Question

In the adjoining Figure, AP 10A HPLOB and AP = BP 1 SAOP ABOP 7 justify by giving reasons​

Answer 1

Answer:

Step-by-step explanation:

In ΔOAP and ΔOBP,

OP = OP    (Common)

∠OAP = ∠OBP  (90°)    (Radius is perpendicular to the tangent at the point of contact)

OA = OB  (Radius of the circle)

∴ ΔOAP ≅ ΔOBP  (RHS congruence criterion)

⇒ ∠OAP = ∠OBP = 120°/2 = 60°

In ΔOAP,

cos ∠OPA = 60° = AP/OP

∴ 1/2 = AP/OP

⇒ OP = 2 AP