Let's analyze the problem:

We are given a quadrilateral $CDEF$ that forms a rhombus, and we know the following information:

• $\angle FCG = 4b + 26^\circ$
• $\angle CFE = 60^\circ$

Since $CDEF$ is a rhombus, its properties are as follows:

1. All sides of a rhombus are equal.
2. The diagonals of a rhombus bisect each other at right angles (90°).
3. The diagonals of a rhombus bisect the angles of the rhombus.

Step 1: Use angle properties of the rhombus

The angle $\angle CFE = 60^\circ$ represents an exterior angle of the triangle $\triangle CFG$. Since the diagonals bisect the angles in a rhombus, the diagonal $CF$ bisects $\angle C$, and we know:

$\angle FCG = \angle GCE = 4b + 26^\circ$

The angles in triangle $\triangle CFG$ must sum up to 180°:

$\angle FCG + \angle CFG + \angle GCF = 180^\circ$

Since $\angle CFG = 90^\circ$ (because the diagonals of a rhombus are perpendicular), we can write:

$(4b + 26^\circ) + 90^\circ + (4b + 26^\circ) = 180^\circ$

Step 2: Solve the equation

Simplifying this:

$2(4b + 26^\circ) + 90^\circ = 180^\circ$ $8b + 52^\circ + 90^\circ = 180^\circ$ $8b + 142^\circ = 180^\circ$ $8b = 180^\circ - 142^\circ$ $8b = 38^\circ$ $b = \frac{38^\circ}{8} = 4.75^\circ$

Thus, the value of $b$ is $4.75^\circ$.

Let me know if you need further details or have any questions.

Here are 5 related questions:

1. What are the properties of diagonals in a rhombus?
2. How do you find the angles of a triangle when given two angles?
3. What is the sum of the interior angles of a triangle?
4. How are exterior angles related to the interior angles of polygons?
5. Can you derive the length of the diagonals using trigonometry if all sides of the rhombus are known?

Tip: In a rhombus, remember that the diagonals always bisect the interior angles, and they intersect at 90°.