In `DeltaABC, angleB=90^(@) and AB : BC = 2:1`. The value of of `sinA+cotC` is

Options

A. `3+sqrt5`
B. `(2+sqrt5)/(2sqrt5)`
C. `2+sqrt5`
D. `3sqrt5`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Triangle**:
- Given \( \angle B = 90^\circ \), we have a right triangle \( ABC \) with \( AB \) as one leg and \( BC \) as the other leg.
- The ratio \( AB : BC = 2 : 1 \) means we can let \( AB = 2k \) and \( BC = k \) for some positive value \( k \).

2. **Finding the Length of \( AC \)**:
- We can use the Pythagorean theorem to find \( AC \):
\[ AC^2 = AB^2 + BC^2 \]
Substituting the values:
\[ AC^2 = (2k)^2 + (k)^2 = 4k^2 + k^2 = 5k^2 \]
Therefore,
\[ AC = \sqrt{5k^2} = k\sqrt{5} \]

3. **Calculating \( \sin A \)**:
- In triangle \( ABC \), \( \sin A \) is defined as the ratio of the length of the side opposite angle \( A \) (which is \( BC \)) to the hypotenuse \( AC \):
\[ \sin A = \frac{BC}{AC} = \frac{k}{k\sqrt{5}} = \frac{1}{\sqrt{5}} \]

4. **Calculating \( \cot C \)**:
- \( \cot C \) is defined as the ratio of the length of the side adjacent to angle \( C \) (which is \( AB \)) to the side opposite angle \( C \) (which is \( BC \)):
\[ \cot C = \frac{AB}{BC} = \frac{2k}{k} = 2 \]

5. **Finding \( \sin A + \cot C \)**:
- Now we can combine the results:
\[ \sin A + \cot C = \frac{1}{\sqrt{5}} + 2 \]

6. **Finding a Common Denominator**:
- To add these fractions, we need a common denominator:
\[ \sin A + \cot C = \frac{1}{\sqrt{5}} + \frac{2\sqrt{5}}{\sqrt{5}} = \frac{1 + 2\sqrt{5}}{\sqrt{5}} \]

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