ABCD is a rhombus. We are asked to find the length of side $AB$ and the measure of angle $\angle ABC$. We are given that $AB = 12y$, $BC = 4x + 15$, $\angle BAC = 12y$, and $\angle BDC = 7x$. Also, we have the equation $12y + 4x + 15 = 180$

1. Problem Description

ABCD is a rhombus. We are asked to find the length of side $AB$ and the measure of angle $\angle ABC$. We are given that $AB = 12y$, $BC = 4x + 15$, $\angle BAC = 12y$, and $\angle BDC = 7x$. Also, we have the equation $12y + 4x + 15 = 180$

2. Solution Steps

Since ABCD is a rhombus, all sides are equal. Therefore,

$AB = BC$

$12y = 4x + 15$

We are also given that

$12y + 4x + 15 = 180$

Since $12y = 4x + 15$, we can substitute this into the equation above:

$12y + 12y = 180$

$24y = 180$

$y = \frac{180}{24} = \frac{30}{4} = \frac{15}{2} = 7.5$

Thus $y = 7.5$

Then $12y = 4x + 15$ becomes

$12(7.5) = 4x + 15$

$90 = 4x + 15$

$75 = 4x$

$x = \frac{75}{4} = 18.75$

$AB = 12y = 12(7.5) = 90$

Therefore, $AB = 90$

Now, let's find the measure of $\angle ABC$. Since ABCD is a rhombus, the diagonals bisect the angles.

So, $\angle ABC = 2 \angle FBC$.

Since diagonals of a rhombus are perpendicular, $\angle BFA = 90^{\circ}$.

In triangle ABF, $\angle BAF = 12y = 12(7.5) = 90^{\circ}$

This is impossible. So there must be something wrong with the original equation. It has to be

$\angle BAC + \angle BCA = 180$ is impossible.

Since $ABCD$ is a rhombus, opposite sides are parallel, so $12y$ and $4x+15$ must be two adjacent angles such that $12y+4x+15 = 180$ gives adjacent angles add to

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0.

In triangle BCD, $\angle DBC = \angle BDC = 7x$.

$\angle BCD = 4x+15$

Then $7x+7x+4x+15 = 180$

$18x+15=180$

$18x = 165$

$x = \frac{165}{18} = \frac{55}{6}$

$x \approx 9.167$

Then $4x+15 = 4(\frac{55}{6})+15 = \frac{220}{6} + \frac{90}{6} = \frac{310}{6} = \frac{155}{3}$

$BC = 4x+15$ and $AB = 12y$ so $4x+15 = 12y$ so we need the length.

Given $12y+4x+15=180$. Since they are not adjacent angles, we need to use adjacent angles are

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0. $\angle BAD + \angle ADC = 180$. $\angle BAC=12y$, so $\angle BAD=24y$ since the diagonals bisect the angles of a rhombus. $\angle BDC=7x$, so $\angle ADC=14x$. $24y+14x = 180$ then divide by 2 gives $12y+7x=90$

$AB=BC$, so $12y=4x+15$, so $12y-4x = 15$.

$12y+7x = 90$

$(12y+7x)-(12y-4x)= 90-15$

$11x=75$

$x=75/11$

$12y = 4(75/11)+15 = (300+165)/11=465/11$

$y=(465/11)/12 = 465/(11*12) = 465/132=155/44$

$AB = 12y = 12 * (155/44) = 3*155/11 = 465/11 \approx 42.27$

$\angle ABC = 180 - 2\angle BDC = 180 - 2(7x) = 180-14x = 180 - 14(75/11) = 180- 1050/11=1980-1050/11=930/11 = 84.55$

3. Final Answer

$AB = \frac{465}{11}$

$m\angle ABC = \frac{930}{11}$