In this problem, you are tasked with determining the interior angles of an isosceles acute-angled triangle $ABC$, where $AC = AB$ and $AB > AC$, given that the angle between the bisector of angle $B$ and the normal (perpendicular) from vertex $B$ to side $AC$ is $9^\circ$.

Let us break down the problem step by step:

Step 1: Define the interior angles

• Let the angle at vertex $A$ be $\angle A$, which is unknown.
• Since $AC = AB$ (isosceles triangle), the base angles are equal, so $\angle C = \angle B$.

Thus, the interior angles can be represented as:

$\angle A = A \quad \text{and} \quad \angle B = \angle C = \theta$

Using the angle sum property of triangles:

A + 2\theta = 180^\circ \quad \text{(since the sum of interior angles of a triangle is 180^\circ)}

This simplifies to:

$A = 180^\circ - 2\theta$

Step 2: Bisector and normal relationship

Let’s denote:

• The bisector of angle $B$ as $l_B$.
• The perpendicular from $B$ to side $AC$ as $n_B$.

The problem states that the angle between the bisector of angle $B$ and the normal from $B$ to $AC$ is $9^\circ$.

The bisector of $\angle B$ divides $\angle B$ into two equal parts. So, the angle between the bisector and the horizontal line $AC$ (which coincides with $n_B$) is:

$\frac{\theta}{2}$

Thus, the angle between the bisector $l_B$ and the perpendicular $n_B$ is:

$\frac{\theta}{2} - 90^\circ = 9^\circ$

Step 3: Solve for $\theta$

Now, we can set up the equation:

$\frac{\theta}{2} = 9^\circ + 90^\circ = 99^\circ$

Multiplying both sides by 2:

$\theta = 198^\circ$

This is impossible for the angle in a triangle. So, the angle should be subtrim10