Question

In the figure,O is the centre of the circle. If angle ADC = 140°. Find x

Answer 1

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°

Answer 2

Answer:

R.E.F image

OA=OC⇒∠OAC=∠ACO=X

∠ADC+∠AMC=180

∘

(cyclic quadrilateral )

∠AMC=180

∘

−140

∘

=40

∘

∠ADC=2∠AMC=2×40

∘

=80

∘

In △AOC

∠ADC+∠OAC+∠OCA=180

∘

x+80

∘

+x=180

∘

2x=100

∘

x=50

∘

Step-by-step explanation:

R.E.F image

OA=OC⇒∠OAC=∠ACO=X

∠ADC+∠AMC=180

∘

(cyclic quadrilateral )

∠AMC=180

∘

−140

∘

=40

∘

∠ADC=2∠AMC=2×40

∘

=80

∘

In △AOC

∠ADC+∠OAC+∠OCA=180

∘

x+80

∘

+x=180

∘

2x=100

∘

x=50

∘