In `DeltaABC, angleA=60^(@), angleB=90^(@), angleC=30^(@)`. Let H be its orthocentre, then :
(where symbols used have usual meanings)

Options

A. `AH=c`
B. `CH=a`
C. `AH=a`
D. `BH=0`

Answer

To solve the problem, we will analyze triangle ABC with the given angles and find the distances from the orthocenter H to the vertices A, B, and C.

### Step-by-Step Solution:

1. **Identify the Angles and Triangle Type**:
- Given angles are \( \angle A = 60^\circ \), \( \angle B = 90^\circ \), and \( \angle C = 30^\circ \).
- Since \( \angle B = 90^\circ \), triangle ABC is a right triangle.

2. **Use the Sine Rule to Find the Sides**:
- According to the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
- Here, \( a \) is the side opposite to angle A, \( b \) is opposite angle B, and \( c \) is opposite angle C.

3. **Calculate the Sides**:
- For side \( a \):
\[ a = 2R \sin A = 2R \sin 60^\circ = 2R \cdot \frac{\sqrt{3}}{2} = R\sqrt{3} \]
- For side \( b \):
\[ b = 2R \sin B = 2R \sin 90^\circ = 2R \cdot 1 = 2R \]
- For side \( c \):
\[ c = 2R \sin C = 2R \sin 30^\circ = 2R \cdot \frac{1}{2} = R \]

4. **Calculate Distances from the Orthocenter (H) to the Vertices**:
- The distances from the orthocenter H to the vertices A, B, and C are given by:
\[ AH = 2R \cos A, \quad BH = 2R \cos B, \quad CH = 2R \cos C \]
- Calculate \( AH \):
\[ AH = 2R \cos 60^\circ = 2R \cdot \frac{1}{2} = R \]
- Calculate \( BH \):
\[ BH = 2R \cos 90^\circ = 2R \cdot 0 = 0 \]
- Calculate \( CH \):
\[ CH = 2R \cos 30^\circ = 2R \cdot \frac{\sqrt{3}}{2} = R\sqrt{3} \]

5. **Relate Distances to Sides**:
- From the calculations:
- \( AH = R \) (which is equal to side \( c \))
- \( BH = 0 \)
- \( CH = R\sqrt{3} \) (which is equal to side \( a \))

6. **Conclusion**:
- The distances from the orthocenter to the vertices are:
- \( AH = c \)
- \( BH = 0 \)
- \( CH = a \)
- Therefore, the verified options are 1, 2, and 4.