The given image presents a rectangle $ABCD$ with an additional diagonal $BD$. The angles and side relationships are labeled as follows:

• $\angle DAB = 27^\circ$
• $AB = 3x$
• $BD = y$

The task asks to find the values of $x$ and $y$.

Step-by-step solution:

Since $ABCD$ is a rectangle, we know that the diagonals divide the rectangle into two right triangles. In the triangle $ABD$, the angle $\angle DAB = 27^\circ$, and we can use trigonometry to express the relationships between the sides.

1. Using Trigonometry:

   In right triangle $ABD$, we can use the trigonometric functions:

   • $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$
   • $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$

   Here, $AB = 3x$, $AD = x$, and $BD = y$.

   Using $\sin(27^\circ)$ and $\cos(27^\circ)$, we set up the equations:

   $\sin(27^\circ) = \frac{AB}{BD} = \frac{3x}{y}$ $\cos(27^\circ) = \frac{AD}{BD} = \frac{x}{y}$

2. Solving for $x$ and $y$:

   Rearranging the first equation:

   $y = \frac{3x}{\sin(27^\circ)}$

   Rearranging the second equation:

   $y = \frac{x}{\cos(27^\circ)}$

   Now we equate both expressions for $y$:

   $\frac{3x}{\sin(27^\circ)} = \frac{x}{\cos(27^\circ)}$

   Simplifying the equation by dividing both sides by $x$ (assuming $x \neq 0$):

   $\frac{3}{\sin(27^\circ)} = \frac{1}{\cos(27^\circ)}$

   Cross-multiplying gives:

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