Question

To draw a pair of tangents to a circle inclined at 40°, the angle at the centre

of the circle between the two radii is​

Answer 1

Answer:

The angle between the two radii, OA and $\mathrm{OB}$ is $140^{o}$

Step-by-step explanation:

Given:

PA and PB are tangents drawn from an external point $P$ to the circle.

$\angle \mathrm{OAP}=\angle \mathrm{OBP}=90^{\circ} \quad$ (Radius is perpendicular to the tangent at point of contact)

To find: The angle between the two radii, OA and $\mathrm{OB}$

Step 1: Calculate the angle between the two radii

In quadrilateral OAPB,

$\angle \mathrm{APB}+\angle \mathrm{OAB}+\angle \mathrm{AOB}+\angle \mathrm{OBP}=360^{\circ}$

$40^{\circ}+90^{\circ}+\angle \mathrm{AOB}+90^{\circ}=360^{\circ}$

$220^{\circ}+\angle \mathrm{AOB}=360^{\circ}$

$\angle \mathrm{AOB}=360^{\circ}-220^{\circ}=140^{\circ}$

Thus, the angle between the two radii, OA and $\mathrm{OB}$ is $140^{\circ}$.