Question

1. lines hx and mb are shown. point w is the intersection point of $overleftrightarrow{hx}$ and $overleftrightarrow{mb}$. find the measure of $angle hwm$ if $mangle hwb = (25x + 2)^circ$ and $mangle bwx = (20x - 2)^circ$.

Explanation:

Step1: Identify angle relationship

Angles \( \angle HWB \) and \( \angle BWX \) are supplementary (form a linear pair), so their sum is \( 180^\circ \).
\[ (25x + 2) + (20x - 2) = 180 \]

Step2: Solve for \( x \)

Combine like terms:
\[ 45x = 180 \]
Divide both sides by 45:
\[ x = \frac{180}{45} = 4 \]

Step3: Find \( m\angle HWB \)

Substitute \( x = 4 \) into \( 25x + 2 \):
\[ 25(4) + 2 = 100 + 2 = 102^\circ \]

Step4: Find \( m\angle HWM \)

\( \angle HWM \) and \( \angle HWB \) are vertical angles? Wait, no—wait, \( \angle HWM \) and \( \angle BWX \)? Wait, no, let's re-examine. Wait, lines \( HX \) and \( MB \) intersect at \( W \). So \( \angle HWM \) and \( \angle BWX \) are vertical angles? Wait, no, \( \angle HWB \) and \( \angle MWX \) are vertical, \( \angle HWM \) and \( \angle B W X \)? Wait, no, let's see: \( \angle HWB + \angle BWX = 180^\circ \) (linear pair). Then \( \angle HWM \) and \( \angle BWX \) are vertical angles? Wait, no, \( \angle HWM \) and \( \angle B W X \): Wait, \( HX \) is a line, \( MB \) is a line. So \( \angle HWM \) and \( \angle B W X \) are vertical angles? Wait, no, \( \angle HWM \) and \( \angle B W X \): Let's label the angles. At point \( W \), \( HX \) is a straight line, \( MB \) is a straight line. So \( \angle HWB \) and \( \angle BWX \) are adjacent, forming a linear pair (sum to \( 180^\circ \)). Then \( \angle HWM \) is vertical to \( \angle BWX \)? Wait, no, \( \angle HWM \) and \( \angle B W X \): Wait, \( \angle HWM \) and \( \angle B W X \) – actually, \( \angle HWM \) and \( \angle B W X \) are vertical angles? Wait, no, \( \angle HWB \) and \( \angle MWX \) are vertical, \( \angle HWM \) and \( \angle B W X \) are vertical? Wait, maybe I made a mistake. Wait, let's correct: \( \angle HWM \) and \( \angle B W X \) are vertical angles? No, wait, \( HX \) is a line, so \( H-W-X \) is straight. \( MB \) is a line, \( M-W-B \) is straight. So the vertical angles are \( \angle HWM \) and \( \angle B W X \), and \( \angle HWB \) and \( \angle MWX \). Wait, but we found \( \angle HWB = 102^\circ \), and \( \angle BWX = 20x - 2 = 20(4) - 2 = 78^\circ \). Then \( \angle HWM \) is equal to \( \angle BWX \)? No, wait, no: \( \angle HWM \) and \( \angle HWB \) – wait, no, \( \angle HWM \) and \( \angle B W X \) are vertical angles? Wait, no, \( \angle HWM \) and \( \angle B W X \): Let's see, \( \angle HWM \) is at \( W \), between \( H \) and \( M \), and \( \angle B W X \) is between \( B \) and \( X \). Since \( HX \) and \( MB \) intersect at \( W \), \( \angle HWM \) and \( \angle B W X \) are vertical angles, so they are equal? Wait, no, that can't be. Wait, no, \( \angle HWB \) and \( \angle MWX \) are vertical angles, and \( \angle HWM \) and \( \angle B W X \) are vertical angles. Wait, but \( \angle HWB + \angle BWX = 180^\circ \), so \( \angle HWM = \angle BWX \)? Wait, no, \( \angle HWM \) and \( \angle HWB \) – wait, maybe I messed up the angle labels. Wait, the problem says "Find the measure of \( \angle HWM \)". Let's re-express: \( HX \) and \( MB \) intersect at \( W \). So \( \angle HWB \) and \( \angle BWX \) are adjacent, forming a linear pair (sum to \( 180^\circ \)). Then \( \angle HWM \) is vertical to \( \angle BWX \)? Wait, no, \( \angle HWM \) is opposite to \( \angle BWX \)? Wait, no, \( \angle HWM \) and \( \angle B W X \): Let's draw it mentally. \( H \)---\( W \)---\( X \) is a straight line. \( M \)---\( W \)---\( B \) is a straight line. So the four angles at \( W \) are \( \angle HWM \), \( \angle HWB \), \( \angle BWX \), \( \angle MWX \). So \( \angle HWM \) and \( \angle BWX \) are vertical angles (opposite each other), and \( \angle HWB \) and \( \angle MWX \) are vertical angles. Also, \( \angle HWM + \angle HWB = 180^\circ \) (linear pair), and \( \angle HWB + \angle BWX = 180^\circ \) (linear pair). Wait, that means \( \angle HWM = \angle BWX \). Wait, but we found \( \angle BWX = 20x - 2 = 78^\circ \), so \( \angle HWM = 78^\circ \)? Wait, no, that contradicts. Wait, no, let's check again. Wait, \( \angle HWB = 25x + 2 = 102^\circ \), \( \angle BWX = 20x - 2 = 78^\circ \). Then \( \angle HWM \) is adjacent to \( \angle HWB \), forming a linear pair? Wait, no, \( H \)---\( W \)---\( X \) is straight, so \( \angle HWM + \angle HWB = 180^\circ \)? No, \( \angle HWM \) is between \( H \) and \( M \), \( \angle HWB \) is between \( H \) and \( B \). So \( \angle HWM \) and \( \angle HWB \) are adjacent, with \( M \) and \( B \) on opposite sides of \( HX \). So actually, \( \angle HWM \) and \( \angle HWB \) are supplementary? Wait, no, \( M \)---\( W \)---\( B \) is straight, so \( \angle HWM + \angle HWB = 180^\circ \)? Wait, no, \( M \)---\( W \)---\( B \) is a straight line, so the sum of angles on a straight line is \( 180^\circ \). So \( \angle HWM + \angle HWB = 180^\circ \)? Wait, no, \( \angle HWM \) and \( \angle BWM \) would be, but \( BWM \) is a straight line. Wait, I think I made a mistake in identifying the angles. Let's start over.

