Question

$\overline{pl}$ and $\overline{km}$ are diameters of $\odot t$. find $m\overarc{pj}$.

Explanation:

Step1: Identify right angle

Since \( PL \) is a diameter and there's a right angle at \( T \) (the right - angle symbol), the angle between \( PT \) and \( LT \) is \( 90^{\circ} \). We know the angle between \( KT \) and \( JT \) is \( 32^{\circ} \), and we need to find the measure of arc \( PJ \). The sum of angles around a point \( T \) on a straight line (diameter \( PL \)) and considering the other angles. Wait, actually, since \( KM \) and \( PL \) are diameters, the sum of angles in a semicircle? Wait, no. Let's look at the angles at the center. The angle for arc \( PJ \) is the angle at \( T \) between \( P \) and \( J \). We know that the angle between \( LT \) and \( KT \): Wait, maybe a better way. The total angle in a circle is \( 360^{\circ} \), but since \( PL \) and \( KM \) are diameters, they divide the circle into four semicircles? No, diameters intersect at \( T \), so the vertical angles are equal. Wait, looking at the diagram, we have angle \( \angle PTN = 42^{\circ} \), \( \angle NT M=48^{\circ} \), \( \angle KTL \) (wait, no, there's a right angle between \( PT \) and \( LT \)? Wait, the diagram has a right - angle symbol between \( PT \) and \( LT \), so \( \angle PTL = 90^{\circ} \). Then, the angle between \( KT \) and \( JT \) is \( 32^{\circ} \), and we need to find \( \angle PTJ \). Since \( \angle PTL = 90^{\circ} \), and \( \angle KTL \) (wait, no, \( KM \) is a diameter, so \( \angle KTM = 180^{\circ} \)? No, let's re - examine. The center angles: We know that \( \angle PTJ+\angle JT K+\angle KTL = 180^{\circ} \)? Wait, no, \( PL \) is a diameter, so the straight line \( PL \) means that the sum of angles on one side of \( PL \) is \( 180^{\circ} \). Wait, the right - angle is \( \angle PTL = 90^{\circ} \), so \( \angle PTK+\angle KTL = 90^{\circ} \)? No, maybe I made a mistake. Wait, the angle between \( PT \) and \( LT \) is \( 90^{\circ} \), so \( \angle PTJ+\angle JT K+\angle KTL=90^{\circ} \)? No, \( \angle KTL \) is not given. Wait, no, looking at the diagram, the angle between \( KT \) and \( JT \) is \( 32^{\circ} \), and we know that the angle between \( PT \) and \( LT \) is \( 90^{\circ} \), and the angle between \( KT \) and \( LT \): Wait, maybe the correct approach is: Since \( \angle PTL = 90^{\circ} \), and \( \angle JT K = 32^{\circ} \), and we need to find \( \angle PTJ \). Wait, no, let's see the angles at \( T \). The angle for arc \( PJ \) is \( \angle PTJ \). We know that \( \angle PTL = 90^{\circ} \), and \( \angle KTL \) (wait, no, \( KM \) is a diameter, so \( \angle KTM = 180^{\circ} \), but maybe the right - angle is \( \angle PTL = 90^{\circ} \), so \( \angle PTJ=90^{\circ}-\angle JT K - \angle KTL \)? No, this is confusing. Wait, maybe the diagram has \( \angle PTJ = 90^{\circ}- 32^{\circ}-(180^{\circ}- 90^{\circ}- 42^{\circ}- 48^{\circ}) \)? No, that's not right. Wait, let's start over.

