Question

question
triangle stu is formed by connecting the midpoints of the side of triangle pqr. the measures of the interior angles of triangle pqr are shown. find the measure of ∠qts.
figures not necessarily drawn to scale.

Explanation:

Step1: Find angle at Q in triangle PQR

In triangle \( PQR \), we know that the sum of interior angles of a triangle is \( 180^\circ \). Given \( \angle P = 49^\circ \) and \( \angle R = 41^\circ \), so \( \angle Q=180^\circ - 49^\circ - 41^\circ \)
\( \angle Q = 90^\circ \) (since \( 49 + 41=90 \), and \( 180 - 90 = 90 \))

Step2: Analyze midpoints and triangle properties

Since \( S \) and \( T \) are midpoints (as \( STU \) is formed by connecting midpoints), \( ST \) is a midline of triangle \( PQR \). By the midline theorem, \( ST \parallel PR \), and \( SQ=\frac{1}{2}PQ \), \( TQ = \frac{1}{2}RQ \) (wait, actually, since \( Q \) is right - angled, and \( S \) is midpoint of \( PQ \), \( T \) is midpoint of \( RQ \)? Wait, no, looking at the diagram, \( Q \) is right - angled, \( S \) is on \( PQ \), \( T \) is on \( RQ \), and \( U \) is on \( PR \). Since \( S \) and \( T \) are midpoints, \( ST \) is midline, so \( ST\parallel PR \), and \( S \) is midpoint of \( PQ \), \( T \) is midpoint of \( RQ \). Also, in right - triangle \( PQR \) with right angle at \( Q \), \( S \) is midpoint of \( PQ \), \( T \) is midpoint of \( RQ \), so \( ST \) is parallel to \( PR \), and \( \triangle QST \) - wait, no, we need to find \( \angle QTS \).

