Question

if angle between two tangents drawn from a point p to a circle of radius a and centre o is 60 degree, thn prove that ap = a root3

Answer 1

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We know that tangent is always
perpendicular to the radius at the point
of contact.
So, ∠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.
So, ∠OPA = 12∠APB = 12×60° = 30°
According to the angle sum property
of triangle-
In ∆AOP,∠AOP + ∠OAP + ∠OPA =
180°⇒∠AOP + 90° + 30° =
180°⇒∠AOP = 60°
So, in triangle AOP
tan angle AOP = AP/ OA
√ 3= AP/a
therefore, AP = √ 3a
hence, proved
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Answer 2

In triangle POA and Triangle POB
OA = OB (Radii of the same circle)
OP = OP (Common)
Angle OAP = Angle OBP (90° each, as tangent is perpendicular to the radius through the point of contact)

Therefore Triangle POA is congruent to triangle POB by RHS.

Therefore angle OPA = angle OPB by cpct.

Therefore Angle OPA=1/2×angle APB
Angle OPA = 30°

Now in triangle OAP,
tan(30°) = OA/AP
$\frac{1}{ \sqrt{3} } = \frac{r}{ap}$
AP=
$r \sqrt{3}$