Question

find m∠bgh. m∠bgh = \boxed{\space}°
(there is a diagram with lines ab and cd parallel, intersected by transversal ef at points g and h. ∠bgh is 3x°, and ∠ghd is (2x + 50)°)

Explanation:

Step1: Identify Angle Relationship

Since \( AB \parallel CD \) and \( EF \) is a transversal, \( \angle BGH \) and \( \angle GHD \) are same - side interior angles? Wait, no, actually \( \angle BGH \) and \( \angle GHD \) are same - side interior angles? Wait, no, looking at the angles \( 3x^{\circ} \) (which is \( \angle BGH \)) and \( (2x + 50)^{\circ} \) (which is \( \angle GHD \)), since \( AB\parallel CD \), these two angles are same - side interior angles? Wait, no, actually, if we consider the parallel lines \( AB \) and \( CD \) cut by transversal \( EF \), \( \angle BGH \) and \( \angle GHD \) are same - side interior angles? Wait, no, maybe they are alternate interior angles? Wait, no, the diagram shows that \( AB \) and \( CD \) are parallel, and \( EF \) intersects them at \( G \) and \( H \). So \( \angle BGH \) and \( \angle GHD \) are same - side interior angles? Wait, no, actually, \( \angle BGH \) and \( \angle GHD \) are same - side interior angles, but wait, no, if we look at the positions, \( \angle BGH \) and \( \angle GHD \) are same - side interior angles, but actually, in the diagram, since \( AB\parallel CD \), \( \angle BGH \) and \( \angle GHD \) are same - side interior angles? Wait, no, maybe they are supplementary? Wait, no, wait, \( \angle BGH \) and \( \angle GHD \) are same - side interior angles, so they should be supplementary? Wait, no, no, wait, actually, \( \angle BGH \) and \( \angle GHD \) are same - side interior angles, so \( \angle BGH+\angle GHD = 180^{\circ} \)? Wait, no, that can't be. Wait, maybe \( \angle BGH \) and \( \angle GHD \) are alternate interior angles? Wait, no, the angle \( \angle BGH = 3x \) and \( \angle GHD=(2x + 50) \). Wait, maybe they are equal? Wait, if \( AB\parallel CD \), then \( \angle BGH \) and \( \angle GHD \) are same - side interior angles? No, wait, let's re - examine. The lines \( AB \) and \( CD \) are parallel, and \( EF \) is a transversal. So \( \angle BGH \) and \( \angle GHD \) are same - side interior angles, so they are supplementary? Wait, no, that would mean \( 3x+(2x + 50)=180 \). Wait, but maybe they are equal? Wait, no, maybe I made a mistake. Wait, actually, \( \angle BGH \) and \( \angle GHD \) are same - side interior angles, so \( 3x+(2x + 50)=180 \)? Wait, let's solve that equation.

Step2: Solve for \( x \)

