(ii) secant

(iii) two

(iv) point of contact

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

(A) 60Â°

(B) 70Â°

(C) 80Â°

(D) 90Â°

(A) 50Â°

(B) 60Â°

(C) 70Â°

(D) 80Â°

[Using Pythagoras Theorem]

OPÂ

In Î”OAP and Î”OAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the circle)

OA = OA (Common side)

Î”OAP â‰… Î”OAS (SSS congruence condition)

âˆ´ âˆ POA = âˆ AOS

â‡’âˆ 1 = âˆ 8

Similarly we get,

âˆ 2 = âˆ 3

âˆ 4 = âˆ 5

âˆ 6 = âˆ 7

Adding all these angles,

âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 + âˆ 5 + âˆ 6 + âˆ 7 +âˆ 8 = 360Âº

â‡’ (âˆ 1 + âˆ 8) + (âˆ 2 + âˆ 3) + (âˆ 4 + âˆ 5) + (âˆ 6 + âˆ 7) = 360Âº

â‡’ 2 âˆ 1 + 2 âˆ 2 + 2 âˆ 5 + 2 âˆ 6 = 360Âº

â‡’ 2(âˆ 1 + âˆ 2) + 2(âˆ 5 + âˆ 6) = 360Âº

â‡’ (âˆ 1 + âˆ 2) + (âˆ 5 + âˆ 6) = 180Âº

â‡’ âˆ AOB + âˆ COD = 180Âº

Similarly, we can prove that âˆ BOC + âˆ DOA = 180Âº

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.