In a `Delta ABC, D` and E are two points on AC & BC such that DE is prependicular to BC. DE=18 cm , `CE=5 cm & tan (angleABC)=3.6`. Find the ratio AC to CD.

Options

A. `BC//2CE`
B. `2BC//CE`
C. `CE//2BC`
D. `2CE//BC`

Answer

To solve the problem step by step, we will use the information provided in the question and the relationships between the angles and sides of the triangles involved.

### Step 1: Understand the Given Information
We have a triangle ABC with points D and E on sides AC and BC respectively. The line segment DE is perpendicular to BC. We are given:
- DE = 18 cm
- CE = 5 cm
- tan(∠ABC) = 3.6

### Step 2: Set Up the Right Triangle
Since DE is perpendicular to BC, we can form two right triangles: triangle ADF and triangle DCE, where F is the foot of the perpendicular from D to BC.

### Step 3: Use the Tangent Function
From the definition of tangent in triangle ABC:
\[ \tan(∠ABC) = \frac{AF}{BF} \]
Given that \(\tan(∠ABC) = 3.6\), we can express this as:
\[ \frac{AF}{BF} = 3.6 \]

### Step 4: Analyze Triangle DCE
In triangle DCE, we can also apply the tangent function:
\[ \tan(∠DCE) = \frac{DE}{CE} = \frac{18}{5} \]
We can calculate this value:
\[ \tan(∠DCE) = \frac{18}{5} = 3.6 \]

### Step 5: Equate the Angles
Since \(\tan(∠ABC) = \tan(∠DCE)\), we have:
\[ ∠ABC = ∠DCE \]
This implies that triangle ABC is isosceles with AB = AC.

### Step 6: Establish Similar Triangles
Triangles ACF and DCE are similar because they share angle C and both have a right angle:
- ∠ACF = 90°
- ∠DCE = 90°
- ∠C is common

### Step 7: Set Up the Proportions
From the similarity of triangles ACF and DCE, we can write:
\[ \frac{AC}{CD} = \frac{CE}{CF} \]

### Step 8: Find CF
Since DE is perpendicular to BC and divides it into two equal parts (because triangle ABC is isosceles), we have:
\[ CF = \frac{BC}{2} \]

### Step 9: Substitute Values
Now we can express the ratio:
\[ \frac{AC}{CD} = \frac{CE}{CF} = \frac{5}{\frac{BC}{2}} = \frac{5 \times 2}{BC} = \frac{10}{BC} \]

### Step 10: Conclusion
Thus, the ratio of AC to CD is given by:
\[ \frac{AC}{CD} = \frac{10}{BC} \]