You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Sum of Opposite Angles of a Cyclic Quadrilateral

In the given ...

Question

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

Open in App

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°

Suggest Corrections

Similar questions

Q. In the figure at right, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that

∠ D B C = 30^{o}

and

∠ B A C = 50^{o}

.
Find (i)

∠ B C D

(ii)

∠ C A D

Q. In the given figure, ABCD is a cyclic quadrilateral , whose diagonals intersect at

E

such that

∠ B A C = 30^{0}

∠ D B C = 70^{0}

and AB=BC,Find

∠ A C D

Q. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If

∠ D B C = 80^{\circ}, ∠ B A C

40^{\circ},

find

∠ B C D

Q. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If

∠ D B C = 80^{\circ}, ∠ B A C

40^{\circ},

find

∠ B C D

in degrees.

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If $∠$ DBC = 70 $^{\circ}$ , $∠$ BAC = 30 $^{\circ}$ , find $∠$ BCD. Further, if AB= BC, find $∠$ ECD.

Join BYJU'S Learning Program

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°