Question

In a circle if the angle between two radii is 120°. Then find the angle of inclination between the pair of tangents drawn to circle​

Answer 1

Answer:

Since tangents and radii are perpendicular at the point of contact, in the quadrilateral formed by the two radii and the tangents at their ends, we have two right angles at the two points of contacts.

Let the angle between the tangents be x

. Then

120+ 90 +90 +x=360

x=60

Answer 2

Answer:

60^\circ \;\) Let us consider a circle with center O and OA and OB are two radii such that \(\angle {\mathrm{AOB }} = {\mathrm{ }}60^\circ \;\). AP and BP are two intersecting tangents at point P at point A and B respectively on the circle .

To find : Angle between tangents, i.e. \(\angle {\mathrm{APB}}\) As AP and BP are tangents to given circle, We have, \({\mathrm{OA\;}} \bot {\mathrm{\;AP and OB\;}} \bot {\mathrm{\;BP}}\) [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, \(\angle {\mathrm{OAP = \;}}\angle {\mathrm{OBP = 90^\circ }}\)

In quadrilateral AOBP, By angle sum property of quadrilateral, we have\(\begin{array}{*{20}{l}} {\angle {\mathrm{OAP + \;}}\angle {\mathrm{OBP + \;}}\angle {\mathrm{APB + \;}}\angle {\mathrm{AOB = 360^\circ }}}\\ {{\mathrm{90^\circ + 90^\circ + \;}}\angle {\mathrm{APB + 120^\circ \; = 360^\circ }}}\\ {\angle {\mathrm{APB = 60^\circ }}} \end{array}\)