Wait, the key is that the measure of an arc is equal to the measure of its central angle. So we need to find the central angle \( \angle PTJ \). We know that \( \angle PTL = 90^{\circ} \) (right angle), \( \angle JT K = 32^{\circ} \), and we can find \( \angle KTL \) from the other angles? Wait, no, the sum of angles around point \( T \) on the line \( PL \) (since \( PL \) is a diameter) should be \( 180^{\circ} \) on each side? No, \( PL \) is a straight line, so the sum of angles on one side of \( PL \) is \( 180^{\circ} \). The right - angle is \( 90^{\circ} \), so the other part (from \( K \) to \( P \) through \( J \)) should also sum to \( 90^{\circ} \)? Wait, no, the right - angle is between \( PT \) and \( LT \), so \( \angle PTL = 90^{\circ} \), which means that \( \angle PTJ+\angle JT K+\angle KTL = 90^{\circ} \)? No, \( \angle KTL \) is not given. Wait, maybe the diagram has \( \angle PTJ = 90^{\circ}-32^{\circ}-(180 - 90 - 42 - 48) \)? No, this is wrong. Wait, let's look at the angles given: \( \angle PTN = 42^{\circ} \), \( \angle NT M = 48^{\circ} \), \( \angle JT K=32^{\circ} \), and \( \angle PTL = 90^{\circ} \). Since \( KM \) is a diameter, \( \angle KTM = 180^{\circ} \), so \( \angle PTJ+\angle JT K+\angle KTL = 180^{\circ}-\angle PTL \)? No, I think I messed up. Wait, the correct way: The measure of arc \( PJ \) is equal to the measure of the central angle \( \angle PTJ \). We know that \( \angle PTL = 90^{\circ} \) (right angle), and \( \angle JT K = 32^{\circ} \), and we can find \( \angle KTL \) as follows: The sum of angles \( \angle PTN+\angle NT M+\angle MT L=180^{\circ} \)? No, \( PL \) is a diameter, so \( \angle PTL = 90^{\circ} \), so \( \angle MT L=90^{\circ}-\angle NT M - \angle PTN=90 - 48 - 42 = 0^{\circ} \)? No, that can't be. Wait, maybe the right - angle is not \( \angle PTL \), but \( \angle KTL \)? No, the diagram shows the right - angle between \( PT \) and \( LT \). Wait, let's start over. Let's list all central angles:

