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 supplementary angles

Since \( \angle HWB \) and \( \angle BWX \) are adjacent and form a linear pair, they are supplementary. So, \( m\angle HWB + m\angle BWX = 180^\circ \).
Substitute the given expressions: \( (25x + 2) + (20x - 2) = 180 \).

Step2: Solve for \( x \)

Simplify the left side: \( 25x + 2 + 20x - 2 = 45x \).
So, \( 45x = 180 \).
Divide both sides by 45: \( x = \frac{180}{45} = 4 \).

Step3: Find \( m\angle HWB \)

Substitute \( x = 4 \) into \( m\angle HWB = (25x + 2)^\circ \):
\( m\angle HWB = 25(4) + 2 = 100 + 2 = 102^\circ \).

Step4: Find \( m\angle HWM \)

\( \angle HWM \) and \( \angle HWB \) are vertical angles? No, wait, \( \angle HWM \) and \( \angle BWX \)? Wait, no. Wait, \( \angle HWM \) and \( \angle HWB \): Wait, actually, \( \angle HWM \) and \( \angle BWX \) are vertical? No, let's look at the diagram. Lines \( HX \) and \( MB \) intersect at \( W \). So \( \angle HWM \) and \( \angle BWX \) are vertical? Wait, no. Wait, \( \angle HWB \) and \( \angle MWX \) are vertical, \( \angle HWM \) and \( \angle B W X \) are vertical? Wait, no. Wait, actually, \( \angle HWM \) and \( \angle B W X \) are supplementary? No, wait. Wait, \( \angle HWB \) and \( \angle HWM \) are adjacent and form a linear pair? Wait, no. Wait, let's re-examine.

Wait, \( HX \) is a line, so \( \angle HWM + \angle M W X = 180^\circ \), but maybe easier: \( \angle HWM \) and \( \angle BWX \) are vertical angles? No, wait, \( \angle HWB \) and \( \angle MWX \) are vertical, \( \angle HWM \) and \( \angle B W X \) are vertical? Wait, no. Wait, actually, \( \angle HWM \) and \( \angle B W X \) are supplementary to \( \angle HWB \)? Wait, no. Wait, let's use the fact that \( \angle HWM \) and \( \angle BWX \) are vertical? No, wait, when two lines intersect, vertical angles are equal. Wait, \( \angle HWM \) and \( \angle BWX \): Let's see, \( HX \) and \( MB \) intersect at \( W \). So the angles: \( \angle HWB \), \( \angle BWX \), \( \angle XWM \), \( \angle MWH \) (which is \( \angle HWM \)). So \( \angle HWB + \angle BWX = 180^\circ \) (linear pair), and \( \angle HWM + \angle HWB = 180^\circ \) (linear pair as well? Wait, no. Wait, \( MB \) is a line, so \( \angle HWM + \angle HWB = 180^\circ \)? No, \( MB \) is a line, so \( \angle HWM + \angle MWB = 180^\circ \), but \( \angle MWB \) is \( \angle HWB \)? Wait, no. Wait, \( MB \) is a straight line, so \( \angle HWM + \angle HWB = 180^\circ \)? No, \( HWM \) and \( HWB \) share the side \( HW \), and \( MB \) is a straight line, so yes, \( \angle HWM + \angle HWB = 180^\circ \)? Wait, no, that can't be. Wait, maybe I made a mistake. Wait, \( \angle HWB \) and \( \angle BWX \) are on line \( HX \), so they are supplementary. Then \( \angle HWM \) and \( \angle BWX \) are vertical angles? Wait, no. Wait, \( \angle HWM \) and \( \angle BWX \): Let's label the angles. When two lines intersect, the vertical angles are equal. So \( \angle HWB \) and \( \angle XWM \) are vertical, \( \angle HWM \) and \( \angle BWX \) are vertical. Wait, yes! So \( \angle HWM = \angle BWX \). Wait, but we found \( x = 4 \), so \( m\angle BWX = 20x - 2 = 20(4) - 2 = 80 - 2 = 78^\circ \). Wait, but that contradicts. Wait, no, maybe I mixed up. Wait, let's start over.

Wait, lines \( HX \) and \( MB \) intersect at \( W \). So \( HX \) is a straight line, so \( \angle HWB + \angle BWX = 180^\circ \) (linear pair). So we solved \( (25x + 2) + (20x - 2) = 180 \), got \( x = 4 \). Then \( m\angle HWB = 25(4) + 2 = 102^\circ \), \( m\angle BWX = 20(4) - 2 = 78^\circ \). Now, \( MB \) is a straight line, so \( \angle HWM + \angle HWB = 180^\circ \) (linear pair), so \( m\angle HWM = 180^\circ - m\angle HWB = 180 - 102 = 78^\circ \). Alternatively, since \( \angle HWM \) and \( \angle BWX \) are vertical angles, they should be equal. Let's check: \( \angle BWX = 78^\circ \), \( \angle HWM = 78^\circ \), which matches. So that's correct.

Answer:

\( 78^\circ \)