In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ∠DBC = 60° and ∠BAC = 40°. Find (i) ∠BCD, (ii) ∠CAD.

Open in App

Solution

(i) ∠BDC = ∠BAC = 40° (Angles in the same segment)
In ΔBCD, we have:
∠BCD + ∠DBC + ∠BDC = 180° (Angle sum property of a triangle)
⇒ ∠BCD + 60° + 40° = 180°
⇒ ∠BCD = (180° - 100°) = 80°

(ii) ∠CAD = ∠CBD (Angles in the same segment)
= 60°

Suggest Corrections

Similar questions

∠ D B C = 30^{o}

∠ B A C = 50^{o}

∠ B C D

∠ C A D

E

∠ B A C = 30^{0}

∠ D B C = 70^{0}

∠ A C D

∠ D B C = 80^{\circ}, ∠ B A C

40^{\circ},

∠ B C D

∠ D B C = 80^{\circ}, ∠ B A C

40^{\circ},

∠ B C D

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If $∠ D B C = 80^{\circ}$ and $∠ B A C$ is $40^{\circ} .$ Find $∠ B C D$ .

Join BYJU'S Learning Program