Question

In the figure, C is the centre of the circle and $\angle ABD={30}^o$
a) What is the measure of $\angle ACD$?
b) If $\angle ABD=\angle CAB$ and $AB=6cm$, find the radius of the circle.

Answer 1

(a) The angle made by an arc at any point on the alternate arc is equal to half the angle made at centre.
$\therefore m\angle ABD=\frac{1}{2}m\angle ACD$
$\therefore m\angle ACD={60}^o$
(b)$\angle ABD=\angle CAB$
$\therefore \angle CAB={30}^o$
In $△CAE,\angle ACE+\angle CAE+\angle CEA={180}^o$
${60}^o+{30}^o+\angle CEA={180}^o$
$\angle CEA={90}^o$
$\therefore CE$ is perpendicular to the chord $AB$.
A perpendicular drawn from the centre of the circle to the chord bisects the cord.
$\therefore AE=EB=\frac{1}{2}AB=3cm$
In $△CAE,\sin C=\frac{AE}{AC}$
$\therefore \sin {60}^o=\frac{3}{AC}$
$\therefore AC=3\times \frac{2}{\sqrt{3}}=2\sqrt{3}cm$
Hence, radius of the circle is $2\sqrt{3}cm.$