Problem Analysis:

We are tasked to find the angle $\angle BYC$ given the following details:

1. $\angle BAC = 40^\circ$.
2. The angular bisectors of $\angle ABC$ and $\angle ACB$ meet at $X$.
3. $BXYC$ is a rhombus.

Step-by-step Solution:

1. Understanding the Geometry:

   • In $\triangle ABC$, $\angle BAC = 40^\circ$.
   • The remaining angles, $\angle ABC$ and $\angle ACB$, must satisfy: $\angle BAC + \angle ABC + \angle ACB = 180^\circ$ Substituting $\angle BAC = 40^\circ$: $\angle ABC + \angle ACB = 140^\circ$

2. Introduce Bisectors:

   • Let $\angle ABC = B$ and $\angle ACB = C$.
   • The angular bisectors of $\angle ABC$ and $\angle ACB$ meet at $X$, which means: $\text{Half of } \angle ABC = \frac{B}{2}, \quad \text{and half of } \angle ACB = \frac{C}{2}.$

3. Geometry of $BXYC$:

   • Since $BXYC$ is a rhombus, all its sides are equal, and opposite angles are equal.
   • The diagonals of the rhombus bisect each other at right angles ($90^\circ$).

4. Find $\angle BYC$:

   • The angle $\angle BYC$ is the external angle at vertex $Y$, formed by the angular bisectors.
   • The formula for the angle between two angular bisectors in a triangle is: $\angle BYC = 90^\circ + \frac{\angle BAC}{2}$
   • Substituting $\angle BAC = 40^\circ$: $\angle BYC = 90^\circ + \frac{40^\circ}{2}$ $\angle BYC = 90^\circ + 20^\circ = 110^\circ$

Final Answer:

$\boxed{\angle BYC = 110^\circ}$

Would you like further clarification or a breakdown of any step?

Related Questions:

1. How does the property of a rhombus' diagonals help in solving this problem?
2. Can you explain why $\angle BYC$ is given by $90^\circ + \frac{\angle BAC}{2}$?
3. What would change if $\angle BAC$ were a different value?
4. How do the properties of angular bisectors simplify problems involving triangles?
5. Could we use trigonometric methods to find the value of $\angle BYC$?

Tip: Always use the triangle angle sum property to find missing angles before applying geometric properties like bisectors or rhombus relationships.