Question

Question 1
If a chord AB subtends an angle of ${60}^{\circ }$ at the centre of a circle, then angle between the tangents at A and B is also ${60}^{\circ }$.

Answer 1

False

Since a chord AB subtends an angle of ${60}^{\circ }$ at the centre of a circle
$i.e\angle AOB={60}^{\circ }$
OA = OB = Radius of the circle
$\angle OAB=\angle OBA={60}^{\circ }$
The tangents at points A and B is drawn, which intersects at C.
We know. OA $\perp$ AC and OB $\perp$ BC
$\therefore \angle OAB={90}^{\circ },\angle OBC={90}^{\circ }\Rightarrow \angle OAB+\angle BAC={90}^{\circ }And\angle OBA+\angle ABC={90}^{\circ }\Rightarrow \angle ABC={90}^{\circ }–{60}^{\circ }={30}^{\circ }In\Delta ABC\angle BAC+\angle CBA+\angle ACB={180}^{\circ }$
$\left[Sincesumofallinterioranglesofatriangleis{180}^{\circ }\right]\Rightarrow \angle ACB={180}^{\circ }-({30}^{\circ }+{30}^{\circ })={120}^{\circ }$