Question

In the given figure, $O$ is center of the circle and $TP$ is the tangent to the circle from an external point $T.$ If $\angle PBT={30}^{\circ }$, prove that $BA:AT=2:1.$

Answer 1

Construction: Join OP.

Proof: AB is the chord passing through the center.

So, AB is the diameter.

Since, angle in a semicircle is a right angle.

$\therefore \angle APB={90}^{\circ }$

By using alternate segment theorem, we have

We have, $\angle ABP=\angle PAT={30}^{\circ }$

In triangle BOP, we have

$OB=OP$ (Radius)

$\angle PBO=\angle OPB={30}^{\circ }$ [Angles opposite to equal sides are equal]

$\Rightarrow \angle APB={90}^{\circ }$

$\Rightarrow \angle OPB+\angle OPA={90}^{\circ }$

$\Rightarrow \angle OPA=90-{30}^{\circ }={60}^{\circ }$

Now, in triangle APB, we have

$\Rightarrow \angle BAP+\angle APB+\angle ABP={180}^{\circ }$ [Angle Sum property ]

$\Rightarrow \angle BAP={180}^{\circ }-{90}^{\circ }-{30}^{\circ }={60}^{\circ }$

$\Rightarrow \angle BAP=\angle APT+\angle PTA$ [Exterior angle property]

$\Rightarrow \angle PTA={60}^{\circ }-{30}^{\circ }={30}^{\circ }$

[$\because OP\perp TP,\angle APT=\angle OPT-\angle OPA={90}^{\circ }-{60}^{\circ }={30}^{\circ }$ ]

We know that sides opposite to equal sides are equal.

$\therefore AP=AT$

In right triangle ABP, we have

$\sin \angle ABP=\frac{AP}{BA}$

$\sin {30}^{\circ }=\frac{AT}{BA}$

$\frac{1}{2}=\frac{AT}{BA}$

$\therefore BA:AT=2:1$