Question

ABCD IS A Rectangle, Solve For X

Answer 1

Answer:

Required value of X is 77°.

Step-by-step explanation:

It is given that the provided figure is a rectangle. From the properties of rectangle we know that the angles of rectangle at the corners are of 90°.

Therefore, $\angle$ ABC = 90°.

In ∆ ABC :

$\angle$ABC = 90°

$\angle$BAE = 45°

From the properties of triangles :

• Sum of angles at vertices = 180°

Thus,

= > $\angle$ABE( or ABC ) + $\angle$BAE + $\angle$BEA = 180°

= > 90° + 45° + $\angle$BEA = 180°

= > 135° + $\angle$BEA = 180°

= > $\angle$BEA = 180° - 135°

= > $\angle$BEA = 45°

We know,

Sum of angles on a straight line = 180

Therefore,

= > $\angle$BEA + X + $\angle$DEC = 180°

= > 45° + X + 58° = 180 { Given, $\angle$ DEC = 58° }

= > X + 103 = 180°

= > X = 180° - 103

= > X = 77°

Hence the required value of X is 77°.

Answer 2

Answer:

☑ quadrilateral ABCD is a rectangle therefor measures of angle A, Angle B , Angle C, Angle D is 90'

☑ In a given figure, angle EAB=45'

so, angle DAE = 90 - angle EAB

angle DAE = 90' - 45'

angle DAE = 45'

we have given, angle DEC = 58'

angleDEC is congruent to angle ADE............(transversal angles)

therefor angleADE = 58'

☑ Now, in triangle AED ,

angle DAE + angle AED +angle ADE=180'

....(measure of sum of angles of triangle is 180')

angle AED = x

45' + x + 58' = 180'

x = 180' - 103'

x = 77'

Therefor x = 77 degree