Advertisements
Advertisements
प्रश्न
In the figure, given below, find: ∠ADC, Show steps of your working.
उत्तर
Now, AB || CD
∴ ∠BAD + ∠ADC = 180°
(Interior angles on the same side of parallel lines is 180°)
`=>` ∠ADC = 180° – 105° = 75°
APPEARS IN
संबंधित प्रश्न
In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC.
If the bisector of angle A meets BC at point E and the given circle at point F, prove that:
- EF = FC
- BF = DF
In a cyclic quadrilateral ABCD , AB || CD and ∠ B = 65° , find the remaining angles.
In triangle ABC, AB = AC. A circle passing through B and c intersects the sides AB and AC at D and E respectively. Prove that DE || BC.
The bisectors of the opposite angles A and C of a cydic quadrilateral ABCD intersect the cirde at the points E and F, respectively. Prove that EF is a diameter of the circle.
In following fig., O is the centre of the circle. Find ∠ CBD.
In the following figure, O is the centre of the circle. Find the values of a, b and c.
In the figure, given below, find: ∠ABC. Show steps of your working.
In the given below the figure, O is the centre of the circle and ∠ AOC = 160°. Prove that 3∠y - 2∠x = 140°.
In the given figure, AB is the diameter of a circle with centre O.
∠BCD = 130°. Find:
- ∠DAB
- ∠DBA
In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.