Wait, another approach: In triangle \( PQR \), \( \angle P = 49^\circ \), \( \angle R=41^\circ \), right - angled at \( Q \). Since \( S \) is midpoint of \( PQ \) and \( T \) is midpoint of \( RQ \) (because \( STU \) is midline triangle), then \( ST \parallel PR \), so \( \angle QST=\angle P = 49^\circ \) (corresponding angles), and \( \angle QTS=\angle R = 41^\circ \)? Wait, no, let's use the fact that in triangle \( QTS \), we can find angles. Wait, actually, since \( S \) and \( T \) are midpoints, \( SQ=\frac{1}{2}PQ \), \( TQ=\frac{1}{2}RQ \), and \( \triangle QST\sim\triangle QPR \) (by SAS similarity, as \( \frac{SQ}{PQ}=\frac{TQ}{RQ}=\frac{1}{2} \) and \( \angle Q \) is common). So the corresponding angles are equal. But we need \( \angle QTS \). Wait, in triangle \( QTS \), we know that \( \angle Q = 90^\circ \), and we can find \( \angle QST \) or \( \angle QTS \). Wait, alternatively, since \( ST\parallel PR \), \( \angle QTS=\angle R \)? Wait, no, \( \angle R = 41^\circ \), \( \angle P = 49^\circ \). Wait, let's look at triangle \( QTS \): \( S \) is midpoint of \( PQ \), \( T \) is midpoint of \( RQ \), so \( ST \) is midline, so \( ST\parallel PR \), so \( \angle QTS=\angle R = 41^\circ \)? No, that's not right. Wait, maybe I made a mistake. Wait, \( \angle Q \) is \( 90^\circ \), \( S \) is on \( PQ \), \( T \) is on \( RQ \). Let's consider triangle \( PQR \): angles are \( \angle P = 49^\circ \), \( \angle R = 41^\circ \), \( \angle Q=90^\circ \). Since \( S \) is midpoint of \( PQ \), \( T \) is midpoint of \( RQ \), then \( ST \) is midline, so \( ST\parallel PR \), so \( \angle TSP=\angle P = 49^\circ \) (alternate interior angles). But we need \( \angle QTS \). Wait, in quadrilateral \( QSTU \) - no, let's think about triangle \( QTS \). \( \angle Q = 90^\circ \), and we know that in triangle \( PQR \), \( \angle P = 49^\circ \), so \( \angle PRQ=41^\circ \). Since \( ST\parallel PR \), \( \angle QTS=\angle PRQ = 41^\circ \)? No, wait, \( \angle QTS \): let's see, \( S \) is midpoint of \( PQ \), so \( SQ=\frac{1}{2}PQ \), \( T \) is midpoint of \( RQ \), so \( TQ=\frac{1}{2}RQ \). Also, \( \triangle QTS \) and \( \triangle QPR \): \( \frac{SQ}{PQ}=\frac{TQ}{RQ}=\frac{1}{2} \), and \( \angle Q \) is common, so by SAS similarity, \( \triangle QTS\sim\triangle QPR \). Therefore, \( \angle QTS=\angle QPR = 49^\circ \)? Wait, no, similarity ratio: corresponding angles. In \( \triangle QPR \), angle at \( P \) is \( \angle QPR = 49^\circ \), angle at \( R \) is \( \angle QRP = 41^\circ \). In \( \triangle QTS \), angle at \( S \) corresponds to angle at \( R \), and angle at \( T \) corresponds to angle at \( P \)? Wait, maybe I mixed up the correspondence. Let's label the triangles: \( \triangle QPR \) with vertices \( Q \), \( P \), \( R \); \( \triangle QTS \) with vertices \( Q \), \( T \), \( S \). So \( Q \) corresponds to \( Q \), \( T \) corresponds to \( P \), \( S \) corresponds to \( R \). So \( \angle QTS=\angle QPR = 49^\circ \)? Wait, no, let's use the right - triangle property. Since \( S \) is midpoint of \( PQ \), in right - triangle \( PQT \) (wait, no, \( Q \) is right - angled, \( PQ \) and \( RQ \) are legs. \( S \) is midpoint of \( PQ \), \( T \) is midpoint of \( RQ \), so \( ST \) is midline, \( ST\parallel PR \), so \( \angle QST=\angle P = 49^\circ \), and in triangle \( QTS \), \( \angle Q = 90^\circ \), so \( \angle QTS=180^\circ - 90^\circ - 49^\circ=41^\circ \)? Wait, no, that's conflicting. Wait, let's start over.

