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Consider the problem

Let us join point $O$ to $C$

In $ΔOPAandΔOCA$

$OP=OC$ (Radii of the same circle)

$AP=AC$ (Tangent from point $A$)

$AO=AO$ (Common side)

$ΔOPA≅ΔOCA$ ($SSS$ congruence criterion)

Therefore, $P↔C,A↔A,O↔O$

$∠POA=∠COA.........(1)$

Similarly,

$∠QOB≅∠OCB$

$∠QOB=∠COB.........(2)$

Since,$POQ$ is a diameter of the circle, it is a straight line.

Therefore, $∠POA+∠COA+∠COB+∠QOB=180_{∘}$

So, from equation $(1)$ and equation $(2)$

$2∠COA+2∠COB=180_{∘}∠COA+∠COB=90_{∘}∠AOB=90_{∘} $

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