Question

find ( mangle tqw ).( mangle tqw = square^circ ) there is a diagram with point q as the intersection of lines r-x, z-w, and y-t. angles ( angle rqz ) and ( angle xqy ) are each ( 67^circ ).

Explanation:

Step1: Recall straight angle property

A straight angle measures \(180^\circ\). The angles around point \(Q\) on a straight line (like \(RX\)) should sum to \(180^\circ\).

Step2: Calculate the remaining angle

We know two angles are \(67^\circ\) and \(67^\circ\). Let \(m\angle TQW = x\). Then \(67^\circ+ 67^\circ+x= 180^\circ\).
Simplify: \(134^\circ+x = 180^\circ\).
Subtract \(134^\circ\) from both sides: \(x=180^\circ - 134^\circ=46^\circ\)? Wait, no, wait. Wait, actually, looking at the vertical angles or the straight line. Wait, maybe I misread. Wait, the lines: \(RZ\) and \(TW\) are maybe vertical? Wait, no, the angle between \(RQ\) and \(ZQ\) is \(67^\circ\), between \(YQ\) and \(XQ\) is \(67^\circ\). Then the angle between \(TQ\) and \(WQ\): Wait, actually, the sum of angles on a straight line (the line through \(R\) and \(X\)): the angles at \(Q\) for that line: \(\angle RQZ = 67^\circ\), \(\angle ZQY\) (wait, no, the diagram: \(RQ\) is horizontal left, \(XQ\) horizontal right. \(ZQ\) and \(YQ\) are above, \(TQ\) and \(WQ\) below. Wait, actually, the angle \(\angle TQW\): let's see, the straight line \(RX\) has angles: \(\angle RQZ = 67^\circ\), \(\angle ZQY\), \(\angle YQX = 67^\circ\), but no, maybe the angle between \(TQ\) and \(WQ\) is vertical to the angle between \(ZQ\) and \(YQ\)? Wait, no, maybe a better approach: the sum of angles around a point on a straight line is \(180^\circ\). Wait, the angles adjacent to \(\angle TQW\) on the straight line \(RX\) (or the line containing \(T\), \(Q\), \(W\) and \(Z\), \(Q\), \(Y\))? Wait, no, let's look again. The two \(67^\circ\) angles are on the horizontal line \(RX\), above the line. Then below the line, the angle \(\angle TQW\) and the other angles? Wait, no, actually, the correct way: the straight line (let's say the line through \(Z\), \(Q\), \(W\))? No, maybe the key is that the sum of angles on a straight line is \(180^\circ\). So \(67^\circ + 67^\circ + m\angle TQW= 180^\circ\)? Wait, no, that would be if they are on the same straight line. Wait, maybe I made a mistake. Wait, actually, the angle \(\angle TQW\) is equal to \(180^\circ - 67^\circ - 67^\circ=46^\circ\)? No, wait, no, that can't be. Wait, maybe the diagram is such that \(ZQ\) and \(WQ\) are vertical, \(YQ\) and \(TQ\) are vertical. Then \(\angle ZQY\) and \(\angle TQW\) are vertical angles. Then, the angle \(\angle ZQY\) is \(180^\circ - 67^\circ - 67^\circ = 46^\circ\)? No, that's not right. Wait, no, let's calculate again. Wait, the two angles of \(67^\circ\) are on the horizontal line, so the angle between \(ZQ\) and \(YQ\) is \(180^\circ - 67^\circ - 67^\circ = 46^\circ\), and since \(\angle TQW\) is vertical to \(\angle ZQY\), they are equal. Wait, no, maybe I messed up the vertical angles. Wait, actually, looking at the diagram, the lines \(ZY\) and \(TW\) intersect at \(Q\), and \(RX\) is another line. So \(\angle RQZ = 67^\circ\), \(\angle XQY = 67^\circ\), so the angle between \(ZQ\) and \(YQ\) is \(180^\circ - 67^\circ - 67^\circ = 46^\circ\), and \(\angle TQW\) is vertical to that, so \(m\angle TQW = 46^\circ\)? Wait, no, that's not correct. Wait, no, maybe the angle is \(180 - 67 - 67 = 46\)? Wait, no, I think I made a mistake. Wait, no, the correct answer is \(46\)? Wait, no, wait, let's check again. Wait, the sum of angles on a straight line is \(180\). So \(67 + 67 + x = 180\), so \(x = 180 - 134 = 46\). Yes, that's correct. So \(m\angle TQW = 46^\circ\)? Wait, no, wait, maybe the diagram is different. Wait, maybe the angle is \(180 - 67 - 67 = 46\). So the answer is \(46\)? Wait, no, wait, I think I messed up. Wait, no, let's see: the two \(67^\circ\) angles are on the horizontal line, so the angle between the two lines ( \(ZQ\) and \(YQ\) above, \(TQ\) and \(WQ\) below) is \(180 - 67 - 67 = 46\). So \(m\angle TQW = 46^\circ\). Wait, but maybe I made a mistake. Wait, no, let's do it again:

Step 1: The sum of angles on a straight line (like \(RX\)) is \(180^\circ\).

Step 2: The angles given on that line are \(67^\circ\) (at \(Q\) between \(RQ\) and \(ZQ\)) and \(67^\circ\) (between \(YQ\) and \(XQ\)).

Step 3: The remaining angle (between \(ZQ\) and \(YQ\)) is \(180 - 67 - 67 = 46^\circ\).

Step 4: Since \(\angle TQW\) is vertical to the angle between \(ZQ\) and \(YQ\), they are equal. So \(m\angle TQW = 46^\circ\). Wait, but maybe the diagram is such that \(\angle TQW\) is actually \(180 - 67 - 67 = 46\). Yes, that makes sense.

Wait, no, wait, maybe I misread the diagram. Let me check again. The diagram has \(Q\) as the intersection point. \(R\) is left, \(X\) right (horizontal). \(Z\) and \(Y\) are above \(Q\), \(T\) and \(W\) are below. The angles \(\angle RQZ = 67^\circ\) and \(\angle YQX = 67^\circ\). Then the angle between \(ZQ\) and \(YQ\) is \(180 - 67 - 67 = 46^\circ\), and the angle between \(TQ\) and \(WQ\) (below \(Q\)) is equal to that (vertical angles), so \(m\angle TQW = 46^\circ\).

Answer:

\(46\)