To draw a pair of tangents to a circle which | vedclass

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(D) Let the center of the circle be $O$ and the intersection point of the two tangents be $P$. Let the points of contact of the tangents be $A$ and $B

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) Let the center of the circle be $O$ and the intersection point of the two tangents be $P$. Let the points of contact of the tangents be $A$ and $B$.In the quadrilateral $OAPB$,the angles at the points of contact $A$ and $B$ are $90^{\circ}$ each (since the tangent is perpendicular to the radius at the point of contact).The sum of the angles in a quadrilateral is $360^{\circ}$.Therefore,$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^{\circ}$.Substituting the known values: $\angle AOB + 90^{\circ} + 60^{\circ} + 90^{\circ} = 360^{\circ}$.$\angle AOB + 240^{\circ} = 360^{\circ}$.$\angle AOB = 360^{\circ} - 240^{\circ} = 120^{\circ}$.Thus,the angle between the two radii should be $120^{\circ}$.

To draw a pair of tangents to a circle which are inclined to each other at an angle of $60^{\circ},$ it is required to draw tangents at the endpoints of those two radii of the circle,the angle between which should be: (in $^{\circ}$)

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