In the figure (i) given below, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E. If ∠ADE = 70° and ∠OBA = 45°, calculate

(i) ∠OCA

(ii) ∠BAC

Join OA, OB, OC and AC as shown in the figure below:

(i) ABCD is a cyclic quadrilateral as all the vertices lie on the circumference.

Since exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

⇒ ∠ADE = ∠ABC = 70°.

Arc AC subtends ∠AOC at center and ∠ABC at point B,

∠AOC = 2∠ABC (∵ angle subtended by an arc at centre is double the angle subtended at any other point of the circle.)

⇒ ∠AOC = 2 × 70° = 140°.

From figure,

OA = OC = Radius of the circle.

So, ∠OCA = ∠OAC = x.

Since sum of angles of triangle = 180°

In △OCA,

⇒ ∠AOC + ∠OCA + ∠OAC = 180°
⇒ 140° + x + x = 180°
⇒ 140° + 2x = 180°
⇒ 2x = 180° - 140°
⇒ 2x = 40°
⇒ x = 20°.

Hence, the value of ∠OCA = 20°.

(ii) From above solution,

∠ABC = 70°

From figure,

∠ABC = ∠OBA + ∠OBC
70° = 45° + ∠OBC
∠OBC = 70° - 45°
∠OBC = 25°.

As, OB = OC = radius of the circle.

∴ ∠OCB = ∠OBC = 25°.

From figure,

⇒ ∠ACB = ∠OCB + OCA
⇒ ∠ACB = 25° + 20° = 45°

Since sum of angles of triangle = 180°

In △ABC,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°
⇒ 70° + 45° + ∠BAC = 180°
⇒ 115° + ∠BAC = 180°
⇒ ∠BAC = 180° - 115°
⇒ ∠BAC = 65°

Hence, the value of ∠BAC = 65°.