Question

Draw a sketch in ABC, |ab|=6cm. A=90degree & b =30degree, |ad|is an altitude. Find |cd|

Answer 1

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
1. Draw a line segment \( AB \) of length \( 6 \, \text{cm} \).
2. At point \( A \), draw a line perpendicular to \( AB \) for line \( AC \).
3. 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
\(|cd| = 6 \, \text{cm}\)