Question

find the values of x and y in parallelogram edyf.
sum of all angles is 180
5y + 85 + 40 = 180
5y + 125 = 180
-125 -125
5y = 55
5y/5 = 55/5
y = 11

Explanation:

Step1: Solve for y (already partially done)

In triangle or parallelogram, using angle sum. For the angles involving \( y \), we have \( 5y + 85 + 40 = 180 \). Simplify: \( 5y + 125 = 180 \). Subtract 125: \( 5y = 55 \). Divide by 5: \( y = 11 \).

Step2: Solve for x

In parallelogram EDYF, opposite angles or alternate angles? Wait, in triangle EFY, angle sum is \( 180^\circ \). So \( (4x - 5) + 40 + (angle at Y related) \)? Wait, actually, in parallelogram, \( ED \parallel FY \), so angle at D (5y) and angle at F should be equal? Wait, no, let's check triangle EFY: angles are \( (4x - 5)^\circ \), \( 40^\circ \), and the angle at Y adjacent to 85°? Wait, maybe the angle at F and angle at D are equal? Wait, 5y when y=11 is 55°, no, wait 85° and (4x -5) +40? Wait, maybe the triangle EFY: sum of angles is 180. Wait, the angle at Y in triangle EFY: the angle adjacent to 85° is supplementary? No, let's re-express. Wait, in parallelogram, \( ED \parallel FY \), so \( \angle D + \angle Y = 180 \)? No, maybe the triangle EFY: angles are \( (4x - 5) \), \( 40^\circ \), and the angle equal to 5y? Wait, 5y when y=11 is 55°, no. Wait, maybe the angle at F (let's call it \( \angle F \)) and angle at D (\( 5y^\circ \)) are equal? Wait, no, let's use the triangle EFY. The sum of angles in a triangle is 180°, so \( (4x - 5) + 40 + (angle at Y) = 180 \). But angle at Y: the angle between FY and EY is 40°, and the angle between EY and DY is 85°, so the angle at Y in triangle EFY is \( 180 - 85 = 95 \)? No, wait, maybe I messed up. Wait, the diagram: points E, D, Y, F. EDYF is a parallelogram, so ED || FY and EF || DY. So \( \angle EFD = \angle EDY \) (alternate interior angles). Wait, in triangle EFY, angles are \( (4x - 5)^\circ \), \( 40^\circ \), and the angle at Y which is \( 180 - 85 = 95^\circ \)? Wait, no, let's calculate the angle at Y in triangle EFY: \( 180 - (4x - 5) - 40 = 145 - 4x \). But also, that angle and 85° should be supplementary? No, maybe the angle at F (4x -5) and the angle at D (5y) are equal? Wait, 5y when y=11 is 55°, no. Wait, maybe the triangle EFY: \( (4x - 5) + 40 + 85 = 180 \)? Wait, that would be \( 4x -5 + 125 = 180 \), \( 4x + 120 = 180 \), \( 4x = 60 \), \( x = 15 \). Wait, let's check: \( 4x -5 = 55 \), 55 + 40 + 85 = 180? 55 + 40 is 95, 95 + 85 is 180. Yes! So that's the triangle EFY: angles are \( (4x -5) \), \( 40^\circ \), and \( 85^\circ \). So sum: \( (4x - 5) + 40 + 85 = 180 \). Simplify: \( 4x + 120 = 180 \). Subtract 120: \( 4x = 60 \). Divide by 4: \( x = 15 \).

Answer:

\( x = 15 \), \( y = 11 \)