Question 6·Hard·Lines, Angles, and Triangles

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In the figure shown, points $Q,\,R,\,S$ , and $T$ lie on line segment $PV$ , and triangle $SQX$ is formed by segments $QX$ and $SX$ . Segment $RU$ intersects segment $SX$ at point $W$ , and segments $RU$ and $TU$ meet at point $U$ .

The measure of $\angle SQX$ is $41^\circ$ , the measure of $\angle SXQ$ is $83^\circ$ , the measure of $\angle SWU$ is $102^\circ$ , and the measure of $\angle VTU$ is $152^\circ$ .

Which choice gives the measure, in degrees, of $\angle TUR$ ?

When several triangles and intersections sit on the same straight line, focus on converting every given angle into an angle a segment makes with the baseline. Use triangle angle sums to find missing interior angles, then use linear pairs (supplements) when a ray is reversed (like switching from $RU$ to $UR$ ). Once both rays in the target angle are expressed relative to the same reference direction (the baseline), the desired angle is found by taking the appropriate difference.

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Start with triangle $SQX$

Use the fact that the angles in a triangle add to $180^\circ$ to find $\angle QSX$ .

Use that $W$ is on $SX$

Because $W$ lies on segment $SX$ , the rays $SW$ and $SX$ point in the same direction from $S$ .

Track directions at $U$ carefully

Angle $\angle TUR$ uses the rays $UT$ and $UR$ . Each of these is the opposite direction of a segment direction you may have found ( $TU$ and $RU$ ).

Desmos Guide

Compute the angle at $S$ in triangle $SQX$

Enter a=180-41-83 to compute $\angle QSX$ .

Compute the acute angle that line $RU$ makes with the baseline

Enter b=102-a to compute the acute angle between $WU$ (and thus $RU$ ) and line $PV$ .

Compute the acute angle that line $TU$ makes with the baseline

Enter c=180-152 to compute the acute angle between $TU$ and line $PV$ .

Combine the angles to get $\angle TUR$

Enter d=(180-b)-c. The value shown for d is the measure of $\angle TUR$ .

Step-by-step Explanation

Find the missing angle in triangle $SQX$

In triangle $SQX$ ,

\angle QSX = 180^\circ - 41^\circ - 83^\circ = 56^\circ.

Relate $\angle QSX$ to line $WS$

Since $W$ lies on segment $SX$ , the ray $SW$ is the same ray as $SX$ . Therefore,

\angle QSW = \angle QSX = 56^\circ.

Because $Q$ and $S$ lie on the straight line $PV$ , the line $SQ$ is collinear with $PV$ . So $WS$ makes a $56^\circ$ angle with the line $PV$ (using the angle at $S$ ).

Use $\angle SWU$ to find the angle between $RU$ and $PV$

At $W$ , the angle from ray $WS$ to ray $WU$ is $102^\circ$ . From the diagram, $WS$ slants up to the right while $WU$ slants down to the right, so the acute angle that $WU$ makes with line $PV$ is the difference:

102^\circ - 56^\circ = 46^\circ.

Thus, line $RU$ makes a $46^\circ$ angle with line $PV$ .

Find the acute angle that $UT$ makes with $PV$

At $T$ , $\angle VTU = 152^\circ$ is the obtuse angle between ray $TV$ (to the right along $PV$ ) and ray $TU$ . So the acute angle between $TU$ and line $PV$ is

180^\circ - 152^\circ = 28^\circ.

That means ray $UT$ makes a $28^\circ$ angle with the direction to the right along $PV$ .

Compute $\angle TUR$ at $U$

From Step 3, line $RU$ makes a $46^\circ$ angle with $PV$ , so the opposite ray $UR$ makes an angle of

180^\circ - 46^\circ = 134^\circ

with the direction to the right along $PV$ .

Ray $UT$ makes a $28^\circ$ angle with that same direction, so

\angle TUR = 134^\circ - 28^\circ = 106^\circ.

Question 6·Hard·Lines, Angles, and Triangles

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Step-by-step Explanation

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Step-by-step Explanation