Bent Lines between Parallel Lines

Draw $XY$ through $P$, parallel to $AB$. Then $XY \parallel AB \parallel CD$.

$\angle BEP$ and $\angle EPX$ are on opposite sides of $EP$: $\angle EPX = 30^{\circ}$ (alt. ∠s, AB // XY)

$\angle DFP$ and $\angle XPF$ are on opposite sides of $PF$: $\angle XPF = 45^{\circ}$ (alt. ∠s, XY // CD)

The two parts make up the angle at $P$: $x = 30 + 45 = 75$

Here the bend points to the left, so the angle at $P$ opens to the right. Draw $XY$ through $P$, parallel to $AB$; $PY$ cuts $x^{\circ}$ into $\angle EPY$ and $\angle YPF$.

$\angle AEP$ and $\angle EPY$ are between $AB$ and $XY$, on opposite sides of $EP$: $\angle EPY = 42^{\circ}$ (alt. ∠s, AB // XY)

$\angle CFP$ and $\angle YPF$ are between $XY$ and $CD$, on opposite sides of $PF$: $\angle YPF = 56^{\circ}$ (alt. ∠s, XY // CD)

The two parts make up the angle at $P$: $x = 42 + 56 = 98$

Draw $XY$ through $P$, parallel to $AB$. Then $XY \parallel AB \parallel CD$.

$\angle AEP$ and $\angle EPX$ are between $AB$ and $XY$, on the same side (the left) of $EP$: interior angles.

$\angle EPX = 180^{\circ} - 125^{\circ} = 55^{\circ}$ (int. ∠s, AB // XY)

$\angle CFP$ and $\angle XPF$ are between $XY$ and $CD$, on the same side (the left) of $PF$: interior angles.

$\angle XPF = 180^{\circ} - 140^{\circ} = 40^{\circ}$ (int. ∠s, XY // CD)

The two parts make up the angle at $P$: $x = 55 + 40 = 95$

Draw $XY$ through $P$, parallel to $AB$. The two given angles are of different kinds, so the two parts need different rules.

$\angle BEP$ and $\angle EPX$ are on opposite sides of $EP$: $\angle EPX = 28^{\circ}$ (alt. ∠s, AB // XY)

$\angle CFP$ and $\angle XPF$ are on the same side (the left) of $PF$: $\angle XPF = 180^{\circ} - 124^{\circ} = 56^{\circ}$ (int. ∠s, XY // CD)

The two parts make up the angle at $P$: $x = 28 + 56 = 84$

Draw $XY$ through $P$, parallel to $AB$. Then $XY \parallel AB \parallel CD$.

$\angle EPX = 47^{\circ}$ (alt. ∠s, AB // XY)

$PX$ cuts $\angle EPF$ into $\angle EPX$ and $\angle XPF$, so $\angle XPF = 112^{\circ} - 47^{\circ} = 65^{\circ}$.

$\angle XPF$ and $x^{\circ}$ are between $XY$ and $CD$, on opposite sides of $PF$: alternate angles.

$x = 65$ (alt. ∠s, XY // CD)

(a) Draw $XY$ through $P$, parallel to $AB$. Then $XY \parallel AB \parallel CD$.

$\angle EPX = 52^{\circ}$ (alt. ∠s, AB // XY)

$\angle XPF = 35^{\circ}$ (alt. ∠s, XY // CD)

The two parts make up the angle at $P$: $x = 52 + 35 = 87$

(b) The angle $x^{\circ}$ and the reflex angle $y^{\circ}$ make a complete turn at $P$.

$y = 360 - 87 = 273$ (∠s at a pt)

Draw $XY$ through $P$, parallel to $AB$. Then $XY \parallel AB \parallel CD$, and $PY$ cuts $x^{\circ}$ into $\angle EPY$ and $\angle YPF$.

Upper part: $\angle BEP$ and $\angle EPY$ are between $AB$ and $XY$, on the same side of $EP$.

$\angle EPY = 180^{\circ} - 145^{\circ} = 35^{\circ}$ (int. ∠s, AB // XY)

Lower part: $\angle DFP$ and $\angle YPF$ are between $XY$ and $CD$, on the same side of $PF$.

$\angle YPF = 180^{\circ} - 130^{\circ} = 50^{\circ}$ (int. ∠s, XY // CD)

The two parts make up the angle at $P$: $x = 35 + 50 = 85$

The reflex angle $EPF$ and $x^{\circ}$ together make a complete turn at $P$.

