Advertisements
Advertisements
प्रश्न
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
Advertisements
उत्तर
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
संबंधित प्रश्न
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
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.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that EB = EC.
ABCD is a cyclic quadrilateral such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to ______.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
