

$a = 180 - 75 = 105$ (int. ∠s, AB // CD)

Now take the line $CD$ as a transversal of $PQ$ and $RS$: $a^{\circ}$ and $b^{\circ}$ are between $PQ$ and $RS$, on the same side of $CD$ (to its left).

$b = 180 - 105 = 75$ (int. ∠s, PQ // RS)

$a^{\circ}$ and $\angle CQP$ are between the handrail and the line of the steps, on the same side of $PQ$ (to its left).

$a = 180 - 118 = 62$ (int. ∠s, AB // CD)

At $Q$: $\angle PQD = 180^{\circ} - 118^{\circ} = 62^{\circ}$ (adj. ∠s on a str. line)

Now take the line $CD$ as a transversal of the balusters: $b^{\circ}$ is in the same position at $S$ as $\angle PQD$ at $Q$ (above $CD$, to the right of the baluster).

$b = 62$ (corr. ∠s, PQ // RS)

The two angles are not one of the three pairs, so first carry $\angle APQ$ across to $Q$. $\angle APQ$ (below $AB$, left of $EF$) and $\angle PQD$ (above $CD$, right of $EF$) are alternate angles.

$\angle PQD = (5x - 37)^{\circ}$ (alt. ∠s, AB // CD)

$\angle PQD$ and $\angle CQF$ are opposite each other where $CD$ and $EF$ cross, so they are equal:

$5x - 37 = 3x + 9$ (vert. opp. ∠s)

Subtract $3x$ from both sides: $2x - 37 = 9$.

Add $37$ to both sides: $2x = 46$. Divide both sides by $2$: $x = 23$.

Check: $5(23) - 37 = 78$ and $3(23) + 9 = 78$, so both angles are $78^{\circ}$.

$y^{\circ}$ and $\angle CQF$ are side by side on the straight line $CD$ at $Q$.

$y = 180 - 78 = 102$ (adj. ∠s on a str. line)

The two angles are not one of the three pairs, so first carry $\angle EPB$ down to $Q$.

$\angle PQD = (3x + 5)^{\circ}$ (corr. ∠s, AB // CD)

$\angle PQD$ and $\angle DQF$ are side by side on the straight line $EF$ at $Q$:

$(3x + 5) + (2x - 15) = 180$ (adj. ∠s on a str. line)

$5x - 10 = 180$

Add $10$ to both sides: $5x = 190$. Divide both sides by $5$: $x = 38$.

$\angle EPB = 3(38) + 5 = 119^{\circ}$ and $\angle DQF = 2(38) - 15 = 61^{\circ}$.

Check: $119^{\circ} + 61^{\circ} = 180^{\circ}$.

(a) $\angle PQD$ is in the same position at $Q$ as $\angle EPB$ at $P$ (above the line, right of $EF$).

$\angle PQD = 64^{\circ}$ (corr. ∠s, AB // CD)

(b) $\angle RSD$ and $\angle CSR$ are side by side on the straight line $CD$ at $S$.

$\angle RSD = 180^{\circ} - 118^{\circ} = 62^{\circ}$ (adj. ∠s on a str. line)

(c) Take the edge $CD$ as a transversal of the two footbridges. $\angle PQD$ and $\angle RSD$ are both above $CD$, each on the right of its footbridge: they are corresponding angles.

If $EF$ were parallel to $GH$, these corresponding angles would be equal. But $64^{\circ} \neq 62^{\circ}$, so the footbridges are not parallel. They look parallel in the figure, but how they look is not a reason.

The two angles are on different bars, so carry the angle at $P$ down, one bar at a time.

$\angle PQD = (3x + 12)^{\circ}$ (corr. ∠s, AB // CD)

$\angle QRF = (3x + 12)^{\circ}$ (corr. ∠s, CD // EF)

(The two steps can be one: $\angle QRF = (3x + 12)^{\circ}$ (corr. ∠s, AB // EF), since $AB \parallel EF$ too.)

$\angle QRF$ and $\angle ERQ$ are side by side on the straight line $EF$ at $R$:

$(3x + 12) + (5x + 40) = 180$ (adj. ∠s on a str. line)

Collect like terms: $8x + 52 = 180$. Subtract $52$ from both sides: $8x = 128$. Divide both sides by $8$: $x = 16$.

Check: $\angle GPB = 3(16) + 12 = 60^{\circ}$ and $\angle ERQ = 5(16) + 40 = 120^{\circ}$, and $60^{\circ} + 120^{\circ} = 180^{\circ}$.