All the vertices of an isosceles triangle lie on a circle and each of the base angles of the triangle is \(65^{\circ}\). The angle subtended at the centre of the circle by the base of the triangle is ____________
A. \(130^{\circ} \) B. \(115^{\circ} \) C. \(100^{\circ}\) D. \(65^{\circ}\)
Correct Answer: C
Explanation
\(x^{\circ}+65^{\circ}+65^{\circ}=180^{\circ}\) (sum of \(\angle s\) in a \(\Delta\) ) \(x=180^{\circ}-130^{\circ}=50^{\circ}\) let \(O\) be the centre, hence, \(A() B=2 \times x^{\circ}=2 \times 50^{\circ}=100^{\circ}\)
