Let ABCD be a square and let P be a point on AB such that `AP:PB=1:2.ifangleAPD=theta`, then what is the value of `costheta`?

Options

A. `(1)/(sqrt(10))`
B. `(1)/(sqrt(5))`
C. `(2)/(sqrt(10))`
D. `(2)/(sqrt(5))`

Explanation

To solve the problem, we will follow a systematic approach.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a square ABCD and a point P on side AB such that the ratio of lengths AP to PB is 1:2. We need to find the value of cos(θ), where θ is the angle APD.

2. **Assigning Lengths**:
Let the length of side AB (and all sides of the square) be denoted as X. Given the ratio AP:PB = 1:2, we can express the lengths as:
- AP = \( \frac{X}{3} \)
- PB = \( \frac{2X}{3} \)

3. **Identifying Points**:
Place the square in the coordinate system:
- A(0, 0)
- B(X, 0)
- C(X, X)
- D(0, X)
- P will be at (AP, 0) = \( \left(\frac{X}{3}, 0\right) \)

4. **Finding the Lengths**:
Now, we need to find the length DP. The coordinates of D are (0, X) and P are \( \left(\frac{X}{3}, 0\right) \).
Using the distance formula:
\[ DP = \sqrt{\left(0 - \frac{X}{3}\right)^2 + (X - 0)^2} \]
\[ DP = \sqrt{\left(\frac{X}{3}\right)^2 + X^2} = \sqrt{\frac{X^2}{9} + X^2} = \sqrt{\frac{X^2}{9} + \frac{9X^2}{9}} = \sqrt{\frac{10X^2}{9}} = \frac{X\sqrt{10}}{3} \]

5. **Using Cosine Definition**:
We know that cos(θ) is defined as the adjacent side over the hypotenuse in the right triangle ADP:
\[ \cos(\theta) = \frac{AP}{DP} \]
Substituting the values we found:
\[ \cos(\theta) = \frac{\frac{X}{3}}{\frac{X\sqrt{10}}{3}} = \frac{1}{\sqrt{10}} \]

6. **Final Answer**:
Thus, the value of cos(θ) is:
\[ \cos(\theta) = \frac{1}{\sqrt{10}} \]

### Conclusion:
The final answer is:
\[ \cos(\theta) = \frac{1}{\sqrt{10}} \]