Advertisements
Advertisements
प्रश्न
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DBA
Also, show that the ΔAOD is an equilateral triangle.
उत्तर
∠ADB = 90°
(Angle in a semicircle is a right angle)
∴ ∠DBA = 90° – ∠DAB
= 90° – 60°
= 30°
APPEARS IN
संबंधित प्रश्न
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC
ABC is a right angles triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Calculate the value of x, the radius of the inscribed circle.
Prove that the parallelogram, inscribed in a circle, is a rectangle.
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°; calculate :
- ∠DAB,
- ∠BDC.
In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°.
Calculate:
- ∠EBA,
- ∠BCD.
In the given figure, AB is the diameter of a circle with centre O.
If chord AC = chord AD, prove that:
- arc BC = arc DB
- AB is bisector of ∠CAD.
Further, if the length of arc AC is twice the length of arc BC, find:
- ∠BAC
- ∠ABC
In the figure, given below, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, fins the distance MN between the two chords of lengths 24 cm and 18 cm respectively.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠ADC
Also, show that the ΔAOD is an equilateral triangle.
In the figure, ∠DBC = 58°. BD is diameter of the circle.
Calculate:
- ∠BDC
- ∠BEC
- ∠BAC
In the figure given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.