Advertisements
Advertisements
प्रश्न
In the figure, given below, find:
- ∠BCD,
- ∠ADC,
- ∠ABC.
Show steps of your working.
उत्तर
i. ∠BCD + ∠BAD = 180°
(Sum of opposite angles of a cyclic quadrilateral is 180°)
`=>` ∠BCD = 180° – 105° = 75°
ii. Now, AB || CD
∴ ∠BAD + ∠ADC = 180°
(Interior angles on the same side of parallel lines is 180°)
`=>` ∠ADC = 180° – 105° = 75°
iii. ∠ADC + ∠ABC = 180°
(Sum of opposite angles of a cyclic quadrilateral is 180°)
`=>` ∠ABC = 180° – 75° = 105°
APPEARS IN
संबंधित प्रश्न
In cyclic quadrilateral ABCD, ∠DAC = 27°; ∠DBA = 50° and ∠ADB = 33°.
Calculate:
- ∠DBC,
- ∠DCB,
- ∠CAB.
In the given figure, SP is bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that : SQ = SR.
In cyclic quadrilateral ABCD, ∠A = 3∠C and ∠D = 5∠B. Find the measure of each angle of the quadrilateral.
In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92°, ∠FAE = 20°; determine ∠BCD. Give reason in support of your answer.
In a square ABCD, its diagonals AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and the bisector of angle ABD meets AC at N and AM at L. Show that:
- ∠ONL + ∠OML = 180°
- ∠BAM + ∠BMA
- ALOB is a cyclic quadrilateral.
Prove that the angles bisectors of the angles formed by producing opposite sides of a cyclic quadrilateral (provided they are not parallel) intersect at right triangle.
In following fig., O is the centre of the circle, prove that ∠x =∠ y + ∠ z.
In following fig., O is the centre of the circle. Find ∠ CBD.
ABCD is a parallelogram. A circle through vertices A and B meets side BC at point P and side AD at point Q. Show that quadrilateral PCDQ is cyclic.
If O is the centre of the circle, find the value of x in each of the following figures
