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°

40