Advertisements
Advertisements
प्रश्न
In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20°, find angle AOD.
उत्तर
Join OB,
In ΔOBC,
BC = OD = OB ...(Radii of the same circle)
∴ ∠BOC = ∠BCO = 20°
And Ext. ∠ABO = ∠BCO + ∠BOC
`=>` Ext. ∠ABO = 20° + 20° = 40° ...(i)
In ΔOAB,
OA = OB ...(Radii of the same circle)
∴ ∠OAB = ∠OBA = 40° ...(From (i))
∠AOB = 180° – ∠OAB – ∠OBA
`=>` ∠AOB = 180° – 40° – 40° = 100°
Since DOC is a straight line
∴ ∠AOD + ∠AOB + ∠BOC = 180°
`=>` ∠AOD + 100° + 20° = 180°
`=>` ∠AOD = 180° – 120°
`=>` ∠AOD = 60°
APPEARS IN
संबंधित प्रश्न
In the following figure, AD is a straight line. OP ⊥ AD and O is the centre of both the circles. If OA = 34 cm. OB = 20 cm and OP = 16cm; find the length of AB.
In the following figure; P and Q are the points of intersection of two circles with centres O and O’. If straight lines APB and CQD are parallel to O O'; prove that:
(i) O O' = `1/2AB` (ii) AB = CD
The length of common chord of two intersecting circles is 30 cm. If the diameters of these two circles be 50 cm and 34 cm, calculate the distance between their centres.
In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O1 and O2 are the centers of two circles.
In the figure, given alongside, AOB is a diameter of the circle and ∠AOC = 110°. Find ∠BDC.
In the figure, chords AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm; find DE.
Two circles are drawn with sides AB, AC of a triangle ABC as diameters. They intersect at a point D. Prove that D lies on BC.
In following figure . O is the centre of the circle. Find ∠ BAC.
In following figure , chord ED is parallel to the diameter AC of the circle. Given ∠ CBE = 65° , calculate ∠DEC .
In the given figure, AB and CD are two equal chords of a circle, with centre O. If P is the mid-point of chord AB, Q is the mid-point of chord CD and ∠POQ = 150°, find ∠APQ.
