Find the slopes of $AC$ and $BD$ using the coordinates. What does that tell you about the relationship between these two segments?

The angle $\angle CAB$ is an acute angle between a horizontal segment and a slanted segment. At $B$, you have the same slanted direction and the same horizontal line, but $\angle ABD$ uses the horizontal ray pointing the opposite way. How are those two angles related along a straight line?

At a point on a straight line, the two angles that share a side and fill the straight line add to $180^\circ$. Which angle at $B$ equals $t^\circ$, and which angle are you asked for?

In Desmos, enter the equations of the lines containing the key segments:

• y = 0 for line $AB$,
• y = 2x for line $AC$ (through $(0,0)$ and $(k,2k)$),
• y = 2x - 2 (or any vertical shift) for a line parallel to $AC$ that can represent $BD$. You will see two slanted parallel lines crossing the same horizontal line.

Zoom near the origin to see the acute angle where y = 2x meets y = 0; that angle corresponds to $\angle CAB$ (labeled $t^\circ$ in the problem). Then look at the intersection point of the horizontal line with the other slanted line y = 2x - 2; notice there is also an acute angle between the rightward horizontal ray and this slanted line, equal to the first one by the parallel-line angle relationships.

At the second intersection, focus on the obtuse angle formed by the slanted line and the horizontal ray pointing the opposite way along the same straight line. In Desmos, you can visually check that this obtuse angle and the nearby acute angle form a straight line, so they add to $180^\circ$. The obtuse angle you need (matching $\angle ABD$) is therefore the straight angle minus the acute angle already associated with $t^\circ$.

Sketch the points:

• $A(0,0)$ and $B(k,0)$ lie on the $x$-axis, so $\overline{AB}$ is horizontal.
• $C(k,2k)$ is directly above $B$, so $\overline{BC}$ is vertical and $\angle ABC = 90^\circ$.
• $D(2k,2k)$ is to the right of $C$, so $\overline{CD}$ is horizontal.

This gives you a right triangle $ABC$ and a larger triangle $ABD$ sharing base $AB$.

Compute slopes to see which segments are parallel:

• Slope of $AC$: from $(0,0)$ to $(k,2k)$ is

• Slope of $BD$: from $(k,0)$ to $(2k,2k)$ is

So $AC$ and $BD$ have the same slope and are therefore parallel lines.

$\angle CAB$ is the angle at $A$ between $\overline{CA}$ and $\overline{AB}$. Since $AC$ is parallel to $BD$ and $AB$ extends straight through $B$ (the line is the same, just reversed direction), the acute angle between the horizontal line and the slanted parallel line is the same at every intersection.

So the acute angle at $B$ between the ray \overrightarrow{B\text{(to the right on }AB)}} and $\overrightarrow{BD}$ is also $t^\circ$ by corresponding/alternate interior angles with parallel lines.

At point $B$, the horizontal line through $A$ and $B$ is a straight line, so the ray to the right (along \overrightarrow{B\text{(to the right on }AB)}}) and the ray to the left ($\overrightarrow{BA}$) form a $180^\circ$ angle.

You already know the acute angle between the rightward horizontal ray and $\overrightarrow{BD}$ is $t^\circ$. The angle $\angle ABD$ is formed by the leftward ray $\overrightarrow{BA}$ and $\overrightarrow{BD}$, which forms a linear pair and must add to $180^\circ$:

Solving for $\angle ABD$ gives