Let A,B,C,D,E represent vertices of a regular pentangon ABCDE. Given the position vector of these vertices be a,a+b,b,`lamda a` and `lamdab` respectively.
Q. AD divides EC in the ratio

Options

A. `"cos"(2pi)/(5):1`
B. `"cos"(3pi)/(5):1`
C. `1:2"cos"(pi)/(5)`
D. `1:2`

Explanation

To solve the problem of finding the ratio in which line segment AD divides line segment EC in a regular pentagon, we will proceed step by step.

### Step 1: Understand the Position Vectors
Given the position vectors of the vertices of the pentagon:
- A = **a**
- B = **a + b**
- C = **b**
- D = **λa**
- E = **λb**

### Step 2: Identify the Points
We need to find the ratio in which point D (λa) divides the line segment EC (from E to C). The position vector of point C is **b**, and the position vector of point E is **λb**.

### Step 3: Collinearity of Points
Since points C, E, and the origin (O) are collinear, we can express the position vectors in terms of the origin:
- Position vector of C (OC) = **b**
- Position vector of E (OE) = **λb**

### Step 4: Finding the Ratio
To find the ratio in which AD divides EC, we need to find the ratio of OE to OC:
\[ \text{Ratio} = \frac{OE}{OC} = \frac{\lambda b}{b} = \lambda \]

### Step 5: Understanding the Geometry
From the geometry of the pentagon, we know that:
- The angle at point A is \( \frac{3\pi}{5} \) (since each internal angle of a regular pentagon is \( \frac{3\pi}{5} \)).
- The triangle OAE can be analyzed using the sine rule.

### Step 6: Applying the Sine Rule
From triangle OAE:
\[ \frac{OE}{OA} = \frac{\sin(\angle OAE)}{\sin(\angle OAO)} \]
Where:
- \( OE = \lambda b \)
- \( OA = a \)
- \( \angle OAE = \frac{\pi}{5} \)
- \( \angle OAO = \frac{2\pi}{5} \)

Thus, we have:
\[ \frac{\lambda b}{a} = \frac{\sin(\frac{\pi}{5})}{\sin(\frac{2\pi}{5})} \]

### Step 7: Finding the Other Ratio
From triangle OCD:
\[ \frac{OC}{OD} = \frac{\sin(\angle OCD)}{\sin(\angle ODC)} \]
Where:
- \( OC = b \)
- \( OD = \lambda a \)
- \( \angle OCD = \frac{2\pi}{5} \)
- \( \angle ODC = \frac{\pi}{5} \)

Thus, we have:
\[ \frac{b}{\lambda a} = \frac{\sin(\frac{2\pi}{5})}{\sin(\frac{\pi}{5})} \]

### Step 8: Equating and Solving
Now, we have two equations:
1. \( \frac{\lambda b}{a} = \frac{\sin(\frac{\pi}{5})}{\sin(\frac{2\pi}{5})} \)
2. \( \frac{b}{\lambda a} = \frac{\sin(\frac{2\pi}{5})}{\sin(\frac{\pi}{5})} \)

Multiplying these two equations gives:
\[ \lambda^2 = \frac{\sin^2(\frac{2\pi}{5})}{\sin^2(\frac{\pi}{5})} \]

### Step 9: Final Ratio
From the above, we can derive:
\[ \lambda = \frac{1}{2 \cos(\frac{\pi}{5})} \]

### Conclusion
Thus, the ratio in which AD divides EC is:
\[ \text{AD divides EC in the ratio } \lambda : 1 \]