IF ANGLE BETWEEN TWO TANGENTS DRAWN FROM A POINT P TO A CIRCLE OF RADIUS A AND CENTER O IS 60 DEGREE THEN PROVE THAT AP = A UNDER ROOT 3.

We have the following situation-

We know that tangent is always perpendicular to the radius at the point of contact.

So, $\angle $OAP = 90

We know that if 2 tangents are drawn from an external point, then they are equally inclined to the line segment joining the centre to that point.

$\mathrm{So},\angle \mathrm{OPA}=\frac{1}{2}\angle \mathrm{APB}=\frac{1}{2}\times 60\xb0=30\xb0$

According to the angle sum property of triangle-

$\mathrm{In}\u2206\mathrm{AOP},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\angle \mathrm{AOP}+\angle \mathrm{OAP}+\angle \mathrm{OPA}=180\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{AOP}+90\xb0+30\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{AOP}=60\xb0$

So, in triangle AOP

Hence proved.

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