In the given figure,AT is a tangent to the ci | vedclass

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(A) In $	riangle OAT$,we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.Therefore,$\angle

Vedclass · Accepted Answer

$2 \sqrt{3}$

Vedclass · Answer

(A) In $	riangle OAT$,we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.Therefore,$\angle OAT = 90^{\circ}$.In the right-angled triangle $	riangle OAT$,we have:$\cos(\angle OTA) = \frac{	ext{Adjacent side}}{	ext{Hypotenuse}} = \frac{AT}{OT}$Given that $OT = 4 \, cm$ and $\angle OTA = 30^{\circ}$,we substitute these values:$\cos(30^{\circ}) = \frac{AT}{4}$Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we get:$\frac{\sqrt{3}}{2} = \frac{AT}{4}$$AT = 4 	imes \frac{\sqrt{3}}{2}$$AT = 2 \sqrt{3} \, cm$.

In the given figure,$AT$ is a tangent to the circle with center $O$ such that $OT = 4 \, cm$ and $\angle OTA = 30^{\circ}$. Then $AT$ is equal to (in $cm$):

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