In the figure, ab∥cd. find x and y.

y=131°

x=53°

Step-by-step explanation:

∠ ABD and ∠ BDC are supplementary.

The sum of the supplementary angles =180 °

thus y-4+x=180i

x and 37° are complementary, that  is, they add up to 90°

Thus, x=90-37=53°

Using this value in equation 1 we obtain:

y-4°+53° =180°

y= 180°-53°+4°

y=131°

x  = 53°

y = 131°

Step-by-step explanation:

From the figure we can see a right angled triangle.

AB∥CD

To find the value of x and y

Consider the large triangle, by using angle sum property we can write,

90 + 37 + x = 180

127 + x = 180

x = 180 - 127

x  = 53°

Since AB∥CD and BD is a traversal on these parallel lines.

Therefore <ABD and < CDB are supplementary

we have x  = 53°,

x + (y - 4) = 180

53 + y - 4 = 180

y = 180 - 49 = 131°

y = 131°

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

addition

step-by-step explanation: