Question

In the diagram, the circle contains the vertices A, B, C of $△ABC$. Now $\angle ABC={30}^{\circ }$ and the length of $AC$ is $5$. The diameter of the circle is

• A. $5\sqrt{3}$
• B. $8$
• C. $10$
• D. $5\sqrt{5}$

Answer 1

The correct option is C $10$

In $△ABC$,
$\angle ABC={30}^{\circ }$
Now, Join OA and OC where O is the centre of the circle.
Then, $\angle AOC=2\angle BAC={60}^{\circ }$ (Angle subtended by the chord at the center is double the angle subtended at the circumference)
In $△OAC$
$OA=OC$, hence $\angle OAC=\angle OCA$
thus, by angle sum property,
$\angle OAC=\angle OCA={60}^{\circ }$
or $△OAC$ is an equilateral triangle.
Hence, $OA=OC=AC=5$
Thus, $diameter=2\times 5=10$