The figure is a rhombus, and i am trying to find the measure of angle adb

Angle ∠ADB is 41°.

It is a polygon that has four sides and four corners. The sum of the internal angle is 360 degrees. In Rhombus opposite sides are parallel and all sides are equal. And its diagonals intersect at the right-angle.

Given

A rhombus ABCD,

∠DCA = 13x - 16

∠ACB = 9x + 4

The measure of ∠ADB.

We know that the angle ∠DCA and  ∠ACB are equal.

13x - 16 = 9x + 4

13x - 9x = 4 + 16

       4x = 20

         x = 5

So the angle ∠DCA and ∠ACB will be

∠DCA = 13x - 16 = 64 - 16 = 49

∠ACB = 9x + 4 = 45 + 4 = 49

The sum of the internal angle is 360 degrees

∠ABC + ∠BCD + ∠CDA + ∠DAB = 360

∠CDA + ∠BCD + ∠CDA + ∠BCD = 360

                            ∠ABC + ∠BCD = 180

                                  ∠CDA + 98 = 180

                                          ∠CDA = 82

So we know

2 x ∠ADB = ∠CDA

2 x ∠ADB = 82

     ∠ADB = 41

Thus, Angle ∠ADB is 41°.

More about the rhombus link is given below.

link

86 - 9x

Step-by-step explanation:

In a rhombus, adjacent angles are supplementary.  So:

∠ADE + ∠EDC + ∠DCE + ∠ECB = 180

Also, the diagonals of a rhombus are perpendicular.  Since the sum of a triangle's angles add up to 180:

∠EDC + 90 + ∠DCE = 180

∠EDC + ∠DCE = 90

Substituting:

∠ADE + 90 + ∠ECB = 180

∠ADE + ∠ECB = 90

∠ADE + 9x + 4 = 90

∠ADE = 86 - 9x

∠ADB is the same as ∠ADE, so the answer is 86 - 9x.

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