Combine like terms: \( 3x+2x + 50=180 \)
\( 5x+50 = 180 \)
Subtract 50 from both sides: \( 5x=180 - 50=130 \)
Divide both sides by 5: \( x = 26 \)? Wait, no, that gives \( 5x=130 \), \( x = 26 \), then \( 3x = 78 \), and \( 2x+50=52 + 50 = 102 \), and \( 78+102 = 180 \), which works. Wait, so \( \angle BGH \) and \( \angle GHD \) are same - side interior angles, so they are supplementary. So we set up the equation \( 3x+(2x + 50)=180 \)
\( 5x+50=180 \)
\( 5x=130 \)
\( x = 26 \)
Then \( m\angle BGH=3x=3\times26 = 78 \)? Wait, no, wait, maybe I mixed up the angles. Wait, actually, \( \angle BGH \) and \( \angle GHD \) are same - side interior angles, so they are supplementary. So the equation is \( 3x+(2x + 50)=180 \)
Solving:
\( 5x+50 = 180 \)
\( 5x=180 - 50=130 \)
\( x = 26 \)
Then \( m\angle BGH = 3x=3\times26 = 78 \)? Wait, but let's check again. Wait, maybe \( \angle BGH \) and \( \angle GHD \) are alternate interior angles, so they are equal. Then \( 3x=2x + 50 \), so \( x = 50 \), then \( m\angle BGH=3\times50 = 150 \), and \( 2x + 50=150 \), which also works. Wait, now I'm confused. Let's look at the diagram again. The lines \( AB \) and \( CD \) are parallel, \( EF \) is a transversal. \( \angle BGH \) is above \( AB \) and to the right of \( EF \), \( \angle GHD \) is above \( CD \) and to the right of \( EF \)? Wait, no, \( \angle GHD \) is below \( CD \)? Wait, no, the diagram: \( AB \) is the top line, \( CD \) is the bottom line. \( EF \) is a line that goes through \( G \) on \( AB \) and \( H \) on \( CD \), with \( E \) above \( AB \) and \( F \) below \( CD \). So \( \angle BGH \) is between \( AB \) and \( EF \), on the right side of \( EF \), and \( \angle GHD \) is between \( CD \) and \( EF \), on the right side of \( EF \). So since \( AB\parallel CD \), these two angles are same - side interior angles, so they should be supplementary. Wait, but if they are on the same side of the transversal and between the two parallel lines, they are same - side interior angles, so they are supplementary. So \( 3x+(2x + 50)=180 \)
\( 5x=130 \)
\( x = 26 \)
Then \( m\angle BGH=3\times26 = 78 \). Wait, but if we consider alternate interior angles, \( \angle BGH \) and \( \angle GHC \) would be alternate interior angles, but \( \angle GHD \) is adjacent to \( \angle GHC \) (they are supplementary). Wait, maybe I made a mistake in the angle relationship. Let's re - analyze.

Wait, \( AB\parallel CD \), transversal \( EF \). \( \angle BGH \) and \( \angle GHD \): \( \angle BGH \) is at \( G \), between \( AB \) and \( EF \), on the right of \( EF \). \( \angle GHD \) is at \( H \), between \( CD \) and \( EF \), on the right of \( EF \). So they are same - side interior angles, so they are supplementary. So the equation \( 3x+(2x + 50)=180 \) is correct.

Solving:

\( 3x+2x+50 = 180 \)

\( 5x=180 - 50 \)

\( 5x = 130 \)

\( x=\frac{130}{5}=26 \)

Then \( m\angle BGH = 3x=3\times26 = 78 \). Wait, but let's check with \( x = 50 \). If \( x = 50 \), \( 3x = 150 \), \( 2x+50=150 \), so \( \angle BGH=\angle GHD = 150^{\circ} \), but then they would be same - side interior angles, and \( 150 + 150=300 eq180 \), so that's wrong. So the correct relationship is that they are supplementary, so \( x = 26 \), and \( m\angle BGH = 78^{\circ} \). Wait, no, wait, maybe I had the angle labels wrong. Maybe \( \angle BGH \) and \( \angle GHD \) are alternate interior angles. Let's see, if \( AB\parallel CD \), alternate interior angles are equal. So \( \angle BGH=\angle GHD \), so \( 3x=2x + 50 \), so \( x = 50 \), then \( m\angle BGH=3\times50 = 150^{\circ} \), and \( \angle GHD=2\times50 + 50=150^{\circ} \), and then same - side interior angles would be \( 150+150 = 300\), which is not 180. So that can't be. Wait, maybe the angle \( (2x + 50)^{\circ} \) is \( \angle GHC \), not \( \angle GHD \). Let's re - label the diagram. \( H \) is on \( CD \), with \( C \) to the left and \( D \) to the right. So \( \angle GHC \) is to the left of \( EF \) and \( \angle GHD \) is to the right. So \( \angle BGH \) (right of \( EF \), above \( AB \)) and \( \angle GHC \) (left of \( EF \), above \( CD \)) are alternate interior angles. So \( \angle BGH=\angle GHC \), and \( \angle GHC+\angle GHD = 180^{\circ} \) (linear pair). So \( \angle BGH+\angle GHD = 180^{\circ} \). So \( 3x+(2x + 50)=180 \), which is what we had before. So \( x = 26 \), \( m\angle BGH = 78^{\circ} \).

Answer:

\( 78 \)