Statement-1 :Perpendicular from origin `O` to the line joining the points `A(ccosalpha,csinalpha)` and `B(ccosbeta,csinbeta)` divides it in the ratio 1:1 Statement-2 Perpendicular from opposite vertex to the base of an isosceles triangle bisects it Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1 Statement-1 is True, Statement-2 is False Statement-1 is False, Statement-2 is True

Options

A. Statement - 1 is True, Statement - 2 is True and Statement - 2 is correct explanation for Statemen - 1.
B. Statement - 1 is True, Statement - 2 is True and Statement - 2 is NOT correct explanation for Statemen - 1.
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To solve the problem, we need to evaluate the two statements provided and determine their truth values and whether one statement serves as a correct explanation for the other.

### Step-by-Step Solution:

1. **Understanding Statement 1**:
- Statement 1 asserts that the perpendicular drawn from the origin \( O(0, 0) \) to the line segment joining points \( A(c \cos \alpha, c \sin \alpha) \) and \( B(c \cos \beta, c \sin \beta) \) divides the segment \( AB \) in the ratio \( 1:1 \).
- This means that the perpendicular from the origin to the line segment \( AB \) will intersect it at its midpoint.

2. **Understanding Statement 2**:
- Statement 2 states that the perpendicular drawn from the opposite vertex of an isosceles triangle to the base bisects the base.
- This is a well-known property of isosceles triangles.

3. **Verifying Statement 2**:
- Let’s denote the isosceles triangle as \( ABC \) where \( AB = AC \). If we drop a perpendicular from vertex \( A \) to the base \( BC \), let’s call the foot of the perpendicular \( D \).
- By the properties of isosceles triangles, \( BD = CD \). Thus, Statement 2 is **True**.

4. **Verifying Statement 1**:
- We need to find the lengths \( OA \) and \( OB \) to check if \( OA = OB \).
- Using the distance formula, we calculate:
\[ OA = \sqrt{(c \cos \alpha - 0)^2 + (c \sin \alpha - 0)^2} = \sqrt{c^2 \cos^2 \alpha + c^2 \sin^2 \alpha} = \sqrt{c^2(\cos^2 \alpha + \sin^2 \alpha)} = \sqrt{c^2} = c \]
- Similarly, for \( OB \):
\[ OB = \sqrt{(c \cos \beta - 0)^2 + (c \sin \beta - 0)^2} = \sqrt{c^2 \cos^2 \beta + c^2 \sin^2 \beta} = \sqrt{c^2(\cos^2 \beta + \sin^2 \beta)} = \sqrt{c^2} = c \]
- Since \( OA = OB \), we conclude that triangle \( OAB \) is isosceles.

5. **Conclusion**:
- Since both statements are true and Statement 2 correctly explains Statement 1, the correct answer is:
- **Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.**