Question

If a chord AB subtends an angle of ${60}^{\circ }$ at the centre of a circle, then the angle between the tangents at A and B is also ${60}^{\circ }$.

• A. True
• B. False

Answer 1

The correct option is B False


Since a chord AB subtends an angle of ${60}^{\circ }$ at the centre of a circle
i.e., $\angle AOB={60}^{\circ }$
As OA = OB = Radius of the circle
$\angle OAB=\angle OBA={60}^{\circ }$
The tangents at points A and B is drawn. Which intersect at C.
We know. OA $\perp$ AC and OB $\perp$ BC
$\therefore$ $\angle OAB={90}^{\circ },\angle OBC={90}^{\circ }$
$\Rightarrow$ $\angle OAB+\angle BAC={90}^{\circ }$
And $\angle OBA+\angle ABC={90}^{\circ }$
$\Rightarrow$ $\angle ABC={90}^{\circ }-{60}^{\circ }={30}^{\circ }$
In $△ABC\angle BAC+\angle CBA+\angle ACB={180}^{\circ }$
[Since sum of all interior angles of a triangle is ${180}^{\circ }$]
$\Rightarrow$ $\angle ACB={180}^{\circ }-({30}^{\circ }+{30}^{\circ })={120}^{\circ }$