To draw a pair f tangents to a circle which are inclined

Solution:

Given a pair of tangents to a circle inclined to each other at angle of 100°

We have to find the angle between two radii of circle joining the end points of tangents that is we have to find  in below figure.

Let O be the center of the given circle

Let AB and AC be the two tangents to the given circle drawn from point A

Therefore ∠BAC = 100°

Now OB and OC represent the radii of the circle

Therefore

 [Since Radius of a circle is perpendicular to tangent]

We know that sum of angles of a quadrilateral = 360°

Therefore in Quadrilateral OBAC

$\angle B A C+\angle A C O+\angle C O B+\angle O B A=360^{\circ}$

$100^{\circ}+90^{\circ}+\angle C O B+90^{\circ}=360^{\circ}$

$\angle C O B=360^{\circ}-280^{\circ}$

$\angle C O B=80^{\circ}$

Hence Option (c) is correct.