Sum of angles in \( \triangle PQR \): \( \angle P + \angle Q+\angle R = 180^\circ \), \( 49 + \angle Q+41 = 180 \), so \( \angle Q = 90^\circ \). So \( \triangle PQR \) is right - angled at \( Q \). \( S \) is midpoint of \( PQ \), \( T \) is midpoint of \( RQ \) (because \( STU \) is formed by connecting midpoints, so \( S \), \( T \), \( U \) are midpoints of \( PQ \), \( RQ \), \( PR \) respectively). Then, in right - triangle \( PQR \), the segment \( ST \) connects midpoints of the legs, so \( ST \) is parallel to \( PR \) (midline theorem), and \( ST=\frac{1}{2}PR \). Also, \( SQ=\frac{1}{2}PQ \), \( TQ=\frac{1}{2}RQ \). Now, consider triangle \( QTS \): \( \angle Q = 90^\circ \), \( SQ=\frac{1}{2}PQ \), \( TQ=\frac{1}{2}RQ \). Also, since \( ST\parallel PR \), \( \angle QST=\angle P = 49^\circ \) (corresponding angles, as \( ST\parallel PR \) and \( PQ \) is a transversal). Then, in triangle \( QTS \), using angle sum property: \( \angle QTS=180^\circ-\angle Q - \angle QST \)
\( \angle QTS = 180^\circ-90^\circ - 49^\circ=41^\circ \)? Wait, no, that's not right. Wait, if \( \angle QST=\angle P = 49^\circ \), and \( \angle Q = 90^\circ \), then \( \angle QTS = 41^\circ \). But wait, another way: since \( T \) is midpoint of \( RQ \) and \( S \) is midpoint of \( PQ \), and \( \triangle PQR \) is right - angled at \( Q \), then \( QT = \frac{1}{2}RQ \), \( QS=\frac{1}{2}PQ \). Also, \( \tan(\angle QTS)=\frac{QS}{QT}=\frac{\frac{1}{2}PQ}{\frac{1}{2}RQ}=\frac{PQ}{RQ}=\tan(\angle R) \). Since \( \angle R = 41^\circ \), \( \tan(\angle R)=\frac{PQ}{RQ} \), so \( \angle QTS=\angle R = 41^\circ \)? No, \( \tan(\angle P)=\frac{RQ}{PQ} \), \( \tan(\angle R)=\frac{PQ}{RQ} \). Wait, I think I messed up the correspondence. Let's use the fact that in \( \triangle PQR \), \( \angle P = 49^\circ \), \( \angle R = 41^\circ \), right - angled at \( Q \). \( S \) is midpoint of \( PQ \), \( T \) is midpoint of \( RQ \). Then \( ST \) is midline, so \( ST\parallel PR \), so \( \angle TSP=\angle P = 49^\circ \) (alternate interior angles). Now, in triangle \( QTS \), \( \angle Q = 90^\circ \), \( \angle QST = \angle P = 49^\circ \) (because \( ST\parallel PR \), and \( PQ \) is transversal), so \( \angle QTS=180 - 90 - 49 = 41^\circ \)? Wait, no, \( \angle QST \): if \( ST\parallel PR \), then \( \angle QST \) and \( \angle P \) are corresponding angles? Wait, \( PQ \) is a transversal cutting \( ST \) and \( PR \). So \( \angle QST \) and \( \angle P \) are corresponding angles, so \( \angle QST = \angle P = 49^\circ \). Then in \( \triangle QTS \), angles sum to \( 180^\circ \), \( \angle Q = 90^\circ \), \( \angle QST = 49^\circ \), so \( \angle QTS=180 - 90 - 49 = 41^\circ \). Wait, but let's check with the other angle. \( \angle R = 41^\circ \), so maybe \( \angle QTS=\angle R = 41^\circ \).

Wait, maybe a simpler way: since \( S \) and \( T \) are midpoints, \( ST \) is midline, so \( ST\parallel PR \), so \( \angle QTS=\angle R = 41^\circ \) (corresponding angles, as \( TQ \) is a transversal cutting \( ST \) and \( PR \)).

Answer:

\( 41^\circ \) (Wait, no, wait, let's re - check. Wait, \( \angle P = 49^\circ \), \( \angle R = 41^\circ \), right - angled at \( Q \). \( S \) is midpoint of \( PQ \), \( T \) is midpoint of \( RQ \). Then \( ST \) is midline, so \( ST\parallel PR \), so \( \angle QTS=\angle P = 49^\circ \)? I think I made a mistake in transversal. Let's take transversal \( RQ \): \( ST\parallel PR \), so \( \angle QTS=\angle QPR = 49^\circ \)? Wait, now I'm confused. Let's use the angle sum in \( \triangle PQR \): \( \angle Q = 90^\circ \), \( \angle P = 49^\circ \), \( \angle R = 41^\circ \). Since \( S \) is midpoint of \( PQ \), \( T \) is midpoint of \( RQ \), then \( QT=\frac{1}{2}RQ \), \( QS = \frac{1}{2}PQ \). In \( \triangle QTS \), \( \tan(\angle QTS)=\frac{QS}{QT}=\frac{\frac{1}{2}PQ}{\frac{1}{2}RQ}=\frac{PQ}{RQ}=\tan(\angle R) \). Since \( \tan(\angle R)=\frac{PQ}{RQ} \) (because in \( \triangle PQR \), \( \tan(\angle R)=\frac{PQ}{RQ} \), as \( \angle R \) is at \( R \), opposite side \( PQ \), adjacent side \( RQ \)), so \( \angle QTS=\angle R = 41^\circ \). Yes, that makes sense. So the measure of \( \angle QTS \) is \( 41^\circ \).