In rhombus ABCD, m∠XCB = 6a − 2 and m∠XBC = 4a + 12. What is the measure of ∠XCB?

Angle XCB = 46 degrees ( ∠XCB = 46°)

Step-by-step explanation: Please refer to the picture attached for details.

A rhombus basically is a parallelogram with a diamond shape. All four sides have equal length, opposite sides are parallel and opposite angles are equal.

The sides have been labelled ABCD as stated in the question, and the point X is extended to point B to form angle XBC and also extended to point C to form angle XCB. Upon careful observation we shall notice that a triangle has been formed with sides BXC and angle X is a right angle, which measures 90 degrees.

The sum of angles in a triangle equals 180, therefore,

Angle XBC + Angle XCB + Angle X = 180

Substituting for the given values we now have the following;

4a + 12 + 6a - 2 + 90 = 180

10a + 10 + 90 = 180

10a + 100 = 180

Subtract 100 from both sides of the equation

10a = 80

Divide both sides of the equation 10

a = 8

If XCB = 6a - 2,

Substitute for the value of a

XCB = 6(8) - 2

XCB = 48 - 2

XCB = 46

Therefore angle XCB measures 46 degrees.

idk

step-by-step explanation: