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.

Advertisements
उत्तर

∠ADB = 90°
(Angle in a semicircle is a right angle)
∴ ∠DBA = 90° – ∠DAB
= 90° – 60°
= 30°
संबंधित प्रश्न
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.
Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠RNM

In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate:
- ∠DAB,
- ∠DBA,
- ∠DBC,
- ∠ADC.
Also, show that the ΔAOD is an equilateral triangle.

In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
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 given figure, O is the centre of the circle and ∠PBA = 45°. Calculate the value of ∠PQB.

In the given figure, BAD = 65°, ABD = 70°, BDC = 45°.
(i) Prove that AC is a diameter of the circle.
(ii) Find ACB.

