In the figure,AB is a chord of the circle and | vedclass

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(B) In the figure,$AOC$ is a diameter of the circle. We know that the angle subtended by a diameter at any point on the circle is $90^{\circ}$.Therefo

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) In the figure,$AOC$ is a diameter of the circle. We know that the angle subtended by a diameter at any point on the circle is $90^{\circ}$.Therefore,$\angle ABC = 90^{\circ}$.In $	riangle ACB$,the sum of all interior angles is $180^{\circ}$.So,$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$.$\angle BAC + 90^{\circ} + 50^{\circ} = 180^{\circ}$.$\angle BAC + 140^{\circ} = 180^{\circ}$.$\angle BAC = 180^{\circ} - 140^{\circ} = 40^{\circ}$.Since $AT$ is the tangent to the circle at point $A$,the radius $OA$ is perpendicular to the tangent $AT$.Therefore,$\angle OAT = 90^{\circ}$.Since $A, O, C$ are collinear,$\angle OAT$ is the same as $\angle CAT = 90^{\circ}$.We can write $\angle CAT = \angle CAB + \angle BAT$.$90^{\circ} = 40^{\circ} + \angle BAT$.$\angle BAT = 90^{\circ} - 40^{\circ} = 50^{\circ}$.Thus,the value of $\angle BAT$ is $50^{\circ}$.

In the figure,$AB$ is a chord of the circle and $AOC$ is its diameter such that $\angle ACB = 50^{\circ}$. If $AT$ is the tangent to the circle at point $A$,then $\angle BAT$ is equal to (in $^{\circ}$)

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