If S is the circumcentre of `DeltaABC` and `angleA=50^(@)`, then the value of `angleBCS` is

Options

A. `20^(@)`
B. `40^(@)`
C. `60^(@)`
D. `80^(@)`

Answer

To find the value of angle \( \angle BCS \) in triangle \( \Delta ABC \) where \( S \) is the circumcenter and \( \angle A = 50^\circ \), we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Circumcenter**:
- The circumcenter \( S \) of triangle \( ABC \) is the point where the perpendicular bisectors of the sides of the triangle intersect. It is also the center of the circumcircle of triangle \( ABC \).

2. **Draw the Triangle and Circumcircle**:
- Draw triangle \( ABC \) inscribed in a circle (the circumcircle). Mark the circumcenter \( S \).

3. **Identify the Angles**:
- We know that \( \angle A = 50^\circ \). We need to find \( \angle BCS \).

4. **Using the Inscribed Angle Theorem**:
- The angle \( \angle BCA \) (which is the angle subtended by chord \( BC \) at point \( A \)) is equal to \( \angle A \) at the circumference. Thus, \( \angle BCA = 50^\circ \).

5. **Finding the Central Angle**:
- The angle at the center \( S \) that subtends the same arc \( BC \) is twice the inscribed angle. Therefore, \( \angle BSC = 2 \times \angle BCA = 2 \times 50^\circ = 100^\circ \).

6. **Using the Triangle Sum Property**:
- In triangle \( BSC \), the sum of angles is \( 180^\circ \). Therefore, we can write:
\[ \angle BCS + \angle BSC + \angle CBS = 180^\circ \]
- We already know \( \angle BSC = 100^\circ \) and \( \angle CBS = \angle BCS \) (since \( SB = SC \) as they are both radii of the circumcircle).

7. **Setting Up the Equation**:
- Let \( \angle BCS = x \). Then, we have:
\[ x + 100^\circ + x = 180^\circ \]
- This simplifies to:
\[ 2x + 100^\circ = 180^\circ \]

8. **Solving for \( x \)**:
- Subtract \( 100^\circ \) from both sides:
\[ 2x = 80^\circ \]
- Divide by 2:
\[ x = 40^\circ \]

9. **Conclusion**:
- Therefore, the value of \( \angle BCS \) is \( 40^\circ \).

### Final Answer:
\[ \angle BCS = 40^\circ \]