Question

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

Answer 1

From the figure:

$AB$ is the chord passing through the center,
So, $AB$ is the diameter.
Since angle in a semi-circle is a right angle
$⟹\angle APB={90}^{\circ }$
We will get $\angle APB=\angle PAT={30}^{\circ }[\because$ using alternates segment right angle].

Now, in $△APB$:
$\angle BAP+\angle APB+\angle PAT={180}^{\circ }$
$\angle BAP={180}^{\circ }-\angle APB-\angle PAT$
$={180}^{\circ }-{90}^{\circ }-{30}^{\circ }$
$\angle BAP={60}^{\circ }$

Now, $\angle BAP=\angle PTA+\angle APT$
${60}^{\circ }=\angle PTA+{30}^{\circ }$
$\angle PTA={60}^{\circ }-{30}^{\circ }$
$\angle PTA={30}^{\circ }$

Now, we know that sides opposite to equal angles are equal $AP=AT$
In right triangle $ABP$
$\sin ABP=\frac{AP}{BA}$
$\sin {30}^{\circ }=\frac{AT}{BA}$
$\frac{BA}{AT}=\frac{2}{1}$

$BA:AT=2:1$.