A,B and C are three points on a circle such that the angles subtended by the chords AB and AC at the centre O are `90^@` and `110^@` respectively . Further suppose that the centre O lies on the interior `angleBAC`. The `angle BAC` is

Options

A. `40^@`
B. `80^@`
C. `160^@`
D. `20^@`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have three points A, B, and C on a circle, with the angles subtended by chords AB and AC at the center O being 90 degrees and 110 degrees, respectively. We need to find the angle BAC.

### Step 2: Draw the Diagram
Draw a circle and mark points A, B, and C on the circumference. Mark the center of the circle as O. Draw the chords AB and AC, and indicate the angles subtended at O: ∠AOB = 90° and ∠AOC = 110°.

### Step 3: Use Properties of Isosceles Triangles
Since OA = OB = OC (radii of the circle), triangles OAB and OAC are isosceles.

### Step 4: Find Angles in Triangle OAB
In triangle OAB:
- The sum of angles is 180°.
- Let ∠OAB = ∠OBA = x.
- Therefore, we have:
\[ x + x + 90° = 180° \]
\[ 2x = 90° \]
\[ x = 45° \]
Thus, ∠OAB = 45°.

### Step 5: Find Angles in Triangle OAC
In triangle OAC:
- The sum of angles is also 180°.
- Let ∠OAC = ∠OCA = y.
- Therefore, we have:
\[ y + y + 110° = 180° \]
\[ 2y = 70° \]
\[ y = 35° \]
Thus, ∠OAC = 35°.

### Step 6: Calculate Angle BAC
Now, we can find angle BAC:
\[ \angle BAC = \angle OAB + \angle OAC = 45° + 35° = 80° \]

### Final Answer
Thus, the angle BAC is 80 degrees.

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