Two tangents TP and TQ are …
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Sia ? 6 years, 1 month ago
We know that, the lengths of tangents drawn from an external point to a circle are equal.
{tex}\therefore {/tex} TP = TQ
In {tex}\triangle{/tex}TPQ,
TP = TQ
{tex}\Rightarrow{/tex} {tex}\angle{/tex}TQP = {tex}\angle{/tex}TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)
{tex}\angle{/tex}TQP + {tex}\angle{/tex}TPQ + {tex}\angle{/tex}PTQ = 180º (Angle sum property)
{tex}\therefore {/tex}2 {tex}\angle{/tex}TPQ + {tex}\angle{/tex}PTQ = 180º (Using(1))
{tex}\Rightarrow{/tex} {tex}\angle{/tex}PTQ = 180º – 2 {tex}\angle{/tex}TPQ ...(1)
We know that, a tangent to a circle is perpendicular to the radius through the point of contact.
OP {tex}\perp{/tex} PT,
{tex}\therefore {/tex} {tex}\angle{/tex}OPT = 90º
{tex}\Rightarrow{/tex}{tex}\angle{/tex}OPQ + {tex}\angle{/tex}TPQ = 90º
{tex}\Rightarrow{/tex} {tex}\angle{/tex}OPQ = 90º – {tex}\angle{/tex}TPQ
{tex}\Rightarrow{/tex} 2{tex}\angle{/tex}OPQ = 2(90º –{tex}\angle{/tex}TPQ) = 180º – 2 {tex}\angle{/tex}TPQ ...(2)
From (1) and (2), we get
{tex}\angle{/tex}PTQ = 2{tex}\angle{/tex}OPQ
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