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:
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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 \( AB \).
- In triangle \( ABC \):
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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 \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \): \[ BC^2 = 16 + 16 - 32 \cdot \frac{\sqrt{3}}{2} \] \[ BC^2 = 32 - 16\sqrt{3} \] \[ BC = \sqrt{32 - 16\sqrt{3}} \]
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Finding the Length of the Altitude \( |CN| \):
- The altitude can be calculated using the sine function: \[ CN = AB \cdot \sin(30^\circ) \] Since \( \sin(30^\circ) = \frac{1}{2} \): \[ CN = 4 \cdot \frac{1}{2} = 2 , \text{cm} \]
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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: \[ |AN| = |BN| = \frac{|AB|}{2} = \frac{4}{2} = 2 , \text{cm} \]
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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 \( 2 , \text{cm} \).
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