To solve this problem and find the length of \( |cd| \), we can begin with the triangle \( ABC \) where given:
- \( |ab| = 6 , \text{cm} \)
- \( \angle A = 90^\circ \)
- \( \angle B = 30^\circ \)
Given that \( A \) is the right angle, we can infer that triangle \( ABC \) is a right triangle where \( AB \) is one side opposite \( \angle C \) and \( AC \) is the other side adjacent to \( \angle C \).
Step 1: Identify angles
In a right triangle, the angles add up to \( 180^\circ \). Thus:
\[ \angle C = 180^\circ - \angle A - \angle B = 180^\circ - 90^\circ - 30^\circ = 60^\circ \]
Step 2: Sketch the triangle
- Draw a line segment \( AB \) of length \( 6 , \text{cm} \).
- At point \( A \), draw a line perpendicular to \( AB \) for line \( AC \).
- Construct \( \angle B \) as \( 30^\circ \) from \( AB \).
The point \( C \) should be constructed such that \( \triangle ABC \) matches the angle measurements.
Step 3: Find the lengths of sides in triangle ABC
We can calculate \( |AC| \) and \( |BC| \) using trigonometric ratios.
- Since \( \angle B = 30^\circ \): \[ |AB| = \text{hypotenuse} = 6 , \text{cm} \]
- Opposite side \( |AC| \): \[ |AC| = |AB| \cdot \sin(30^\circ) = 6 \cdot \frac{1}{2} = 3 , \text{cm} \]
- Adjacent side \( |BC| \): \[ |BC| = |AB| \cdot \cos(30^\circ) = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} , \text{cm} \]
Step 4: Determine the altitude \( |AD| \) to side \( BC \)
Since we established that point \( D \) is the foot of the altitude \( |AD| \) from point \( A \) to line \( BC \), we can find \( |CD| \).
Step 5: Using area to find \( |CD| \)
The area \( K \) of triangle \( ABC \) can be calculated as:
\[ K = \frac{1}{2} \times |AB| \times |AC| = \frac{1}{2} \times 6 \times 3 = 9 , \text{cm}^2 \]
Now, using the base \( |BC| \) to find the height (or altitude) \( |AD| \):
\[ K = \frac{1}{2} \times |BC| \times |AD| \] \[ 9 = \frac{1}{2} \times 3\sqrt{3} \times |AD| \] \[ 9 = \frac{3\sqrt{3}}{2} \times |AD| \] \[ |AD| = \frac{18}{3\sqrt{3}} = 2\sqrt{3} , \text{cm} \]
Step 6: Find \( |CD| \)
In triangle \( ACD \):
\[ \tan(30^\circ) = \frac{|AD|}{|CD|} = \frac{2\sqrt{3}}{|CD|} \implies |CD| = \frac{2\sqrt{3}}{\tan(30^\circ)} = 2\sqrt{3} \cdot \sqrt{3} = 6 , \text{cm}. \]
Thus, the length of \( |CD| \) is \( 6 , \text{cm} \).
Final Answer
\(|cd| = 6 , \text{cm}\)
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