Question

In the given figure , o is the centre of the circle ,PA and PB are tangents to the circle then find AQB

Answer 1

Since angle APB=40
ad angle OAP and OPB = 90 
Therefore angle AOB = 140. (sum of angles of quadrilateral)

We know that 
angle AOB = 2angle AQB
140 = 2× AQB 
140/2 = AQB 
angle AQB = 70

Answer 2

∠AQB   = 70°   if In the given figure , o is the centre of the circle ,PA and PB are tangents to the circle

Step-by-step explanation:

OAPB is  a quadrilateral

Sum of all angles of a quadrilateral = 360°

=> ∠OAP  + ∠APB + ∠PBO + ∠BOA  = 360°

∠APB  = 40°  given

∠OAP = ∠PBO  = 90°   as PA & PB are tangents

=> 90° + 40° + 90° + ∠BOA  = 360°

=> ∠BOA  = 140°

∠BOA = 2 ∠AQB  ( angle subtended by Chord AB at center = 2 * at arc)

=> 140° = 2 ∠AQB  

=>  ∠AQB   = 70°

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