In the given figure, O is the center of a circle, AB is a chord and AT is the tangent at A. If ∠AOB = 100° then ∠BAT is equal to
In △AOB
OA = OB [radii of same circle]
∠OBA = ∠OAB [Angles opposite to equal sides are equal]
Also, By Triangle sum Property
∠AOB + ∠OBA + ∠OAB = 180°
100 + ∠OAB + ∠OAB = 180°
2 ∠OAB = 90°
∠OAB = 40°
As AT is tangent to given circle,
We have,
OA ⏊ AT
[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
So, ∠OAT = 90°
∠OAB + ∠BAT = 90°
40° + ∠BAT = 90°
∠BAT = 50°