Question

Draw a sketch in the isosceles triangle acb, |ab|=|ac|=4cm,bac=30degree & |cn| is an altitude. Find |bn|

Answer 1

To solve for the length of \( |bn| \) in the isosceles triangle \( ABC \) where \( |AB| = |AC| = 4 \, \text{cm} \) and \( \angle BAC = 30^\circ \), we will proceed with the following steps:

1. **Understanding the Problem**:
- In triangle \( ABC \):
- \( AB = AC = 4 \, \text{cm} \)
- \( \angle BAC = 30^\circ \)
- \( CN \) is the altitude, which means \( N \) is the point where the altitude meets \( AB \).

2. **Finding Side Length \( |BC| \)**:
- Since we have an isosceles triangle, we can use the Law of Cosines to find the length of \( BC \):
\[
BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)
\]
Plugging in the values:
\[
BC^2 = 4^2 + 4^2 - 2 \cdot 4 \cdot 4 \cdot \cos(30^\circ)
\]
Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[
BC^2 = 16 + 16 - 32 \cdot \frac{\sqrt{3}}{2}
\]
\[
BC^2 = 32 - 16\sqrt{3}
\]
\[
BC = \sqrt{32 - 16\sqrt{3}}
\]

3. **Finding the Length of the Altitude \( |CN| \)**:
- The altitude can be calculated using the sine function:
\[
CN = AB \cdot \sin(30^\circ)
\]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[
CN = 4 \cdot \frac{1}{2} = 2 \, \text{cm}
\]

4. **Finding Height Intersection at \( N \)**:
- Triangle \( ABC \) can be divided into two right triangles, \( ANC \) and \( BNC \). Since \( CN \) is the height, the length of \( AN = BN \).
- The base \( AB \) can be bisected (because of the isosceles property), and thus:
\[
|AN| = |BN| = \frac{|AB|}{2} = \frac{4}{2} = 2 \, \text{cm}
\]

5. **Final Length \( |BN| \)**:
- By now, we completely formalized all segments leading to \( |BN| \).
- Therefore, the length of \( |BN| \) is:
\[
|BN| = 2 \, \text{cm}
\]

In summary, the length of \( |BN| \) is \( 2 \, \text{cm} \).