Advertisements
Advertisements
Question
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
Advertisements
Solution
It is given that, ∠BCD = 100° and ∠ABD = 70°
As we know that sum of the opposite pair of angles of cyclic quadrilateral is 180°.
\[\angle ADC + \angle ABC = 180° \]
\[ \Rightarrow 180° - 80° + 180° - x = 180°\]
\[ \Rightarrow x = 100°\]
In ΔABD we have,
\[\angle DAB + \angle ABD + \angle BDA = 180° \]
\[ \Rightarrow \angle BDA = 180° - 150° = 30° \]
Hence, `angle ABD = 30°`
APPEARS IN
RELATED QUESTIONS
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90°-A, 90° − `1/2 A, 90° − 1/2 B, 90° − 1/2` C.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that EB = EC.
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
In the following figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.
