Question

In the given figure, o is the centre of the circle. If angle adc =140, then what is the value of x

Answer 1

Where is the figure????

Answer 2

I think it's the figure

Angle ACB = 90° (Angle on the semi-circle) ___(i)

since points A,B,C,D are lying on the circle therefore it is a cyclic quad...so,

In quad. ABCD ,

angle D + angle B = 180° (Sum of opposite angle of a quad. is 180°)

140° + angle B = 180°

angle B = 40°

In ∆ACB,

angle A + angle B + angle C = 180° (Sum of all angles of a triangle is 180°)

x + 40° + 90° = 180° ( angle C = 90° ...from eq.(i))

x = 180° - 130°

x = 50°