In `DeltaABC,angleB=90^(@)andAB:BC=2:1` .Then value of sin A+cot C is

Options

A. `3+sqrt(5)`
B. `(2+sqrt(5))/(2sqrt(5))`
C. `2+sqrt(5)`
D. `3sqrt(5)`

Answer

To solve the problem, we need to find the value of \( \sin A + \cot C \) in triangle \( \Delta ABC \) where \( \angle B = 90^\circ \) and the ratio \( AB:BC = 2:1 \).

### Step-by-Step Solution:

1. **Identify Triangle Dimensions**:
Since \( AB:BC = 2:1 \), we can assign lengths to the sides. Let:
- \( AB = 2k \) (opposite to angle C)
- \( BC = k \) (opposite to angle A)

2. **Calculate the Hypotenuse**:
Using the Pythagorean theorem, the hypotenuse \( AC \) can be calculated as:
\[ AC = \sqrt{(AB)^2 + (BC)^2} = \sqrt{(2k)^2 + (k)^2} = \sqrt{4k^2 + k^2} = \sqrt{5k^2} = k\sqrt{5} \]

3. **Find \( \sin A \)**:
The sine of angle A is given by the ratio of the opposite side to the hypotenuse:
\[ \sin A = \frac{BC}{AC} = \frac{k}{k\sqrt{5}} = \frac{1}{\sqrt{5}} \]

4. **Find \( \cot C \)**:
The cotangent of angle C is given by the ratio of the adjacent side to the opposite side:
\[ \cot C = \frac{AB}{BC} = \frac{2k}{k} = 2 \]

5. **Combine \( \sin A \) and \( \cot C \)**:
Now we can find \( \sin A + \cot C \):
\[ \sin A + \cot C = \frac{1}{\sqrt{5}} + 2 \]

6. **Combine into a Single Fraction**:
To combine these, we can express \( 2 \) as a fraction:
\[ 2 = \frac{2\sqrt{5}}{\sqrt{5}} \]
Therefore,
\[ \sin A + \cot C = \frac{1}{\sqrt{5}} + \frac{2\sqrt{5}}{\sqrt{5}} = \frac{1 + 2\sqrt{5}}{\sqrt{5}} \]

### Final Answer:
Thus, the value of \( \sin A + \cot C \) is:
\[ \frac{1 + 2\sqrt{5}}{\sqrt{5}} \]