Question 39·Hard·Lines, Angles, and Triangles

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In the $xy$ -plane, point $A$ is $(0,0)$ and point $B$ is $(k,0)$ , where $k$ is a positive constant. Point $C$ is $(k,2k)$ and point $D$ is $(2k,2k)$ .

Triangles $ABC$ and $ABD$ are drawn. The measure of $\angle CAB$ is $t^\circ$ .

What is the measure of $\angle ABD$ expressed in terms of $t$ ?

For coordinate-geometry angle questions, first sketch or visualize the points to identify which segments are horizontal or vertical and where right angles occur. Then compute slopes to detect parallel lines, since parallel lines create equal acute angles with a given transversal and supplementary obtuse ones. Once you know one angle (like $t^\circ$ ) between a transversal and a parallel line, use linear-pair and supplementary relationships (angles on a straight line sum to $180^\circ$ ) to quickly express any related angle in terms of $t$ without doing trigonometry.

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Sketch and mark right angles

Draw the points and connect them. Which segments are horizontal, which are vertical, and where do you see a right angle in triangle $ABC$ ?

Look for parallel lines

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

Compare the acute and obtuse angles at $A$ and $B$

$\angle CAB$ is an acute angle between a horizontal segment and a slanted segment. At $B$ , the slanted segment is parallel to $AC$ , but $\angle ABD$ uses the horizontal ray pointing in the opposite direction. How are the two angles related along a straight line?

Think about linear pairs

At a point on a straight line, two adjacent 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?

Desmos Guide

Use a convenient value of $k$

Because multiplying all coordinates by the same positive factor does not change any angles, set $k=1$ for a visual check. In Desmos, enter the points $A=(0,0)$ , $B=(1,0)$ , $C=(1,2)$ , and $D=(2,2)$ .

Graph the lines containing the key segments

Enter y=0, y=2x, and y=2x-2. The first line contains $AB$ , the second contains $AC$ , and the third contains $BD$ . The two slanted lines are parallel.

Compare the angles at the intersections

The acute angle between y=0 and y=2x corresponds to $t^\circ$ . At $B$ , the angle between the rightward horizontal direction and y=2x-2 is the same acute angle. The angle formed with the leftward ray $\overrightarrow{BA}$ is its supplement, so subtract the acute angle from $180^\circ$ .

Step-by-step Explanation

Visualize and classify the segments

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.

Thus, $ABC$ is a right triangle, and triangles $ABC$ and $ABD$ share base $\overline{AB}$ .

Show that $AC$ and $BD$ are parallel

Compute the slopes:

\begin{aligned} m_{AC} &= \frac{2k-0}{k-0}=2,\\ m_{BD} &= \frac{2k-0}{2k-k}=2. \end{aligned}

Since $AC$ and $BD$ have the same slope, they are parallel.

Relate $\angle CAB$ to the acute angle at $B$

$\angle CAB$ is the angle between $\overline{AC}$ and the horizontal ray from $A$ through $B$ . Because $AC\parallel BD$ , the corresponding acute angle at $B$ between the rightward horizontal ray and $\overrightarrow{BD}$ also measures $t^\circ$ .

Use a linear pair at $B$

The rightward horizontal ray from $B$ and $\overrightarrow{BA}$ are opposite rays. Therefore, the $t^\circ$ acute angle and $\angle ABD$ form a linear pair:

\angle ABD+t^\circ=180^\circ.

Thus,

\angle ABD=(180-t)^\circ.

Question 39·Hard·Lines, Angles, and Triangles

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation