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Question

In fig. if $$ TP $$ and $$ TQ $$ are the tangents to a circle with centre $$ O $$ so that $$ \angle POQ =110^o $$, then find $$ \angle PTQ $$.
1786571_5b0113e8aef642d4aa5c45fbfde10436.png


Solution

Since $$ TP $$ and $$ TQ $$ are the tangents to the circle with center $$ O $$ 

So, $$ OP \bot PT $$ and $$ OQ \bot QT $$

$$ \Rightarrow \angle OPT  = 90^0 , \angle OQT  = 90^0 $$ and $$ \angle POQ = 110^0 $$

So, in quadrillateral $$ OPTQ $$ we have

$$ \angle POQ  + \angle OPT + \angle PTQ + \angle TQO  = 360^0 $$

$$ \Rightarrow 110^0 + 90^0 + \angle PTQ  + 90^0 = 360^0 \Rightarrow \angle PTQ + 290^0 = 360^0 $$

$$ \therefore \angle PTQ  = 360^0 - 290^0 \Rightarrow \angle PTQ = 70^0 $$

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