The problem describes a triangle $ABC$ where all angles are the same, meaning it's an equilateral triangle.

Key observations:

1. Triangle $ABC$ is equilateral, so all angles are $60^\circ$.
2. Point $C$ is due east of point $B$, meaning the line $BC$ is horizontal.
3. You are asked to find the bearing of point $B$ from point $A$. Bearings are measured clockwise from the north.

Steps to solve:

1. Since $ABC$ is equilateral, the angle at each vertex is $60^\circ$. Specifically, the angle $\angle BAC = 60^\circ$.
2. We also know that the line from $B$ to $C$ is horizontal (due east). This means the angle from north at point $A$ to the line $AB$ is the same as the interior angle of the equilateral triangle.
3. To calculate the bearing of $B$ from $A$, note that the bearing is measured clockwise from north. Since the angle at $A$ is $60^\circ$ and north points directly upward (vertically), the bearing is: $\text{Bearing of } B \text{ from } A = 360^\circ - 60^\circ = 300^\circ$

Thus, the bearing of point $B$ from point $A$ is 300°.

Do you need more details or have any questions?

Related questions:

1. What is the bearing of point $C$ from point $B$?
2. If point $A$ was directly north of point $B$, what would the bearing be?
3. How would the solution change if the triangle was not equilateral?
4. Can you calculate the distance between points $A$ and $B$ if given the length of side $BC$?
5. How does knowing that $C$ is due east of $B$ help in determining the solution?

Tip:

When working with bearings, always remember that they are measured clockwise from north.