In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that $∠ D B C = 60^{\circ} a n d ∠ B A C = 40^{\circ} . Find (i) ∠ B C D, (i i) ∠ C A D$ .

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Solution

ANSWER:
(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°

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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.

ANSWER: (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°

In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that $∠ D B C = 60^{\circ} a n d ∠ B A C = 40^{\circ} . Find (i) ∠ B C D, (i i) ∠ C A D$ .

ANSWER:
(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°