Question

$AB$ is a diameter and $AC$ is a chord of a circle such that $\angle BAC={30}^{\circ }$. The tangent at $C$ intersects $AB$ produced at $D$. The value of $\frac{BC}{BD}$ equals:

• A. $1$
• B. $2$
• C. $0.5$
• D. None

Answer 1

The correct option is A $1$

Since $AB$ is the diameter of the circle.
$\therefore \angle ACB={90}^{\circ }$ [Angle in a semi circle]
$\therefore \angle ABC={90}^{\circ }-{30}^{\circ }={60}^{\circ }.....\left(i\right)$
Again, $\angle BCD=\angle CAB={30}^{\circ }$ [$\angle$s in the alternate segments]
Now in $\Delta CBD$,
$ext.\angle CBA=\angle BCD+\angle BDC$
${60}^{\circ }={30}^{\circ }+\angle BDC$
$\therefore \angle BDC={30}^{\circ }$
Hence, $\angle BCD=\angle BDC$
$\Rightarrow BC=AD$.