Reflex $\angle EPF = 360^{\circ} - 85^{\circ} = 275^{\circ}$ (∠s at a pt)

First the angle inside the bend: it and the reflex angle make a complete turn at $P$.

$\angle EPF = 360^{\circ} - 280^{\circ} = 80^{\circ}$ (∠s at a pt)

Draw $XY$ through $P$, parallel to $AB$. Then $XY \parallel AB \parallel CD$.

$\angle EPX = 35^{\circ}$ (alt. ∠s, AB // XY)

$PX$ cuts $\angle EPF$ into $\angle EPX$ and $\angle XPF$, so $\angle XPF = 80^{\circ} - 35^{\circ} = 45^{\circ}$.

$\angle XPF$ and $x^{\circ}$ are on opposite sides of $PF$: $x = 45$ (alt. ∠s, XY // CD)

Draw $XY$ through $P$, parallel to $AB$. Then $XY \parallel AB \parallel CD$.

$\angle EPX = x^{\circ}$ (alt. ∠s, AB // XY)

$\angle XPF = (2x - 6)^{\circ}$ (alt. ∠s, XY // CD)

The two parts make up $\angle EPF$:

$x + (2x - 6) = 99$

$3x - 6 = 99$. Add $6$ to both sides: $3x = 105$. Divide both sides by $3$: $x = 35$.

Check: $\angle BEP = 35^{\circ}$ and $\angle DFP = 2(35) - 6 = 64^{\circ}$, and $35^{\circ} + 64^{\circ} = 99^{\circ}$.

Draw $XY$ through $P$ and $UV$ through $Q$, both parallel to $AB$. Then $AB$, $XY$, $UV$ and $CD$ are all parallel.

At $P$: $\angle AEP$ and $\angle EPX$ are between $AB$ and $XY$, on the same side of $EP$.

$\angle EPX = 180^{\circ} - 128^{\circ} = 52^{\circ}$ (int. ∠s, AB // XY)

$PX$ cuts $\angle EPQ$ into $\angle EPX$ and $\angle XPQ$, so $\angle XPQ = 97^{\circ} - 52^{\circ} = 45^{\circ}$.

From $P$ to $Q$: $\angle PQV = 45^{\circ}$ (alt. ∠s, XY // UV)

At $Q$: $\angle VQF$ and $\angle DFQ$ are between $UV$ and $CD$, on the same side of $QF$.

$\angle VQF = 180^{\circ} - 130^{\circ} = 50^{\circ}$ (int. ∠s, UV // CD)

The two parts make up the angle at $Q$: $x = 45 + 50 = 95$

Here the unknown is at the upper bend, so work up from $CD$ as well as down from $AB$. Draw $XY$ through $P$ and $UV$ through $Q$, both parallel to $AB$.

At $P$: $\angle EPX = 50^{\circ}$ (alt. ∠s, AB // XY)

At $Q$: $\angle VQF = 35^{\circ}$ (alt. ∠s, UV // CD)

$QV$ cuts $\angle PQF$ into $\angle PQV$ and $\angle VQF$, so $\angle PQV = 70^{\circ} - 35^{\circ} = 35^{\circ}$.

From $Q$ to $P$, along $PQ$: $\angle XPQ = 35^{\circ}$ (alt. ∠s, XY // UV)

The two parts make up the angle at $P$: $x = 50 + 35 = 85$

The angle inside the bend: $\angle EPF = 360^{\circ} - 255^{\circ} = 105^{\circ}$ (∠s at a pt)

Draw $XY$ through $P$, parallel to $AB$. Then $XY \parallel AB \parallel CD$.

$\angle EPX = 2x^{\circ}$ (alt. ∠s, AB // XY)

$\angle XPF = (x + 15)^{\circ}$ (alt. ∠s, XY // CD)

The two parts make up $\angle EPF$:

$2x + (x + 15) = 105$

$3x + 15 = 105$. Subtract $15$ from both sides: $3x = 90$. Divide both sides by $3$: $x = 30$.

Check: $\angle BEP = 2(30) = 60^{\circ}$ and $\angle DFP = 30 + 15 = 45^{\circ}$; $60^{\circ} + 45^{\circ} = 105^{\circ}$, and $360^{\circ} - 105^{\circ} = 255^{\circ}$.