In `DeltaABC, /_ABC = 70^@, /_BCA = 40^@`. O is the point of intersection of the perpendicular bisectors of the sides, then the angle `/_BOC` is

Options

A. `100^@`
B. `120^@`
C. `130^@`
D. `140^@`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the angles in triangle ABC
We know that:
- Angle ABC = 70°
- Angle BCA = 40°

### Step 2: Calculate angle A
Using the property that the sum of angles in a triangle is 180°, we can find angle A:
\[ \text{Angle A} = 180° - \text{Angle ABC} - \text{Angle BCA} \]
Substituting the known values:
\[ \text{Angle A} = 180° - 70° - 40° = 70° \]

### Step 3: Identify the circumcenter O
Since O is the point of intersection of the perpendicular bisectors of the sides of triangle ABC, it is the circumcenter of the triangle.

### Step 4: Use the property of angles subtended by an arc
We know that the angle subtended by an arc at the center (angle BOC) is twice the angle subtended at the circumference (angle BAC). Therefore:
\[ \text{Angle BOC} = 2 \times \text{Angle A} \]
Substituting the value of angle A:
\[ \text{Angle BOC} = 2 \times 70° = 140° \]

### Step 5: Conclusion
Thus, the angle BOC is:
\[ \text{Angle BOC} = 140° \]

### Final Answer
The angle \( \angle BOC \) is \( 140° \).

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