If tangent PA and PB from …

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Sia ? 6 years, 4 months ago
{tex}\angle{/tex}OPQ = 90o
[The tangent at any point of a circle is {tex}\perp{/tex} to the radius
through the point of contact]
{tex}\angle{/tex}OPA = {tex}\frac 12{/tex}{tex}\angle{/tex}BPA [Centre lies on the bisector of the
angle between the two tangents]
In {tex}\triangle{/tex}OPA,
{tex}\angle{/tex}OAP + {tex}\angle{/tex}OPA + {tex}\angle{/tex}POA = 180o [Angle sum property of a triangle]
{tex}\Rightarrow{/tex} 90o + 40o + {tex}\angle{/tex}POA = 180o
{tex}\Rightarrow{/tex} 130o + {tex}\angle{/tex}POA = 180o
{tex}\Rightarrow{/tex} {tex}\angle{/tex}POA = 50o
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