- \( \angle PTN = 42^{\circ} \)
- \( \angle NT M = 48^{\circ} \)
- \( \angle MT L \): Since \( PL \) is a diameter, \( \angle PTL = 180^{\circ} \)? No, a diameter divides the circle into two semicircles, each \( 180^{\circ} \). So the sum of angles on one side of \( PL \) is \( 180^{\circ} \). Wait, the right - angle symbol is between \( PT \) and \( LT \), so \( \angle PTL = 90^{\circ} \), which means that \( \angle PTJ+\angle JT K+\angle KTL = 90^{\circ} \). But \( \angle KTL \) is equal to \( \angle PTN+\angle NT M \)? No, \( \angle KTL \) and \( \angle PTN+\angle NT M \) are vertical angles? Wait, \( KM \) and \( PL \) are diameters, so they intersect at \( T \), so vertical angles are equal. So \( \angle KTL=\angle PTN+\angle NT M = 42^{\circ}+48^{\circ}=90^{\circ} \). Ah! That's the key. Since \( KM \) and \( PL \) are diameters, the vertical angles \( \angle KTL \) and \( \angle PTN+\angle NT M \) are equal. So \( \angle KTL = 42^{\circ}+48^{\circ}=90^{\circ} \). Now, since \( \angle PTL = 90^{\circ} \) (right angle) and \( \angle KTL = 90^{\circ} \), and we know \( \angle JT K = 32^{\circ} \), then \( \angle PTJ=\angle KTL-\angle JT K \)? Wait, no. Wait, \( \angle PTJ+\angle JT K=\angle KTL \)? No, \( \angle KTL = 90^{\circ} \), and \( \angle PTJ+\angle JT K = 90^{\circ} \)? Wait, no, \( \angle PTJ+\angle JT K+\angle KTL = 180^{\circ} \)? No, I'm getting confused. Wait, let's use the fact that vertical angles: When two diameters intersect, the opposite angles are equal. So \( \angle KTL=\angle PTN+\angle NT M = 42 + 48=90^{\circ} \). Then, since \( \angle PTJ+\angle JT K=\angle KTL \)? No, \( \angle PTJ+\angle JT K = 90^{\circ} \) because \( \angle KTL = 90^{\circ} \) and \( \angle PTJ \) and \( \angle JT K \) are angles that add up to \( \angle KTL \)? Wait, no, \( \angle PTJ+\angle JT K+\angle KTL = 180^{\circ} \) is wrong. Wait, the correct approach: The measure of arc \( PJ \) is the central angle \( \angle PTJ \). We know that \( \angle KTL = 42^{\circ}+48^{\circ}=90^{\circ} \) (vertical angles from the diameters), and \( \angle JT K = 32^{\circ} \), so \( \angle PTJ=\angle KTL-\angle JT K \)? No, \( \angle PTJ = 90^{\circ}-\angle JT K \)? Wait, no, \( \angle PTJ+\angle JT K=\angle KTL \), so \( \angle PTJ = 90^{\circ}-32^{\circ}=58^{\circ} \)? Wait, no, that doesn't seem right. Wait, no, let's look at the straight line \( PL \). Since \( PL \) is a diameter, the sum of angles on one side of \( PL \) is \( 180^{\circ} \). We have \( \angle PTN = 42^{\circ} \), \( \angle NT M = 48^{\circ} \), \( \angle MT L \) (wait, no), and on the other side, \( \angle PTJ+\angle JT K+\angle KTL = 180^{\circ} \). But \( \angle KTL = 42 + 48 = 90^{\circ} \), \( \angle JT K = 32^{\circ} \), so \( \angle PTJ=180 - 90 - 32=58^{\circ} \)? Wait, no, \( PL \) is a diameter, so the sum of angles from \( P \) to \( L \) through \( J \) and \( K \) is \( 180^{\circ} \). So \( \angle PTJ+\angle JT K+\angle KTL = 180^{\circ} \). We know \( \angle KTL = 42 + 48 = 90^{\circ} \), \( \angle JT K = 32^{\circ} \), so \( \angle PTJ=180-(90 + 32)=58^{\circ} \)? Wait, but that contradicts the earlier thought. Wait, no, the right - angle is \( \angle PTL = 90^{\circ} \), so \( \angle PTJ+\angle JT K = 90^{\circ} \), because \( \angle KTL = 90^{\circ} \) (vertical angle). So \( \angle PTJ=90 - 32 = 58^{\circ} \). Yes, that makes sense. Because \( \angle KTL \) is equal to \( \angle PTN+\angle NT M=42 + 48 = 90^{\circ} \) (vertical angles), and since \( \angle PTJ+\angle JT K=\angle KTL \) (because they are adjacent angles forming \( \angle KTL \)? No, \( \angle PTJ \) and \( \angle JT K \) are adjacent angles with \( \angle KTL \) as their sum? Wait, no, \( \angle PTJ \) and \( \angle JT K \) are on one side of the diameter \( PL \), and \( \angle KTL \) is on the other side, but since \( PL \) is a diameter, the sum of angles on each side is \( 180^{\circ} \), but the right - angle shows that \( \angle PTL = 90^{\circ} \), so \( \angle PTJ+\angle JT K = 90^{\circ} \), and \( \angle KTL = 90^{\circ} \). So \( \angle PTJ=90 - 32 = 58^{\circ} \). Therefore, the measure of arc \( PJ \) is equal to the measure of the central angle \( \angle PTJ \), which is \( 58^{\circ} \). Wait, but let's check again. The central angle for arc \( PJ \) is \( \angle PTJ \). We know that \( \angle PTN = 42^{\circ} \), \( \angle NT M = 48^{\circ} \), \( \angle JT K = 32^{\circ} \), and \( \angle PTL = 90^{\circ} \). Since \( KM \) is a diameter, \( \angle KTM = 180^{\circ} \), so \( \angle PTJ+\angle JT K+\angle KTL = 180^{\circ} \). And \( \angle KTL=\angle PTN+\angle NT M = 42 + 48 = 90^{\circ} \), \( \angle JT K = 32^{\circ} \), so \( \angle PTJ=180-(90 + 32)=58^{\circ} \). Yes, that's correct.

Step1: Calculate the vertical angle \( \angle KTL \)

Since \( PL \) and \( KM \) are diameters, the vertical angles \( \angle KTL \) and \( \angle PTN+\angle NT M \) are equal. Given \( \angle PTN = 42^{\circ} \) and \( \angle NT M = 48^{\circ} \), we have:
\( \angle KTL=\angle PTN+\angle NT M = 42^{\circ}+48^{\circ}=90^{\circ} \)

Step2: Calculate the central angle \( \angle PTJ \)

We know that the sum of angles on a straight line (diameter \( PL \)) is \( 180^{\circ} \). Let \( \angle PTJ=x \), \( \angle JT K = 32^{\circ} \) and \( \angle KTL = 90^{\circ} \). Then:
\( x+\angle JT K+\angle KTL = 180^{\circ} \)
Substitute \( \angle JT K = 32^{\circ} \) and \( \angle KTL = 90^{\circ} \) into the equation:
\( x + 32^{\circ}+90^{\circ}=180^{\circ} \)
\( x=180^{\circ}-(32^{\circ}+90^{\circ}) \)
\( x = 58^{\circ} \)
Since the measure of an arc is equal to the measure of its central angle, \( m\widehat{PJ}=\angle PTJ = 58^{\circ} \)

Answer:

\( 58^{\circ} \)