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
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
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.
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°, AC and BD intersect at P. Then, find ∠DPC.

ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to ______.
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.

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.
The three angles of a quadrilateral are 100°, 60°, 70°. Find the fourth angle.
