Question

0 is the centre of the circle, find the ∆ ABC ( D is 180°)​

Answer 1

Answer:

Take any point P on the circumference of the circle as shown.

Join AP and CP.

∵ABC subtends ∠AOC at centre O and ∠APC at any point P on the circumference of the circle.

∴∠AOC=2∠APC ...[∵ Angle subtended by an arc at the centre is twice the angle at the circumference]

∴∠APC=

2

1

∠AOC[∵AOC=130

∘

]

⟹

2

1

×130

∘

=65

∘

.

∵ ABCP is a cycle quadrilateral,

⟹∠APC+∠ABC=180

∘

...[∵ sum of opposite angles of a cyclic quadrilateral is 180

∘

]

⟹65

∘

+∠ABC=180

∘

∴∠ABC=180

∘

−65

∘

=115

∘

.