Solution:
Given:
X is the center of the circle,
AB and AC are two tangents to the circle.
∠BXC = 105°
To find:
The value of ∠BAC
Here,
- ∠XBA = 90° and ∠XCA = 90° [because, a tangent to a circle is perpendicular to the radius of the circle drawn at the point of contact]
- ∠XBA + ∠BXC + ∠XCA + ∠BAC = 90° [because, sum of the angles of quadrilateral ABXC]
Or, 90° + 105° + 90° + ∠BAC = 360°
Or, 285° + ∠BAC = 360°
Or, ∠BAC = 75°
∴ The value of ∠BAC is 75°.