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
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.
In the figure m(arc LN) = 110°,
m(arc PQ) = 50° then complete the following activity to find ∠LMN.
∠ LMN = `1/2` [m(arc LN) - _______]
∴ ∠ LMN = `1/2` [_________ - 50°]
∴ ∠ LMN = `1/2` × _________
∴ ∠ LMN = __________
In the given figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
ABCD is a cyclic quadrilateral such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
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 non-parallel sides of a trapezium are equal, prove that it is cyclic.
The three angles of a quadrilateral are 100°, 60°, 70°. Find the fourth angle.
