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प्रश्न
In the figure, given below, find: ∠ABC. Show steps of your working.
उत्तर
∠ADC + ∠ABC = 180°
(Sum of opposite angles of a cyclic quadrilateral is 180°)
`=>` ∠ABC = 180° – 75° = 105°
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संबंधित प्रश्न
In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40° and ∠ABD = 60°, find;
- ∠DBC
- ∠BCP
- ∠ADB
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In a cyclic quadrialteral ABCD , if m ∠ A = 3 (m ∠C). Find m ∠ A.
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 Δ DEF are 90° - `"A"/2` , 90° - `"B"/2` and 90° - `"C"/2` respectively.
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If angle BCG=108° and O is the centre of the circle, find: angle DOC
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate : ∠BAC
In the given below the figure, O is the centre of the circle and ∠ AOC = 160°. Prove that 3∠y - 2∠x = 140°.
The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backward bisects the opposite side.
In the adjoining figure, AB is the diameter of the circle with centre O. If ∠BCD = 120°, calculate:
(i) ∠BAD (ii) ∠DBA
An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle, to prove the theorem complete the activity.
Given: ABCD is cyclic,
`square` is the exterior angle of ABCD
To prove: ∠DCE ≅ ∠BAD
Proof: `square` + ∠BCD = `square` .....[Angles in linear pair] (I)
ABCD is a cyclic.
`square` + ∠BAD = `square` ......[Theorem of cyclic quadrilateral] (II)
By (I) and (II)
∠DCE + ∠BCD = `square` + ∠BAD
∠DCE ≅ ∠BAD
