if O is the centre of …
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Sia ? 4 years, 10 months ago
OP {tex}\perp{/tex} PR [{tex}\because{/tex} Tangent and radius are {tex}\perp{/tex} to each other at the point of contact]
{tex}\angle{/tex}{tex}OPQ = 90° - 50° = 40°{/tex}
{tex}OP = OQ{/tex} [By isosceles triangle’s property]
{tex}\angle{/tex}OPQ = {tex}\angle{/tex}{tex}OQP = 40°{/tex}
In {tex}\triangle{/tex}OPQ,
{tex}\Rightarrow{/tex} {tex}\angle{/tex}O + {tex}\angle{/tex}P + {tex}\angle{/tex}Q = 180°
{tex}\Rightarrow{/tex} {tex}\angle{/tex}{tex}O + 40° + 40° = 180°{/tex}
{tex}\angle{/tex}{tex}O = 180°-80° = 100°.{/tex}
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