Lines \( HX \) and \( MB \) intersect at \( W \). So \( HX \) is a straight line, so \( \angle HWB + \angle BWX = 180^\circ \) (linear pair). We solved \( (25x + 2) + (20x - 2) = 180 \), got \( x = 4 \), so \( \angle HWB = 102^\circ \), \( \angle BWX = 78^\circ \). Now, \( \angle HWM \) and \( \angle BWX \) are vertical angles? Wait, no, \( \angle HWM \) is opposite \( \angle BWX \)? Wait, no, \( \angle HWM \) is formed by \( H \), \( W \), \( M \), and \( \angle BWX \) is formed by \( B \), \( W \), \( X \). Since \( HX \) and \( MB \) intersect at \( W \), vertical angles are \( \angle HWM \) and \( \angle BWX \), and \( \angle HWB \) and \( \angle MWX \). Wait, but vertical angles are equal. So \( \angle HWM = \angle BWX \)? But \( \angle BWX = 78^\circ \), so \( \angle HWM = 78^\circ \)? Wait, no, that can't be, because \( \angle HWB + \angle HWM = 180^\circ \) (since \( M \)---\( W \)---\( B \) is straight). Wait, yes! \( M \)---\( W \)---\( B \) is a straight line, so the sum of angles on that line is \( 180^\circ \). So \( \angle HWM + \angle HWB = 180^\circ \). We found \( \angle HWB = 102^\circ \), so \( \angle HWM = 180^\circ - 102^\circ = 78^\circ \)? Wait, no, that's the same as \( \angle BWX \). Wait, that makes sense because \( \angle HWM \) and \( \angle BWX \) are vertical angles, so they should be equal. So yes, \( \angle HWM = 78^\circ \)? Wait, no, wait, \( \angle HWB \) is \( 102^\circ \), and \( \angle HWM \) is adjacent to it on the straight line \( MB \), so \( \angle HWM = 180^\circ - 102^\circ = 78^\circ \). Alternatively, since \( \angle BWX = 78^\circ \) and \( \angle HWM \) is vertical to \( \angle BWX \), they are equal. So either way, \( \angle HWM = 78^\circ \). Wait, but let's confirm with \( x = 4 \). \( \angle BWX = 20(4) - 2 = 78^\circ \), and \( \angle HWM \) is vertical to \( \angle BWX \), so \( \angle HWM = 78^\circ \). Or, since \( \angle HWB = 102^\circ \) and \( M-W-B \) is straight, \( \angle HWM = 180 - 102 = 78^\circ \). Yep, that works.

Answer:

\( 78^\circ \)