Solve this question... Of ncert class 10 page 214
Answers
↪️Since , tangents drawn from an external point to a circle are equal.
>•• AP = AC
↪️Thus, in ∆APO and ∆ ACO,
>••. AP = AC
AO = AO ( Common )
OP = OC ( Radius of circle )
↪️So , by SSS criterion of congruence , we have :
∆APO ~= ∆ ACO.
( ~= Congruency sign )
=> angle PAO = angle OAC
=> angle PAC = 2angle CAO
↪️Similarly , we can prove that ;
angle CBO = angle OBQ
=> angle CBQ = 2angle CBO
↪️Since , XY || X'Y'
angle PAC + angle QBC = 180°.
[ Sum of interior angles on the same side
of transversal is 180°]
=> 2angle CAO + 2angle CBO = 180°
=> angle CAO + angle CBO = 90° .......(i)
↪️In ∆ AOB ,
=> angle CAO + angle CBO + angle AOB = 180° =>angle CAO + angle CBO = 180° - angle AOB ..(ii)
Since , from equation (i) and (ii) , we get ;
180° - angle AOB = 90°
angle AOB = 90 °
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