in an isosceles acute-angled triangle ABC (AC=AB and AB>AC), the measure of the angle between the bisector of the interior angle at vertex B and the normal from the same vertex to the opposite side is 9 degrees. Determine the measures of the interior angles of that triangle

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

In an isosceles acute-angled triangle ABC (AC = AB and AB > AC), the measure of the angle between the bisector of the interior angle at vertex B and the normal from the same vertex to the opposite side is 9 degrees. Determine the measures of the interior angles of that triangle.

In this geometry problem, we solve for the interior angles of an isosceles acute-angled triangle where AC = AB and AB > AC. Given that the angle between the bisector of angle B and the normal from B to AC is 9°, explore the step-by-step solution using triangle properties and geometric theorems. Suitable for high school students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation