Question

In the given fig. Angle bto =30° find angle ato where o is center if circle and ta and tb are tangents

Answer 1

Answer: $30^{\circ}$

Step-by-step explanation:

Here, T is a point outside the circle having the center O,

While A and B are the points on the circumference of the circle,

Such that, TA and TB are the  tangent lines,

By the property of tangents,

$m\angle TBO = m\angle TAO = 90^{\circ}$

$\implies \angle TBO\cong \angle TAO$

By the property of a tangent on a circle by the common point,

$\angle TAO\cong TBO$

Also, $TO\cong TO$ ( Reflexive)

Thus by AAS postulate of congruence,

$\triangle ATO\cong \triangle BTO$

By CPCTC,

$m\angle ATO=m\angle BTO = 30^{\circ}$

Answer 2

In Δ ato and Δ bto

oa=ob →Radii of circle

∠oat=∠obt=90°→→Line from the center of the circle to the point of contact of tangent is perpendicular to the tangent.

at=bt →→Length of tangents from external point to a circle are equal.

Δ ato ≅ Δ bto→→[SAS Congruency criterion]

∠ ato = ∠ bto→→→[CPCT]

As, ∠ b t o= 30°

So, ∠ a t o= 30°