Question

10. Prove that the angle between the two tangents drawn from an external point to a circle
is supplementary to the angle subtended by the line-segment joining the points of
contact at the centre
​

Answer 1

Step-by-step explanation:

Draw a circle with center O and take a external point P. PA and PB are the tangents.

As radius of the circle is perpendicular to the tangent.

OA⊥PA

Similarly OB⊥PB

∠OBP=90

o

∠OAP=90

o

In Quadrilateral OAPB, sum of all interior angles =360

o

⇒∠OAP+∠OBP+∠BOA+∠APB=360

o

⇒90

o

+90

o

+∠BOA+∠APB=360

o

∠BOA+∠APB=180

o

It proves the angle between the two tangents drawn from an external point to a circle supplementary to the angle subtented by the line segment

Answer 2

we congruent triangleOAP and triangle OBP

OA=OB(radius)

PA=PB(tangents)

angleA=angleB(each 90dgree)

thanks